Since 1dA_uv=dA_xy, the total area is not changed at all. Therefore the area inside x²–10xy+26y²=1 is exactly equal to Π1². I think if I worked diligently with dxdy integrals and used trig substitutions as they are taught in 152 I might, after a while, be able to get this result. But a whole heck of a lot of work would need to be done.

So what's going on ...
This result is not used as in the past examples. That is, people don't decide, "Hey, let's look at u and v and see ..." Rather, what happens is that sometimes folks realize they need to evaluate some (horrible) double/triple/whatever integral. They look at it, and see, somehow, some sort of links between the integrand and the region. They see, somehow, that everything could be described in terms of other variables. Then they reach in and use the result that follows. Note that no one I know uses this result "casually" -- they use it only if they really need it.

The theorem
Suppose x and y are written as functions of u and v. Then JAC, the area distortion factor, is the absolute value of a certain determinant:

    | ∂x/∂u ∂x/∂v | 
det |             | 
    | ∂y/∂u ∂y/∂v |

∫∫_RuvFUNC_uv (JAC) dA_uv=∫∫_RxyFUNC_xydA_xy

Names
JAC is called the Jacobian. The result above, discussed in section 15.4, and particularly stated on pages 928 and 929 of the text, is called the Change of Variables Formula.

Maybe I should give a slight indication where the result comes from. So here is some heuristic reasoning: I want to describe how a tiny Δu by Δv rectangle in the u,v-plane gets distorted in the xy-plane. JAC is this distortion factor. I would like to compare the areas. Here x=x(u,v) and y=y(u,v) are some functions, and I don't know much about them.

I was only able to suggest the following information during class. Suppose one corner of the small uv box is at the point (u,v). Then the horizontal edge of the box goes from (u,v) to (u+Δu,v). What happens in the xy-plane? If Δu is very small, we can hope that the image of the horizontal line segment is some curve, and maybe it is also straight. This almost (?) line segment starts at (x(u,v),y(u,v)) and goes to ... well, what happens to x(u,v) and y(u,v) if we "kick" u to u+Δu? The linear approximation idea is that this becomes (except for higher order errors which are very small if Δu is small) x(u+Δu,v)≈x(u,v)+(∂x/∂u)Δu and y(u+Δu,v)≈y(u,v)+(∂y/∂u)Δu. So the "edge" in the xy-plane is nearly a vector with tail at (x(u,v),y(u,v) and head at (x(u,v)+(∂x/∂u)Δu,y(u,v)+(∂y/∂u)Δu). This is the vector (in the diagram shown, this is v) <(∂x/∂u)Δu,(∂y/∂u)Δu> which is Δu (a scalar) multiplying the vector <(∂x/∂u),(∂y/∂u)>. The vector for the other edge (which is in the diagram shown, this is w) is Δv multiplying <(∂x/∂v),(∂y/∂v)>. To get the area of the parellelogram of the rectangle determined by these vectors, we find the magnitude of their cross product. Let's compute the cross product. It will be (Δu)(Δv) multiplied by: | i j k | det| (∂x/∂u) (∂y/∂u) 0 | = [(∂x/∂u)(∂y/∂v)-(∂y/∂u)(∂/∂v)]k | (∂x/∂v) (∂y/∂v) 0 | To get the magnitude of this vector which only has a k component, we take the absolute value of the coefficient, which is exactly what I called JAC before. This sort of explains a piece of the formula above. I hope it helps you "swallow" the change of variables formula above.

So ...
This theorem is difficult to work with but the results can wonderful when you can use it. Here are two computations I tried to show in class.

This example is artificial but useful as a start
Compute ∫∫_R(x–y)⁴⁰(x+y)⁵⁰dA where R is the rectangular region with corners (1,–1), (2,0), (0,2), and (–1,1). This is an irritating integral. But there is some not well concealed symmetry. The boundaries of rectangle can be written as x+y=2, x+y=0, x–y=–2, and x–y=2.
It almost seems as if the integrand and the region are begging us to rewrite everything in terms of u and v where u=x–y and v=x+y. Then the region of integration can be described –2≤u≤2 and 0≤v≤2. The integrand becomes u⁴⁰v⁵⁰. Notice that if we add the equations u=x–y and v=x+y and divide by 2 we get x=(1/2)(u+v). If we subtract the first equation from the second and divide by 2 we get y=(1/2)(v–u).

What's JAC?
Since x=(1/2)(u+v) and y=(1/2)(v–u) we compute

    | ∂x/∂u ∂x/∂v |       | 1/2  1/2 | 
det |             | = det |          | = -1/4 -1/4 = -1/2 
    | ∂y/∂u ∂y/∂v |       | 1/2 -1/2 |

We have in effect parameterized the xy plane with (1/2)(u+v)=x and (1/2)(v–u)=y. So everything in x and y could be written in terms of u and v. The "General Change of Variables" result becomes what follows in this case:

∫∫_Rxy(x–y)⁴⁰(x+y)⁵⁰dA_xy=∫_v=0^v=1∫_u=–2^u=2u⁴⁰v⁵⁰(1/2)du dv. This can be evaluated exactly easily because it is just a mess of powers of u and v. The answer is: (1/2)2·(2⁴¹/41)(2⁵¹/51).

Crazy people all over ... Or you could just try it in Maple as it is. But we will need to break up the integral into three pieces (in either dxdy or dydx). Also, I want to learn how much time and space the computation takes, so I will use showtime. The instruction showtime(true); has this effect (from the Help page): Any Maple statement entered is evaluated normally, its result returned followed by a line numbered O1, O2, .. with the time taken and the amount of memory used being displayed. Here we go. > showtime(true); O1 := func:=(x-y)^40*(x+y)^50; 40 50 (x - y) (x + y) time = 0.00, bytes = 7382 O2 := A:=int(int(func,x=-y..2+y),y=-1..0); 618970019642690137449562112 --------------------------- 1173 time = 0.08, bytes = 1394659 O3 := B:=int(int(func,x=-y..2-y),y=0..1); 41125671617232447642991204624847361028540479941115904 ----------------------------------------------------- 62654905899056975234831847747 time = 0.08, bytes = 1275540 O4 := C:=int(int(func,x=-2+y..2-y),y=1..2); 74187486054615395748140995710384329611242731900764160 ----------------------------------------------------- 62654905899056975234831847747 time = 0.13, bytes = 1880463 O5 := A+B+C; 4951760157141521099596496896 ---------------------------- 2091 time = 0.00, bytes = 3963 O6 := %-(2^(41)*2^(51))/(41*51); 0 time = 0.00, bytes = 3988 So our result is the same as the rather painful direct computation, which took a total of .29 seconds. That is not terrible (but not a computation that one would want to do in "real time" applications). More important to me is that this "direct" computation gave no insight. "The purpose of computing is insight, not numbers."

What's going on? If there is something common among the algebraic and geometric specifications of a double (or a triple!) integral, then we can sometimes take advantage. That's what's going on. Another example, but this one more realistic The following example could arise in thermodynamics or physical chemistry. Suppose R is the region in the first quadrant bounded by y=2x, y=4x, y=1/x, and y=3/x. Let's compute ∫∫Rx4y dA. Here a neat "change of variables" is a bit hidden, but maybe you can see that the boundary curves of the region are y/x=2 and y/x=4 and xy=1 and xy=3. Then you might (!) think to define u=y/x and v=xy. If you do, then uv=(y/x)(xy)=y2 so that y=u1/2v1/2. Then v=xy becomes v=x(u1/2v1/2) so that x=u–1/2v1/2. Here's where I had to stop in class because I ran out of time. A complete solution follows.

With the equations x=u–1/2v1/2 and y=u1/2v1/2 the original integrand x4y becomes u–3/2v5/2. The Jacobian computation is: | ∂x/∂u ∂y/∂u | | -(1/2)u-3/2v1/2 (1/2)u-1/2v1/2 | det | | = det | | | ∂x/∂v ∂y/∂v | | (1/2)u-1/2v-1/2 (1/2)u1/2v-1/2 | and this is –(1/4)u–1–(1/4)u–1. We want the absolute value so we have (1/2)(1/u). In this case, which is considerably more complicated than the others above, the amount of stretching depends on the value of u. In the other cases we looked at previously, the stretching was the same at all points. In the real world, non-uniform stretching is more likely. (Take either a piece of taffy or a steel bar and pull at the ends. I bet that the part near the center stretches more than the parts near the ends.) The double integral which results is ∫13∫24u–3/2v5/2(1/2)(1/u)du dv. The region of integration has become a rectangle, the integrand is not horrible, and the Jacobian factor is also not too bad. I won't compute this, but I hope that you see it is easy enough. Possible QotD I was going to ask students about the following transformation and its consequences. Suppose we change (u,v) to (x,y) using the equations x=u+v2 and y=v. Then here is how some points are changed: (u,v) coords (x,y) coords (1,1) (2,1) (1,2) (5,2) (3,1) (4,1) (3,2) (7,2) The line segment from (1,1) to (1,2) has u fixed as 1, and v varying between 1 and 2. Therefore y varies from 1 to 2, and x=1+v2=1+y2. This is part of a sideways parabola. The Jacobian of this transformation is | xu yu | | 1 0 | det | | = det | | = 1 | xv yv | | 2v 1 | so area doesn't change with this transformation -- to me this is a bit surprising again. People sometimes call this sort of mapping a non-linear shear. Polar and spherical ... The factors involved in integration in polar/cylindrical (r drd&theta) and spherical (ρ2sin(φ) dρdφdθ) all come from Jacobian calculations. The spherical factor is the absolute value of a determinant of a 3-by-3 matrix. The computation is not fun. I've done it several times and have managed to make mistakes each time. Proof? Who needs a proof? Verification of the change of variables formula (a proof) for double and on up integrals is very difficult. What's needed is knowledge of linear algebra and lots of comfort with limit processes, since the two sides of the formula are quite complicated Riemann sums. Working through the proof in detail takes about two weeks in an upper-level math course. Please believe this result, and try a few examples on your own.

A juggling act ...
I discussed the difficulties of using this method with a number of students after each lecture. There are three aspects: how the region can be described, how the integrand can be described, and how the Jacobian can be computed. Somehow successful use is done by noticing nice coincidences. This can take a lot of experimentation. In examples for this class, an introductory course, I'll try to be sure that things, the problems I'll ask you to do, will be relatively straightforward.

Today we will discuss systematically methods which will help with integrals over regions with axial or central symmetry. Both coordinate systems are ways of extending polar coordinates to R³.

Cylindrical coordinates This is a coordinate system that augments the r and θ of polar coordinates with z. Any problem with an axis of symmetry may be easier to understand in cylindrical coordinates. In words, the position of a point in the cylindrical coordinate system is described by its height, z, from the base coordinate plane. The foot of a perpendicular from the point to the plane then has a description in terms of an angle, θ, from an initial ray (usually the positive x-axis) and a distance, r, from the origin.

Some basic axially symmetric surfaces

r=5 is the collection of points in R3 whose distance to the "axis" is 5. The axis is the z-axis, so this will be a right circular cylinder of radius 5 having the z-axis as axis of symmetry.	z=7r gives a right circular cone whose axis of symmetry is the z-axis. How can you "see" this? Well, if we restrict ourselves to the slice of this surface through the xz-plane (with y=0) we get a picture sort of like what is shown. Why? Because if y=0, r=sqrt(x2+y2)=x (at least for x>0), so the result is the line shown. In general, since θ is not restricted, we get all the points shown as we revolve the "profile" curve around the z-axis. And this is a cone with vertex at the origin.	z=3r2 is a paraboloid, because r2=x2+y2 and you should see, I hope, that the result is what happens when the profile curve, a parabola through the origin, is revolved around the z-axis.

(Almost a real problem!) Moment of inertia of a cone
Suppose a right circular cone with base radius R and height H is filled with a homogeneous substance with constant density, C. What is the moment of inertia of the cone about its axis of symmetry?
Let me be more clear about some vocabulary.

Right circular cone
Take a circle (to be called the base). Put a line perpendicular to the plane of the circle through the circle's center. Pick a point on this line which is not on the plane of the circle. Connect that point (called the vertex) with the edge of the circle. The solid interior to the collected line segments and the circle is called a right circular cone. The "right" refers to the right angle that the axis of symmetry makes with the base.

Moment of inertia
Take a little piece of mass, m, external to a line, L. The moment of inertia of m about L is defined to be Q²m where Q is the distance from m to L.
There are many discussions of the moment of inertia on the web. One link declares that it is the "inertia with respect to rotational motion" and another reads "... the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation."
I think of a small merry-go-round in a playground, and trying to push the seats around (with many noisy, small children on them). The moment of inertia measures the resistance of the merry-go-round to being pushed.

Beginning the analysis
Take a little piece of volume, dV, inside the cone. (Note: this is a piece of volume inside the cone. We are not just considering the surface of the cone -- the cone is filled.) The mass of this volume is C dV. Suppose Q is the distance of the piece from the axis of symmetry. Then the moment of inertia of this chunk of mass about the axis of symmetry is Q²C dV. To get the moment of inertia of the whole cone we need to add up the pieces of the moment of inertia. So we need ∫∫∫_{The whole cone}Q²C dV.
A major decision in this and many other geometric/physical problems is where/how to put a coordinate system on the objects involved. Here almost surely people would agree that the axis of symmetry should be the z-axis. Sane human beings can disagree about where the origin should be. Some would put it at the vertex of the cone, with the base "up", and some would put the origin at the center of the base of the cone, with the vertex "up". I'll do the first alternative because I think some of the algebra will be simpler. As I mentioned in class, I drew the cone in this awkward way because I wanted people to think about how they would prefer to see it.

Coordinates?
Now the cone is sitting correctly (?) in the picture. The chunk of volume is at (x,y,z), and the closest point on the axis of symmetry is (x,y,0). The distance between (x,y,z) and the axis must therefore be sqrt(x²+y²). We should convert the triple integral into an iterated integral. What should be the order? Actually, it is possible to do this in any order, but the simplest way has dz on the outside. Then the z limits are clear: from 0 to H, and the slices with z=CONSTANT are also simple shapes: circles. The triple integral ∫∫∫_{The whole cone}Q²C dV becomes the triply iterated integral ∫_z=0^z=H(∫∫(sqrt(x²+y²))²C dA_xy)dz. I wrote dA_xy to remind myself that the double integral is in the xy-plane.

The inside double integral is: ∫∫(x²+y²)C dA_xy.

Recognition: polar
Things are in red so that a bell will ring in your head and you will think, polar!!!. Certainly x²+y²=r² and dA_xy=r dr dθ. The limits on θ for a whole circle are 0 and 2π. The limits on r are 0 (the center of the circle) out to the radius of the circle, which I will cleverly call RAD. The double integral is then ∫_θ=0^θ=2π∫_r=0^r=RAD(r²)C r dr dθ.

RADius
Look at the cone sideways and see some expected right triangles, so RAD/z=R/H and RAD=(R/H)z. The double integral becomes
∫_θ=0^θ=2π∫_r=0^r=(R/H)zC r³ dr dθ.

We compute the moment of inertia. Here it is:
∫_r=0^r=(R/H)zC r³ dr=(C/4)r⁴]_r=0^r=(R/H)z=(C/4)R⁴z⁴/H⁴.
∫_θ=0^θ=2π(C/4)R⁴z⁴/H⁴dθ=[(π C)/2]R⁴z⁴/H⁴. (Easy: no θ in the integrand so just multiply by 2π.)
∫_z=0^z=H[(π C)/2]R⁴z⁴/H⁴dz=[(π C)/10]R⁴z⁵/H⁴]_z=0^z=H=[(π C)/10]R⁴H.

Is this correct? The units of moment of inertia should be mass·(length)2. Since C is a density, C's units are mass/(length)3. And R4H is length5 so the units are correct. Sigh. What about the crazy constants (π and 10)? An engineering student who took Math 251 in a previous semester sent me e-mail about this, and here is part of his message: Upon consultation with my statics text, I present to you: ... I_x = 3/10 * m * a^2 where a is the radius of the base of the cone. Let's see: the statics text refers to the mass, m, of the cone. That is the density, C, multiplied by the volume. The volume of this right circular cone is (π/3)R2H. The student's "a" is our R. So the formula 3/10 * m * a^2 becomes (3/10)C[(π/3)R2H]R2 which is indeed [(π C)/10]R4H. We have confirmation by high authority: "my statics text".

The location of Hill Center
I enlightened students with these facts about Hill Center, Rutgers building #3752:
    HC's latitude is 40.523193^oN and HC's longitude is 74.464012^oW.
    Or, in more antique fashion (degrees/minutes/seconds), the latitude is 40^o31´23´´ and the longitude is 74^o27´50´´. Or maybe .707263 radians and 1.299642 radians. Sigh.
Do not be as confused as I am. This is not about stalactites (the down-dropping things) and stalagmites (the up-growing things).
We discussed what latitude and longitude are. The prime meridian is a great circle (a circle whose center is the center of the earth) and it goes through Greenwich, England and the north/south poles. The longitude is the angle between that great circle and the great circle connecting HC and the north/south poles. The angle has vertex at the center of the earth. W=west in the latitude, and it means the the angle opens to the west of the prime meridian. Latitude is the angle from the intersection of the great circle describing HC's longitude with the plane of the equator, again with the vertex at the center of the earth. N=north means that we look in the northern hemisphere. Constant latitude means a "small" circle. Constant longitude means a great circle (actually semicircle). HC is located at the unique intersection on the surface of the earth of these two curves.
I presume you know that the "23 and a half" degree tilt of the axes (north/south pole line) from the ecliptic (the plane of the earth's orbit about the sun) is responsible for seasonal variation. Nature is terrific!

Spherical coordinates
Take a point in space. We describe its position with one length and two angles. The length is the distance of the point to the origin: the length of the radius vector. The first angle, φ, is the angle from the positive z-axis to the radius vector. The second angle, θ, is the angle from the positive x-axis to the projection of the radius vector on the xy-plane. Spherical coordinates are very useful in problems with central symmetry.

I deduced the following formulas:
      x=ρ sin(φ)cos(θ)
      y=ρ sin(φ)sin(θ)
      z=ρ cos(φ)
It is useful to know that such formulas exist, but that I rarely use them. One result that I have used frequently comes from the fact that ρ represents the distance from (x,y,z) to the origin.
      x²+y²+z²=ρ².

Standard restrictions on spherical coordinates
Because the angles sort of fold over when Π's and 2Π's are added, most people who use spherical coordinates put some restrictions on how big/small θ and φ can be. If we only allow ρ>0, θ to be between 0 and 2Π and φ to be between 0 and Π, then there will be unique spherical coordinates for every point in R³. So I will generally work with these restrictions.

Some shapes in spherical coordinates

ρ=constant gives a sphere centered at the origin. So, for example, ρ=5 is a sphere centered at the origin of radius 5.	φ=constant gives a right circular cone whose axis of symmetry is the z-axis. For example, φ=Π/6 is a cone with vertex at the origin and whose axis of symmetry is the positive z-axis. The angle between the positive z-axis and any of the cone's "generators" (lines from the vertex on the surface of the cone) iw Π/6 (yes, 30o). The bottom half of the cone is not included because that is where φ is between Π/2 and Π.	θ=constant gives a half-plane, with the z-axis being the edge of the half-plane. For example, θ=Π/4 gives a half-plane which is perpendicular to the half-line y=x (x>0) in the xy-plane. The other half of the plane is where θ is 3Π/2, and so it is not included in this object.

Integral #1
A spherical region of radius R is filled with material whose density is directly proportional to the distance from the origin. What is its mass?
This is not very realistic. The center is light and fluffy and the outer edge is heavy and tough (my kind of cooking?). The density is supposed to interpolate linearly between these extremes. Maybe the appropriate assignment would be to build an object of this type.

The math setup
Take a small piece of volume, dV, in the sphere. The corresponding piece of mass, dm, is related to dV by dm=(density)dV. We know that the "density is directly proportional to the distance from the origin." Place the origin of the coordinate system at the center of the sphere. So there is some constant C>0 so density=C ρ. And the total mass is the sum of the dm's. This "sum" should be a triple integral:
Total mass=∫∫∫_{The whole ball}C ρ dV.
In spherical coordinates, a description of a sphere of radius R centered at the origin is easy: ρ goes from 0 to R, θ goes from 0 to 2Π, and φ goes from 0 to Π. We just use the agreed upon ranges for the angles to sweep out a whole sphere. There is one sticky point, however.

dV in spherical coordinates
We need to convert dV to spherical coordinates. In fact,
      dV=ρ²sin(φ)dρdθdφ
I know this is true (both true and absurd!). First, there is a discussion which is supposed to be convincing in the text (on pages 915 and 916). Second, I said it in class. Third, I actually can give an understandable argument if there is enough time later in the course. This strange multiplier is an example of what is called the Jacobian, a factor used to convert volume in one coordinate system to another. I may have time to discuss the computation later. In any case, when I use spherical coordinates, I almost never bother thinking about this weird mess, but I just write it. You can think of the Jacobian as the algebraic equivalent of a "penalty" for using spherical coordinates. As the possible user, you need to decide whether using spherical coordinates is worth the trouble. Sometimes the description of the region is so darn simple that the dV formula is clearly so bad enough.

The computation
So we have
Total mass=∫_φ=0^φ=Π∫_θ=0^θ=2Π∫_ρ=0^ρ=R(C ρ)ρ²sin(φ)dρdθdφ.
The inner integral
∫_ρ=0^ρ=R(C ρ)ρ²sin(φ)dρ=∫_ρ=0^ρ=RCρ³sin(φ)dρ=Csin(φ)ρ⁴/4]_ρ=0^ρ=R=Csin(φ)R⁴/4.
The middle integral
∫_θ=0^θ=2ΠCsin(φ)R⁴/4 dθ=(2Π C)sin(φ)R⁴/4=[(Π C)/2]sin(φ)R⁴. (Just multiply by 2Π, since there is no θ in the integrand.)
The outer integral
∫_φ=0^φ=Π[(Π C)/2]sin(φ)R⁴dφ=–[(Π C)/2]cos(φ)R⁴]_φ=0^φ=Π=(Π C)R⁴.
I don't know any way to check this answer. Build a model? Weigh it?

Is this silly?
Well, yes, it is silly. The problem is invented and certainly designed exactly for spherical coordinates. But I would not use spherical coordinates, which definitely have peculiarities (look at the pictures above and look at the expression for dV) unless both the region and the integrand can both be described in a nice fashion with spherical coordinates. I won't use this coordinate system otherwise. (Could you imagine using spherical coordinates to describe a cube?)

Integral #2
Consider the region in the first octant consisting of points whose distance to the origin is at least 1. Imagine that this is filled with material whose density is inversely proportional to the fifth power of the distance to the origin. What is the mass of this object?

Translating All of R3 is divided into eight parts by the coordinate planes: x=0, y=0, and z=0. Each part is called an octant. While the corresponding regions in the plane (the quadrants) have individual designations, the only octant that is named is the first: the octant where x>0 and y>0 and z>0. In this first octant, I'm excluding points whose distance to the origin is less than 1. What does the remaining region look like? Here are several possible pictures of the region. In this picture (sort of the corner of a rectangular box), a spherical "bite" has been taken out of the corner. The bite is centered at the vertex (the origin) and has radius 1. Wow!

To the right is a more oblique view of the octant with the bite. The nice thing about this region is that it can be described very briefly in terms of spherical coordinates. Certainly, ρ will go from 1 (as close to the origin as the bite will let us get) out to ... out to ... infinity (an improper integral!). What about θ and φ? Here students should look closely at the definitions of θ and φ. Each of them will go from 0 to Π/2. This is best confirmed by taking "angles" with vertex at the origin and a side along the x- (respectively, z-) axis and then opening the second side of the angle to an aperture of Π/2 (I think "aperture" means the angle's opening). By the way, as I remarked in class, this is actually a more realistic example than the first integral. Things like this do occur in electricity and magnetism.

The computation
Again dm=(density)dV=[C/ρ⁵]dV because "density is inversely proportional to the fifth power of the distance to the origin." And we know the limits from the discussion above, so the total mass is
∫_φ=1^φ=Π/2∫_θ=0^θ=Π/2∫_ρ=1^ρ=∞[C/ρ⁵]ρ²sin(φ)dρdθdφ.
The integrand is [C/ρ³]sin(φ) after cancelling some powers.
The inner integral
This is an improper integral, so I will be careful.
∫_ρ=1^ρ=BIGC/ρ³sin(φ)dρ=–C/(2ρ²)sin(φ)]_ρ=1^ρ=BIG=–C/(2(BIG)²)sin(φ) +C/(2(1)²)sin(φ). As BIG→∞, the term –C/(2(BIG)²)sin(φ)→0 so the improper integral ∫_ρ=1^ρ=∞[C/ρ³]sin(φ)dρ converges and its value is (C/2)sin(φ).
The middle integral
∫_θ=0^θ=Π/2(C/2)sin(φ)dθ=(C/2)sin(φ)(Π/2)=[(C Π)/4]sin(φ).
The outer integral
∫_φ=0^φ=Π/2[(C Π)/4]sin(φ)dφ=–[(C Π)/4]cos(φ)]_φ=0^φ=Π/2=[(C Π)/4]
Again, I will admit that I don't know any way to check this answer. When such an integral comes from a real physical problem, there is frequently some way to see if the final answer is reasonable.

Further defense of silly (the same defense)
I would only use this technique, I hope!, where both the region and the integrand are suitable. So, although the problems may have seemed silly, they are the sort of applications which might occur. We will need integration in spherical coordinates a few times later in the course.

QotD
Try to set up in spherical coordinates the triple integral of z over the lower half of the sphere of radius 5 centered at the origin. Everything should be written in terms of spherical coordinates!

• What do we integrate?
  Since z=ρ cos(φ) and dV=ρ²sin(φ)dρdθdφ we should be integrating ρ³cos(φ)sin(φ)dρdθdφ.
• What about the limits?
  For ρ we go from 0 to 5 (the origin to the farthest away point). For θ we got all around the z-axis, so from 0 to 2Π. For φ, which is to me the trickiest, we go from a right angle (on the xy-plane) to a straight angle (on the negative z-axis), so from Π/2 to Π.

A student's request The planes x=0 and y=0 and z=0 divide R3 into eight chunks. Differently put, if you remove these planes from space, you'll have eight pieces left. Each of the eight pieces is characterized by requiring that the variables x, y, and z have specific (non-zero!) signs. I was asked by a student the last time I taught 251 how the spherical coordinates φ and θ relate to these sign restrictions. This is not a silly question. The answer is a bit complicated with details, but maybe looking at it will help you. Look to the right. There is a very bare diagram with the spherical angles φ and θ sketched. Of course, φ is the angle between the radius vector and the positive z-axis. People usually request that φ be in the interval [0,Π]. If this angle is acute, so 0<φ<Π/2, then the radius vector will be above the xy-plane, no matter what the value of θ. This means z>0 exactly coincides with 0<φ<Π/2. If we push the radius vector below the xy plane, then z<0 and φ will be larger than Π/2. So z<0 is the same as Π/2<φ<Π. θ does not affect the sign of z at all. It interacts with the signs of x and y. So we can just look "downwards" in R3 from high up on the positive z axis. Then we might understand what we're seeing as something like usual polar coordinates (remember, the z information is carried by φ so we don't need that here). Certainly we can just read off the sign combinations of x and y by the usual quadrant information of θ. Sign of x Sign of y Sign of z Interval of θ Interval of φ x>0 y>0 z>0 0<θ<Pi/2 0<φ<&pi/2 x<0 y>0 Π/2<θ<Π x<0 y<0 Π/2<θ<3Π/2 x>0 y<0 3Π/2<θ<2Π x>0 y>0 z<0 0<θ<Pi/2 Π/2<φ<Π x<0 y>0 Π/2<θ<Π x<0 y<0 Π/2<θ<3Π/2 x>0 y<0 3Π/2<θ<2Π I hope this is helpful to other people who are trying to understand spherical coordinates.

Today begins three lectures where I will attempt to describe other, more sneaky methods to compute multiple integrals. These methods generally are used when specific computations are given and the regions or the integrand (the function to be integrated) share some kinds of symmetries. All of today's examples will be relevant to engineering education. I'll begin with the following problem. I want to compute a double integral, something like ∫∫_Rf(x,y) dA. I will describe an f and an R.

Let's let f(x,y)=x²+y². That's certainly a simple enough function, just a degree 2 polynomial in x and y.

Suppose R is the region in the xy-plane defined by these restrictions: it is in the upper half plane where y>0, and it has boundary given by y=x and y=–x (two straight lines or, actually, since they are inside a half plane, just two rays) and the circles x²+y²=2 and x²+y²=4. I think, or I hope, that this region is shown in the picture to the right.

I hope that there are enough accidents (?) and coincidences (??) so that you are a bit suspicious. Of course, this example is totally arranged. I hope that it makes you think of polar coordinates.

dA in polar coordinates
Here's a mostly emotional argument for how dA should be described in polar coordinates. Later I will be able to give a more precise derivation. Or you can look in the textbook (section 15.4) for a more careful discussion.
Suppose I want to compute the area obtained by changing r to r+dr and θ to θ+dθ. The picture displays this area, dA, magnified a lot. As mentioned, dA is an area and has dimensions length². If dθ and dr are very small, the area dA is approximately rectangular, and maybe the area is the product of the length of its sides. Well, one side is dr but the other side is not dθ. Angles don't have dimensions (they are ratios!) and, anyway, if you move circles centered at the origin in and out, you can see that the intercepted arcs change in length. These arcs are very short close to the origin and are longer as the radius of the circle gets bigger. In fact, the length of the intercepted arc is directly proportional to r. This length is also directly proportional to dθ: if the angle at the origin is doubled, the length of the intercepted arc is also doubled. Well, "directly proportional" means that there is some constant, uhhh ..., let's call it K, so that the length is K r dθ. What is K? In the nicest world, K would be 1 because then I would not have to worry about it any more. Well, golly, that is exactly why radian measure was invented: so this darn constant would be 1 and would not need attention.

Comment: so what is K and what about those words? Why is K=1 in radians? Well, the circumference of a circle of radius r is 2π r. Here the dθ is 2π. So apparently the K is indeed 1. If you insisted on using degrees in all of calculus, then the angle for a whole circle would be 360, and for Kr dθ=K(360)r to be 2π r, you would need K=2π/360, which is approximately the obnoxious number .01745. I looked on the web, and the only other candidate for angle measurement I found was the grad, introduced in France as part of the metric system (my calculator permits angle computations in grads). There are 400 grads in a circle (I didn't know that) and therefore the constant K, if we used grads in calculus, would be 2π/400 which is approximately .01571, also obnoxious. Yes, things would be better if π were equal to 3. Euphemism: The expression of an unpleasant or embarrassing notion by a more inoffensive substitute The word "golly" is a euphemism for "God" and the word "darn" is a euphemism for "damn".

Computing the integral
Let's return to computing ∫∫_Rx²+y²dA, if R is the region shown to the right (in the upper half plane, with the curves arcs of circles centered at the origin).
How does one recognize that the integral is "polarish"? It is a classroom example, but the integrand has central symmetry, and so does the region. You may be helped if you recall the conversion formulas

From r, θ to x, y      From x, y to r, θ   
-------------------    ------------------- 
   x=r cos θ              r2=x2+y2
   y=r sin θ              tan θ=y/x

I recognize (primarily from the picture, but I can also use the formulas) that R is described by π/4≤θ≤3π/4 and by 2≤r≤4. We can convert the integral into polar coordinates:

∫∫_Rx²+y²dA= ∫_π/4^3π/4∫₂⁴r² r dr dθ= ∫_π/4^3π/4∫₂⁴r³dr dθ= ∫_π/4^3π/4(1/4)r⁴]_r=2^r=4dθ=(60)θ]_π/4^3π/4=30π.

Certainly the computation is easy. It was arranged so that after conversion to polar coordinates things would work out well. The computation in rectangular coordinates, including finding the boundaries of the integrals (there would have to be two of them) and then computing the antiderivatives, would be very tedious. This is not an entirely artificial example: it is the computation of the moment of inertia about the origin of a thin homogeneous plate in the shape of the region R.

The earth is flat
So here I will try to convince you by combining a valuable and truthful computation with extremely dubious logic, that the earth is flat. Please be reassured: the earth is probably not flat.

Newton's Law of Universal Gravitation
Suppose I have two "point masses", m₁ and m₂, which are a distance d apart. The magnitude of the force attracting them together is directly proportional to the product of their masses and inversely proportional to the square of the distance separating them. The constant of proportionality is usually called G (alas, not to recognize the lecturer!). Therefore the magnitude of the force is G m₁m₂/d².

A very good estimate for the actual value of G was found as a result of a remarkably precise experiment done by Henry Cavendish in 1797 and 1798. Here is part of a description of the experiment: The apparatus constructed by Cavendish was a torsion balance made of a six-foot (1.8 m) wooden rod suspended from a wire, with a 2-inch (51 mm) diameter 1.61-pound (0.73 kg) lead sphere attached to each end. Two 12-inch (300 mm) 348-pound (158 kg) lead balls were located near the smaller balls, about 9 inches (230 mm) away, and held in place with a separate suspension system.[8] The experiment measured the faint gravitational attraction between the small balls and the larger ones. The currently accepted value of G is about 6.67259 x 10–11 Nm2/kg2. Gravity is actually much weaker than, say, magnetism. There is just a great deal of mass around, and very few magnetic monopoles.

The plate: from description to integral
Let me assume that the "universe" consists of an infinite flat homogeneous plate, and an external small object with a mass of m whose distance to the plate is D. What is the gravitational attraction of the object to the plate? A major part of such a problem is setting it up. The correct location of the origin and the axes can make problems much easier. In this case, I believe there are two reasonable locations for the origin: the object, or the closest point on the plane to the object. I'll use that closest point to be the origin. Of course, the xy-plane will be the plate, and therefore the coordinates of the object will be (0,0,D). The plate is homogeneous and thin. To avoid having too many letters around, let me assume that the plate is 1 unit thick (otherwise I'll just have to carry around the thickness in all of the computations, and I have a hard enough time with my own thickness, both mental and physical). Since the plate is homogeneous (the same at every point), it has a density, ρ. A small chunk of the plate ("dA") located at the point (x,y) will have mass equal to ρ dA (remember the thickness is 1, and so it is already in the formula).

Now let us convert the ideas into more rigid "mathspeak". The magnitude of the force from the external mass to the dA piece of the plate is Gmρ dA/d². The piece is located at (x,y), and (x,y), (0,0), and the location of the external mass are at the vertices of a right triangle. The hypotenuse of the right triangle is d, and the leg of the triangle from the external mass to (0,0) is D. The distance from (0,0) to (x,y) is sqrt(x²+y²). Therefore d²=D²+(sqrt(x²+y²))². The square root and the square cancel. The magnitude of the force is Gmρ dA/(D²+x²+y²). Several students noticed a surprising symmetry. Since we are dealing with the whole plane, R², the chunk of dA at (x,y) has an antipodal chunk at (–x,–y), having the same mass and the same distance to the external object. Therefore the "lateral" parts of the forces (parallel to the plane) exactly cancel out. We only need to compute the vertical component of the force.

The vertical component of the force is the magnitude of the force multiplied by the cosine of the angle, φ, between the vertical line and the line connecting the external object to dA. But cos(φ) is D/d, which is D/sqrt(D²+x²+y²). The function to be integrated is the vertical component of the gravitational attraction between the external object and dA. This is GmρD dA/(D²+x²+y²)^3/2. Since the plate is infinite, we want ∫∫_All of R²Gmρ dA/(D²+x²+y²)^3/2.

Computing the integral
Many of the letters are constants: G and m and ρ and D. We can pull them out of the integral (but we will remember them for the final result). We need to compute:
∫∫_R²dA/(D²+x²+y²)^3/2.
Since this comes immediately after a discussion of polar coordinates, the student alert to pedagogical plans (how folks teach) will immediately think of converting to polar coordinates. Indeed, even those who are not so ... prescient might think: the region has symmetry around (0,0) and the integrand has that same x²+y², so let's try polar coordinates!

Then dA=r dr dθ, and r²=x²+y², and all we need are the limits on the integral. For the whole plane, r should go from 0 to ∞, and θ should go from 0 to 2π. The appearance of ∞ forces me to finally acknowledge that this is an improper integral.

∫₀^2π∫₀^∞r dr dθ/(D²+r²)^3/2.

The inner (improper) integral
I will be careful, since I am supposed to be teaching a math course.
∫₀^∞r dr dθ/(D²+r²)^3/2=lim_B→∞∫₀^Br dr dθ/(D²+r²)^3/2
The r accompanying the dr is exactly what's needed to do the substitution u=D²+r² with du=2r dr. We sort out the constant by guessing (maybe).
∫₀^Br dr dθ/(D²+r²)^3/2=–1/sqrt(D²+r²)]_r=0^r=B= –1/sqrt(D²+B²)–{–1/sqrt(D²+0²)}
As B→∞, the term –1/sqrt(D²+B²)→0. The other term has minus signs which cancel, and (let's say D>0) square/sqrt which cancel, so the limit is 1/D.

The outer integral is easy: ∫₀^2π(1/d)dθ=(1/D)θ]_θ=0^&theta=2π=2π/D.

But we need to multiply by the factors we pulled out. The whole answer is:
GmρD(2π/D)=Gmρ2π.

And, therefore ...
There is no D in the answer!. The gravitational attraction of a flat earth is constant! Now the lecturer discussed the fact that he weighs the same standing on the floor and standing on a chair. Therefore ... therefore ... the earth is flat. (And even more supporting argument: wouldn't people who wanted to lose weight climb Mt. Everest, because they would lose weight when ...).

Discussion of the claim

Capacitor
This is still an interesting and useful computation. An electron is very small. If we try to analyze the attraction an electron might have to a small charged plate, even, say, 1/4 inch square, then, to the electron, the plate might as well be infinite. That is, if the electron is near the center of the plate, the edge effect hardly matters at all. And the force on the electron does not depend on distance. Such considerations occur in the design of classical capacitors, used in many devices.

HOMEWORK
Please read about triple integrals and cylindrical and spherical coordinates in 12.7 and 15.4. If you do this, you will find the next lecture much more comprehensible. And, otherwise, there will just be too many formulas! This is almost guaranteed.

QotD
Compute the double integral of 1/r⁴ over those points in the first quadrant of the plane which have distance from the origin at least 1. A picture of this domain is shown to the right. Again, a polar description of this domain is quite simple, while a rectangular description would be irritating and complicated.

The "first quadrant" implies that the restriction on θ is from 0 to Π/2. Now "at least 1" means that r's should start from 1 and then increase without any bound, so the upper limit of the r integral is r=∞. This is the background for my asserting that the quantity which should be computed is
∫_θ=0^θ=Π/2∫_r=1^r=∞(1/r⁴)(r dr dθ).
You should not forget that dA is r dr dθ.

The computation is straightforward, I hope. The inner integral is ∫_r=1^r=∞(1/r³)dr (combine the r with the other power!). Now we can see this as a limit for Q→∞ of ∫_r=1^r=Q(1/r³)dr which is –1/{2r²}|_r=1^r=Q=–1/{2Q²}–(–1/2)=1/2–1/{2Q²}. As Q→∞ this integral's value →1/2.
The computation is completed by observing ∫_θ=0^θ=Π{1/2}dθ=Π/4.

The ocean
People who study the ocean (people such as those affiliated with the Rutgers Institute of Marine & Coastal Sciences) are interested in such aspects as the temperature and salinity and pressure and flow of the water. They employ remote sensing devices to try to record data at various depths. Then analysis of the data together with theory is used to try to predict interesting things: the weather, fishing prospects, etc. I'm going to look at a very simple model and try to link it up with what we study in this course.

The average temperature of a box of ocean
Consider a "box of ocean", say the region between x=a and x=b, y=c and y=d, and z=e and z=f (here a<b, c<d, and e<f). We might put some sort of measuring device at a point in this box and measure the temperature of the water at that point. One or a few temperature measurements are probably not going to give good information. If the economics (!) and the equipment and time (!) are available, many measurements should be made. One representation of the measurements might be the average: so the computation would be

SUM of all of the temperature measurements
------------------------------------------
  The number of temperature measurements

Going abstract: the "limit"
Let me look at the average a bit more. The discussion that follows seems very clever to me.
Suppose that I assume that the number of observations is n³ where n is a large positive integer. Then I would have something like this:

SUM of all of the temperature measurements
------------------------------------------
                    n3

SUM of all of the temperature measurements   (b–a)(d–c)(f–e)
------------------------------------------ · ---------------
                    n3                       (b–a)(d–c)(f–e)

∫∫∫box of oceanT(x,y,z) dV
-------------------------
    Volume of the box

A specific example
What if our box was bounded by x=0 and x=2, y=0 and y=3, and z=0 and z=5, and the temperature at (x,y,z) was given by the formula T(x,y,z)=x²+7yz? Then if we wanted to compute the average temperature we would convert a triple integral into a (triply) iterated integral. In this case, I see no advantage in any one of the six possible orders, so:
∫₀²∫₀⁵∫₀³ x²+7yz dy dz dx
Let's compute, from the inside out: ∫₀³ x²+7yz dy dz dx=yx²+(7/2)y²z]_y=0^y=3=3x²+(63/2)z.
∫₀⁵3x²+(63/2)z dz=3x²z+(63/4)z²]_z=0^z=5=15x²+(63/4)(25).
∫₀²15x²+(63/4)(25) dx=5x³+(63/4)(25)x]_x=0^x=2=40+(63/2)(25).

If this were the 21^st century instead of 1872, we could type:

> int(int(int(x^2+7*y*z,y=0..3),z=0..5),x=0..2);
                                    1655/2

This isn't the average temperature. For that we need to divide by the volume of the box which is 2·3·5=30. The result is (331/6).

The "moral" of this: computation of triple iterated integrals
I don't think that there are any essential new difficulties introduced when we move from evaluating double iterated integrals to evaluating triple iterated integrals. Yes, there are more opportunities for error (50% more?) but they are not new in type. So I won't devote too much time to actual evaluation, at least in this lecture.

Describing a volume in space
Since the difficulties involved in computation of a triple iterated integral really are just those we've seen already with double interated integrals, I want to illustrate something that definitely seems more complicated to me: going from a description of a region in space over which we want to compute a definite integral to the corresponding iterated integrals (and there are 6=3! possible orders for the iterated integral). Let me "integrate" (convert to iterated integrals) the function SQUIRREL over the region in space (R³) defined by y=0, z=3, and z=x²+y.
I want to begin by sketching the region. The planes y=0 (the xz-plane) and z=3 (push the xy-plane up three units) are easy enough. The surface z=x²+y cut by y=0 and z=3 is maybe not so obvious. When y=0 we get a parabolic arc cut off at z=3 in the xz-plane. As y increases, the parabolic arc is translated up, but still cut off at z=3. In the yz-plane, when x=0, the slice is a segment of the line z=y from (0,0) (with the coordinates being y and z) to (3,3). The surface cuts the plane z=3 with the parabola 3=x²+y or y=3–x², which opens "downward" (in the standard orientation of xy-planes).

I've attempted to sketch the surface to the right of this description. The colors are meant to show some of the curviness. There are some extreme points which turn out to be useful in setting up iterated integrals. Those are the points (0,0,0), (0,3,3), (sqrt(3),0,3), and (–sqrt(3),0,3). These points are where each of the coordinates (x and y and z) attain maximum and minimum values on the solid regions whose boundary curves were given.

Nomenclature
The surface z=x²+y is called a tilted parabolic cylinder. It is a parabolic cylinder because it results from a family of parallel lines in space which all meet the parabola z=x² in the xz-plane. It is "tilted" because these lines are not perpendicular to the xz-plane.

Now to the right is Maple's attempt to draw the tilted parabolic cylinder in the region of interest to us. The picture to the right is the result of using the command:
implicitplot3d(z=x^2+y,x=-1.75..1.75,y=0..3,z=0..3,
     grid=[40,40,40],axes=normal,labels=[x,y,z]);
This command did not display an immediate result on my home computer. It requested that Maple to check a three-dimensional grid of 40³=64,000 points, and then compute the light and the angle, etc. I rotated and chose lighting so that I got the image displayed here. That's why "supercomputers" are needed to draw the lighting effects for Pixar, etc.

Maple can draw ... ... some useful pictures for us when we want to look at double and triple integrals. Last time we looked at the iterated integral ∫02∫x=0x=1–(1/2)y3–3x–(3/2)y dx dy The command plot3d(3-3*x-(3/2)*y,x=0..1-(1/2)*y,y=0..2); produces (after putting in the axes and making the view constrained) the graph shown to the right. I did not know until fairly recently that Maple had the capacity to show only pictures corresponding to double integral limits. This could be very helpful.

A region in space
Now back to the triple integral. There are six different orders that are possible when converting a triple integral to an iterated integral. I did three of the six orders. Let's convert ∫∫∫_This regionSQUIRREL dV into various iterated triple integrals.

dx dy dz
I'll try this order first: ∫( ∫∫ SQUIRREL dx dy)dz.
I've mentioned that my personal inclination in finding limits of iterated integrals is working from the outside-most limit "in". There are definitely people who are successful and do the exact opposite. I would recommend that you find your own "natural" style and try to follow that path. For me, I would look at the z limits first. For this shape, I would try to find the highest and lowest z's in the spatial region. This is not a complicated region, and we've already sketched it quite well. The highest and lowest z's are, respectively, z=0 and z=3. So we've got ∫_z=0^z=3(∫∫SQUIRREL dx dy)dz.

Now let's try slicing the region by z=CONSTANT, where the CONSTANT is some unknown number between 0 and 3. This horizontal slice of the original spatial region gets us something in the xy-plane. If you were in class, you may recall that there was some effort involved in sketching the slices that are shown here. But one boundary of the sliced region is y=0, along the x-axis. The other, curved boundary, is "inherited" from z=x²+y. Now z=CONSTANT so as a curve in the xy-plane, if we write it in the standard y=function of x format, we get y=z–x². Therefore this is a parabola (the square on x!) opening down (the minus sign). The top of the parabola (the vertex) occurs when x=0, and there y=z. The intersection(s) of the parabola with the x axis occur when y=0, and there 0=z–x², so that x=±sqrt(z). The inner double integral is ∫∫ SQUIRREL dx dy. What are the bounds on the dy integral? We must look at the slice, and see what the highest and lowest values of y are on the slice. The lowest value is 0 and highest value is z: but on the slice, z is a CONSTANT. The highest value depends on z. Now we know: ∫_y=0^y=z∫ SQUIRREL dx dy
Now in the region pictured, I will slice with y=CONSTANT and see how big and how small x can be. This is a slice of a slice (maybe [slice]²?). So the boundary is given by z=x²+y, and with both z and y CONSTANT, I get x²=z–y, so that x=&plusnmn;sqrt(z–y). These will be the limits on the dx integral.

So the answer is:
∫_z=0^z=3∫_y=0^y=z∫_{x=–sqrt(z–y)}^{x=+sqrt(z–y)} SQUIRREL dx dy dz.

dy dz dx
Now ∫( ∫∫ SQUIRREL dy dz)dx.
Examine the original picture and the limits on the outermost variable, x, should be revealed. The largest and smallest x's in this region are ±sqrt(3), and therefore we get ∫_x=–sqrt(3)^x=sqrt(3)( ∫∫SQUIRREL dy dz)dx. Our task is now to slice with x=CONSTANT and try to get the other integrals' bounds.

Again, once the "picture" is presented then much of the remainder of the work is made much easier. We spent some time in class drawing this picture. When x=CONSTANT, then certainly the slice goes through the side (on the xz-axis) so that y=0 becomes the left boundary, if we have z assigned to be the vertical coordinate and take y to be the horizontal coordinate. Also the top of the region is still z=3. The other edge is "inherited" again as the effect of the equation z=x2+y. As I mentioned in class, it is this edge which irritates my highly trained mathematical psychology (is there such a thing?). Notice that x=CONSTANT, so that z=x2+y is a straight line in the yz-plane. The slope of this line is 1. And, when y=0, z must be x2.
The limits on the outside of the double iterated integral ∫∫SQUIRREL dy dz can now be "read off" from the picture, since the smallest value of z is x2 and the largest value is 3. Therefore we have the limits on the outside of the double iterated integral: ∫z=x2z=3∫SQUIRREL dy dz. Finally, the bounds on the dy integral are obtained by slicing the slice. So now z=CONSTANT also, and y goes from y=0 to the right side, which is a point on the line (it still hurts to write this when there is a square in the equation!) z=x2+y, and therefore the upper bound is y=z–x2.

So the answer is:
∫x=–sqrt(3)x=sqrt(3)∫z=x2z=3∫y=0y=z–x2 SQUIRREL dy dz dx.

dz dx dy
My last attempt: ∫( ∫∫ SQUIRREL dz dx)dy.
Again, the picture shows that y in the solid region varies from 0 to 3, and we've got 2 of the 6 limits (o.k., the easiest of them): ∫_y=0^y=3( ∫∫ SQUIRREL dz dx)dy. The y=CONSTANT slice should give the other information.

Again, the picture gives much of the information we need. Drawing the picture was some work. Here with y=CONSTANT, the top of the slice is caused by the plane z=3. The bottom of the slice is z=x²+y. Now since x is a variable, this is indeed a parabola. The parabola opens up (positive coefficient on the square term) and has vertex (0,y): the first coordinate is x and the second coordinate in this slice is z. The parabola intersects the line z=3 when 3=x²+y. Since y=CONSTANT, this occurs when x=±sqrt(3–y). The outer, x limits, on the double integral will be x=–sqrt(3–y) and x=+sqrt(3–y). Now slice the slice, for make x=CONSTANT also. z will vary. The highest value of z will be 3 on the [slice]². The lowest value of z is given by z=x²+y.

The final way the poor SQUIRREL is chopped up and then summed is
∫_y=0^y=3∫_{x=–sqrt(3–y)}^{x=+sqrt(3–y)}∫_z=x²+y^z=3SQUIRREL dz dx dy.

Comments
First, this is a classroom example. The solid region is actually not very complicated. It is a convex region (nothing jutting out at an angle) with boundaries given by low-degree polynomials. The word convex means that line segments whose ends are in the region always have the whole line segment in the region. The problem would be much more complicated if the functions defining the boundary weren't so simple, or if some of the slices weren't convex (then we'd need to split up the integrals, etc.). I remarked in class and I'll repeat here that the process of finding these limits seems to be difficult, and hard to describe -- I don't know yet of a computer program which can do it reliably.
Here are the "answers" again:
∫z=0z=3∫y=0y=z∫x=–sqrt(z–y)x=+sqrt(z–y) SQUIRREL dx dy dz
∫x=–sqrt(3)x=sqrt(3)∫z=x2z=3∫y=0y=z–x2 SQUIRREL dy dz dx.
∫y=0y=3∫x=–sqrt(3–y)x=+sqrt(3–y)∫z=x2+yz=3SQUIRREL dz dx dy.
I can't immediately see that the darn limits describe the same volume in R3. Maybe you can. But you should see, just looking at the patterns of the answers, what sorts of limits are "legal" and what are not. You can only have variables in the limits if they haven't been integrated yet. For example, in the last answer, the lower limit of the innermost integral is z=x2+y, and the outside two integrals are dx and dy. I could not have a limit in, say, the middle integral of the form z=x2+y because there would be only one variable left to be integrated, and there isn't any way to "kill" both x and y. So there is a rough guide to the grammar (?) of the bounds on iterated integrals.

How can you check this kind of "computation"? Generally checking these things can be difficult and tedious. Luckily, we are in the 21st century and I have powerful friends. Well, I guess I can ask some electrons to run around. Look at the following: > W:=x^6*y^8*z^2; 6 8 2 W := x y z > int(int(int(W,x=-sqrt(z-y)..sqrt(z-y)),y=0..z),z=0..3); 1/2 417942208512 3 ----------------- 5763232475 > int(int(int(W,y=0..z-x^2),z=x^2..3),x=-sqrt(3)..sqrt(3)); 1/2 417942208512 3 ----------------- 5763232475 > int(int(int(W,z=x^2+y..3),x=-sqrt(3-y)..sqrt(3-y)),y=0..3); 1/2 417942208512 3 ----------------- 5763232475 I specified a "random" function, W, to replace SQUIRREL. I wanted the antiderivatives not to be a problem, so I just specified some powers of x and y and z. I asked Maple to compute the triple iterated integrals using the three ways we found. The answers are shown. They are such large and silly numbers, and they all agree exactly. This makes me fairly confident the bounds on the iterated integrals are correct. How clever? Not very clever ... I could have specified W=0, and then I bet the computation would returned 0 for all of the setups. This answer would not be very helpful. I could have specified W=1 and the result would be (24/5)sqrt(3), the volume, for all of the setups. That would be fine. In fact, I confess that the first W that I tried was actually x3y7z2. Why wasn't this a very clever choice, and what was the answer I got? Hint: 3 is odd and this region in space is ... The other orders You may wish to check your "skill" by finding the limits on the other three iterated integrals for SQUIRREL. Here you can check your answers: ∫y=0y=3∫z=yz=3∫x=–sqrt(z–y)x=sqrt(z–y)SQUIRREL dx dz dy ∫z=0z=3∫x=–sqrt(z)x=sqrt(z)∫y=0y=z–x2 SQUIRREL dy dx dz ∫x=–sqrt(3)x=sqrt(3)∫y=0y=3–x2∫z=x2+yz=3SQUIRREL dz dy dx These can all be (sort of!) read from the pictures above, and these pictures were on the board at the end of class.

Volume of a tetrahedron
A tetrahedron is an object with flat sides having four corners. I want to compute the volume of a tetrahedron whose corners are at (0,0,0), (1,0,0), (0,2,0), and (0,0,3). There are several ways to compute this volume, including some which need no "calculus", just vector manipulation (using the triple product formula, for example with a cross product and a dot product). To the right is an attempt at a picture. It shows the corners (the four vertices) and the faces, and I've made some attempt to show the four sides with differently decorated "stripes" on each. There are four flat sides. One side is on each coordinate plane (xy-, yz-, xz-) and there is a tilted face. The equation of the tilted face (the points (1,0,0) and (0,2,0) and (0,0,3) are on the tilted face) is x+(y/2)+(z/3)=1. I got this equation by guessing (well, if Ax+By+Cz=1 is the equation, plug in the various points).

Here we use a double integral to find the volume of the tetrahedron. I think of this solid as lying over a triangle in the xy-plane. The triangle is determined by (0,0), (1,0), and (0,2).The height of the solid over this triangle is z=3–3x–(3/2)y, which the equation for the tilted face gives by solving for z.

As a double integral
So the volume is ∫∫_BaseHeight dA, and this is ∫∫_{The triangle}3–3x–(3/2)y dA. I'll convert this to an iterated integral to compute it.

47 second break for theory
In one variable calculus, the initial glimpse at the theory in back of the definite integral assumes that the function doesn't have any jumps. But real functions can jump! The functions which are met in mechanical engineering (just hit something!) can certainly look like what's shown to the right. And similarly, functions met in digital signal processing really can look like that also. They certainly can be integrated. The secret is that the jumps really aren't very important. They can be put inside little boxes where the variation doesn't matter very much (the red boxes in the picture). So the sums defining the definite integral still approach a limit, the "correct" limit.

Here I am apparently not even worrying about the domain. Well, this is what we could do if we had another 30 minutes to fritter away on details. I could define a function piecewise in this way:
F(x,y)=3–3x–(3/2)y if (x,y) is in the triangle, and F(x,y)=0 if (x,y) is not in the triangle. Suppose R is any rectangle in the xy-plane which contains the triangle. Then the volume of the tetrahedron would be ∫∫_RF(x,y) dA. I hope that you will see this double integral is the same as the double integral over the triangle that I'll compute by looking at iterated integrals. The discontinuities of the piecewise-defined function turn out to give a perturbation of the Riemann sums which →0 as the size of the pieces →0. If the Riemann sum is gotten from an n-by-n partition, the discontinuities would be located in at most 3n pieces, and n² is much bigger than 3n when n is large.

Converting to iterated integrals
Let's write ∫∫_{The triangle}3–3x–(3/2)y dA as a dx dy iterated integral. That means figuring out the bounds on the integrals.
I will work from the outside in. So first I need to get the lowest and highest values of y in the triangular base:
∫_Lowest y^Highest y∫_?^??3–3x–(3/2)y dx dy
There's a sketch of the base to the right, and the sketch declares that the Lowest y is 0 and the Highest y is 2. Now I imagine (and frequently draw, as shown on the sketch!) a very thin collection of dx by dy rectangles being added up in a row across the region. It is so thin that y is almost constant and the x's range from the leftmost edge to the rightmost edge. The left edge is certainly 0 always. But the right edge depends on y. When y is very near the bottom (y=0), the right edge is very near 1. When y is near the top (y=1), the right edge is near 0. What is the relationship between x and y on this edge? Of course the edge reflects the tilted face of the tetrahedron, which has the equation x+(y/2)+(z/3)=1. On the base, z=0, so the equation giving the tilted side of the triangular base must be x+(y/2)=1. Therefore x on the rightmost edge is given by x=1–(1/2)y. Here is the resulting iterated integral:
∫₀²∫_x=0^x=1–(1/2)y3–3x–(3/2)y dx dy
Even thought it is not logically necessary (because the dx dy notation does determine what variable is integrated first), I do tend to write "x=" on the limits of the inner integrals. This may save me from confusion and error as I compute.

Computing the iterated integral
I'll first compute the inner integral:
∫_x=0^x=1–(1/2)y3–3x–(3/2)y dx= (antidifferentiate with respect to x, so y is a constant here!) 3x–(3/2)x²–(3/2)yx]_x=0^x=1–(1/2)y= 3{1–(1/2)y}–(3/2){1–(1/2)y}²–(3/2)y(1–(1/2)y)–0. The –0 comes from the lower limit, x=0. I tend to expand and "simplify" here. So we get:
3–(3/2)y–(3/2){1–(1/2)y}²–(3/2)y(1–(1/2)y)=3–(3/2)y–(3/2){1–y+(1/4)y²}–(3/2)y+(3/4)y²= (3/2)–(3/2)y+(3/8)y²
Now the outer integral:
∫₀²(3/2)–(3/2)y+(3/8)y²dy=(3/2)y–(3/4)y²+(1/8)y³]₀²=(3/2)(2)–(3/4)(4)+(1/8)(8)=1.
Maybe it should be "clear" to me that the volume is 1, but it isn't.

The other iterated integral
Now, just to practice, we'll write ∫∫_{The triangle}3–3x–(3/2)y dA as a dy dx iterated integral.
Again, I will work from the outside in. So first I need to get the leftest (leftmost) and rightest (rightmost) values of x in the triangular base:
∫_Leftmost x^Rightmost x∫_?^??3–3x–(3/2)y dy dx
Now the base triangle is again shown to the left, but with the kind of "doodles" that I would make suitable to finding the limits of a dy dx iterated integral. The leftmost value of x is 0 and the rightmost value of x is 1. Now my dx by dy triangles form a vertical strip where x is just about constant. For the inner limits on y I need to know that the strip goes from the bottom, where y=0, to the top. The top will vary, depending on x. The equation of the boundary line for the top is the same: x+(y/2)=1. Now we need to know y as a function of x. So solve for y and get y=2–2x. That's the upper limit on the dx integral. Here is the resulting iterated integral:
∫₀¹∫_y=0^y=2–2x3–3x–(3/2)y dy dx.

Computing the iterated integral
I'll first compute the inner integral:
∫_y=0^y=2–2x3–3x–(3/2)y dy= (antidifferentiate with respect to y, so x is a constant here!) 3y–3xy–(3/4)y²]_y=0^y=2–2x= 3(2–2x)–3x(2–2x)–(3/4)(2–2x)²–0. Again, the –0 comes from the lower limit, y=0. Now expand and simplify:
6–6x–6x+6x²–(3/4){4–8x+4x²}=3–6x+3x²
The outer integral:
∫₀¹3–6x+3x²dx=3x–3x²+x³]₀¹=3–3+1=1.
Thank goodness, we got 1 again.

Possible sources of error in these computations
I'm looking ahead a little bit here. We will discuss triple integrals next time, and these are also usually computed by a transition to triple iterated integrals. There are six possible orders for iterated triple integrals. I make errors frequently. The prominent sources of error include: antidifferentiating with respect to the wrong variable, substituting for the wrong variable, and, well, general confusion. Please try to guard against these. You may make these errors, and just do the computation again, and try to keep your composure intact ("Keep cool, y'know!").
By the way, although I wanted our first example to be as easy as possible, when I was typing up the diary notes above, I ... made several errors and had to go back and redo things. Oh well.

Another one
The base of a solid is the region in the first quadrant of the xy-plane bounded by the curve y=x² and the line y=3x. The height over the xy-plane is given by z=x⁶y⁷. Find the volume of this solid.
The double integral is ∫∫_BaseHeight dA, and this is ∫∫_The shapex⁶y⁷ dA.

One iterated integral, with its computation
We will convert this first to a dy dx integral. The outside limits come first. The most left x gets on the base is x=0. The most right x gets is x=3. We know this because we graphed the base, and found the intersection points of y=x2 and y=3x by solving 3x=x2, which has roots at x=0 and x=3. Therefore the iterated integral looks like:
∫03∫???x6y7dy dx
What about the limits on y? Here the sketch of the base, together with my doodles, may be useful. The vertical strip of boxes tells me that I should add up things from y=x2, the lower bound, to y=3x, the upper bound. Therefore this iterated integral is:
∫03∫y=x2y=3xx6y7dy dx
Now to compute the integral. The inner integral:
∫y=x2y=3xx6y7dy=(1/8)x6y8]y=x2y=3x=(1/8)x6(3x)8–(1/8)x6(x2)8.
This "simplifies" to (1/8)38x14–(1/8)x22. (I am using the powerful rules of exponential manipulation here!) And now the outer integral:
∫03 (1/8)38x14–(1/8)x22dx= (1/8)38(1/15)x15–(1/8)(1/23)x23]03=
(1/8)38(1/15)315–(1/8)(1/23)323=(1/8)323([1/15]–[1/23]). Wow!

The other iterated integral, with its computation
Now for the dx dy integral. The highest and lowest values for y are 0 and 9. Therefore the integral must be:
∫₀⁹∫_?^??x⁶y⁷dx dy
Now we need to consider a (fixed) y slice through the base. The left-hand side of that fixed y slice is determined by y=3x and the right-hand side of the slice is determined by y=x². We need to know the limits on x in terms of y. So we need to know x=Left(y) and x=Right(y). That means "solving for x" in the boundary equations. This (here, in this classroom example!) is not too hard. y=3x becomes x=(1/3)y on the left, and y=x² becomes x=sqrt(y) on the right. The positive square root gets used here because (picture!) we're in the first quadrant. The iterated integral is:
∫₀⁹∫_x=(1/3)y^x=sqrt(y)x⁶y⁷dx dy
The computation begins with the inner integral.
∫_x=(1/3)y^x=sqrt(y)x⁶y⁷dx=(1/7)x⁷y⁸]_x=(1/3)y^x=sqrt(y)=
(1/7){sqrt(y)}⁷y⁷–(1/7){(1/3)y}⁷y⁷=(1/7)y^7/2y⁷–(1/7)(1/3)⁷y⁷y⁷
This now "simplifies" (what a silly word!) to (1/7)y^(21)/2–(1/7)(1/3)⁷y¹⁴. Now the outside:
∫₀⁹(1/7)y^(21)/2–(1/7)(1/3)⁷y¹⁴dy=
(1/7)(2/(23))y^(23)/2–(1/7)(1/3)⁷(1/15)y¹⁵]₀⁹=
(1/7)(2/(23))9^(23)/2–(1/7)(1/3)⁷(1/15)9¹⁵= (1/7)(2/(23))3²³–(1/7)(1/3)⁷(1/15)3³⁰=
(1/7)(2/(23))3²³–(1/7)(1/15)3²³=(1/7)3²³([2/(23)]–[1/15])

Theorem

 1     / 1      1  \     1    / 2      1  \
--- · | ---- - ---- | = --- ·| ---- - ---- |
 8     \ 15     23 /     7    \ 23     15 /

This "theorem" may be the silliest statement of the course.

More on difficulties and on the psychology of the individual
Please don't panic. If you want to compute a double integral, you don't need to do both iterated integrals -- just one of them. I chose to do both to show you how (I hope!).
Notice that we needed to go from one description of the boundary curves: {y=3x, y=x²}, to another: {x=(1/3)y,x=sqrt(y)}, when we did dx dy after dy dx. "Solving" (finding a convenient form for inverse functions) may be difficult (or even impossible in terms of familiar functions).
One last remark: I almost always try to find the bounds on iterated integrals going from the outside-most integral to the inside-most integral. Some people may find the transition from inside to outside more easy (this difference in approach will be more emphatic when we do triple integrals). You should try a series of examples and settle upon what you find most comfortable. And remember that you can always "change" to the other way.

By the way ... > int(int(x^6*y^7,x=(1/3)*y..sqrt(y)),y=0..9); 31381059609 ----------- 115 > int(int(x^6*y^7,y=x^2..3*x),x=0..3); 31381059609 ----------- 115 So Maple gets the same answer both ways, also. Integrating Frog over a region If the integrand (the function to be integrated) is not too weird, then I hope you should be convinced that the actual antidifferentiations probably aren't the essential difficulty. The difficulty is more in setting up the iterated integrals: finding the bounds. Here is a more complicated example. The region R is bounded by y=x+2 and y=x2. What does this region look like? Well, this is a problem in a calculus course, so solving x2=x+2 shouldn't be impossible. In fact, this leads to x2–x–2=0 which is (x–2)(x+1)=0 so intersections occur when x=–1 (so y=(–1)2=1) and when x=2 (so y=22=4). I would like to integrate Frog over the region R: ∫∫RFrog dA. More precisely, I'd just like to set up the bounds of the iterated integrals which are equal to this double integral. Totally randomly, dx dy first The dx dy order introduces an additional kind of complexity. Consider these limits: ∫Bottom of yTop of y∫x=Left(y)x=Right(y)Frog dx dy. The Top of y and Bottom of y are easy enough. In the region R, the smallest y value is 0 and the largest y value is 4. Now think about x=Left(y) and x=Right(y). The thick horizontal blue line in the sketch separates different formulas for the Left limit of x as a function of y. Below it, the Left limit is determined by the left-hand side of the parabola. Above it, the Left limit is determined by the straight line. Theoretically this does not cause any problems. But when you're actually trying to compute everything, what people usually do is separate the pieces: ∫01∫x=Left(y)x=Right(y)Frog dx dy +∫14∫x=Left(y)x=Right(y)Frog dx dy. In the first iterated integral, as y goes from 0 to 1, the left and right boundaries are both given by formulas related to y=x2. Here x=+/–sqrt(y), so Left(y)=–sqrt(x) and Right(y)=+sqrt(x). In the second iterated integral, y goes from 1 to 4. Here also Right(y)=+sqrt(x), but Left(y) comes from y=x+2, so Left(y)=y–2. So the dx dy iterated integrals which are equal to the double integral are: ∫01∫x=–sqrt(y)x=sqrt(y)Frog dx dy +∫14∫x=y–2x=sqrt(y)Frog dx dy. Now dy dx This one is much easier. I can read off the left and right extreme values of x, and then the y boundary values are given by the equations which already "present" the region R. I don't need to split up things. Here it is: ∫–12∫y=x2y=x+2Frog dy dx. Almost surely, unless circumstances were very strange, I would set up the iterated integral this way and not in the dx dy way. The New Jersey Chorus Frog, Pseudacris feriarum kalmi, is an endangered species in some of its range.

A problem from a calculus textbook
This problem shows one further "wrinkle" that can occur with double or triple or any kind of "multiple" integral. Here's the statement:

Evaluate the double integral ∫∫_De^x/ydA, where D={(x,y)|1≤y≤2, y≤x≤y³}.

The dx dy double integral is easy enough to write, because the region D is described suitably:
∫₁²∫_x=y^x=y3e^x/ydx dy
The inner antidifferentiation gives ∫_x=y^x=y3e^x/ydx=ye^x/y]_x=y^x=y3=ye^y3/y–ye¹=ye^y2–ey and then ∫₁²ye^y2–ey dy is just (1/2)e^y2–(e/2)y²]₁²=(1/2)e⁴–2e.

Comment I do know some examples, in "real" applications, not textbooks, where looking at the order of the iterated integrals changes something really nasty into a function which can be handled routinely. So, although this is a textbook/class example, it does show an idea which may be useful. I did "ask" Maple for the antiderivative of e1/y. Its reply contained something I wasn't familiar with. When I asked for help, I essentially was told that this was a function whose derivative was e1/y. In other words, Maple responded to the question, "What's the antiderivative of e1/y" with the statement, "The antiderivative of e1/y." Such a reply may not be useful.

QotD
Consider the iterated double integral ∫_x=0^x=2∫_y=0^y=4–x2TOAD dy dx. I'd like two tasks done:

1. Sketch the region in the plane over which this integral is computed.
2. Write the integral as an iterated double integral in dx dy order. You can't "compute" it since I haven't told you anything about TOAD, so the answer to this part is something of the form ∫_y=CONSTANT1^{y=CONSTANT₂}∫_x=ALGEBRA1^x=ALGEBRA₂ TOAD dy dx where ALGEBRA₁ and ALGEBRA₂ may include y's (but they don't have to). CONSTANT₁ and CONSTANT₂ must just be numbers with no x's or y's.

To the right is Maple's answer to the first question. The answer was obtained with several commands. First, I used plot3d(0,x=0..2,y=0..4-x^2,scaling=constrained,axes=normal, color=yellow,style=patchnogrid,orientation=[-90,0]); to get a "yellow" (looks more mustard-color to me!) region. Then I used spacecurve(<t,4-t^2,0>, t=0..2,color=black,thickness=2,orientation=[-90,0]); to get a black boundary curve. I combined these pictures with a display3d command. I believe that the answer in dx dy order is ∫y=0y=8∫x=0x=sqrt(4–y) TOAD dx dy. The equation y=4–x2 becomes x2=4–y which becomes in turn x=sqrt(4–y).

Check, please! How could I possibly check this answer, since no one is looking at my work? Well, a preliminary check might be to ask my friend this: > int(int(1,y=0..8-x^3),x=0..2); 12 > int(int(1,x=0..(8-y)^(1/3)),y=0..8); 12 Maybe, though, I was just lucky: there aren't many small integers, so it is possible that the answers are just accidentally equal (really: such things do happen). But if I see what follows, my emotional (?) confidence (??) in my answer is considerable strengthened, because it seems highly unlikely that such bizarre answers could "accidentally" be the same. I am considerably encouraged that my answer is correct: > int(int(3*x^5+5*y^7,y=0..8-x^3),x=0..2); 371507663104 ------------ 40755 > int(int(3*x^5+5*y^7,x=0..(8-y)^(1/3)),y=0..8); 371507663104 ------------ 40755

Theorem For any choices of sample points, as n→∞, the Riemann sums→a unique limit, the definite integral of f from a to b, and written ∫_a^bf(x) dx.

Of course the integral sign and the dx's are notation to remind people of the approximating sums. We could approximate the definite integral with Riemann sums, and of course there are many other numerical approximation schemes. But the champion method for computing the definite integral is:

FTC If F´=f, then ∫_a^bf(x) dx=F(b)–F(a).

Defining the double integral
The definition more or less parallels the single integral definition. I'll follow the text closely here. We begin with a nice function (say, continuous) defined on a rectangle R in R² with boundaries x=a, x=b, y=c, and y=d. Chop up the area of the rectangle into a bunch of chunks. In each chunk, choose a sample point. Compute the corresponding Riemann sum, the sum over all the chunks of f's value at the sample point multiplied by the area of the chunk. If f(x,y)>0 on R, then this Riemann sum approximates the volume under z=f(x,y) and over R. Then it's true that as the maximum size of the chunks→0 (here the best way to measure "size" is by diameter rather than, say, area), the Riemann sums→a unique limit. This limit is the double integral over R of f(x,y): ∫∫_Rf(x,y)dA. This is a mathematical abstraction of the volume. The volumes computed as a result of formulas in earlier calculus (solids of revolution, solids with simple cross-sections, etc.) take advantage of symmetry. The theoretical tool defined here allows us to compute volumes without any simple kinds of symmetry. As I mentioned in class, numerical computations of double integrals (and other higher-dimensional creatures) are sometimes necessary, but the computations get much more intricate than the simple ideas shown in calc 2.
Almost all the computations of volumes that I've made in my life have occurred as a result of teaching third semester calculus. Maybe the following is a bit more interesting.

Mass of a plate Maybe this is a more realistic "scenario". Suppose you are given a thin rectangular metal plate with an unknown density distribution. Therefore this is not necessarily a homogeneous thin plate. The plate is too heavy or too unwieldy to weigh directly, and you need to estimate the total mass. Also the mass distribution -- the density -- is not necessarily given by a simple formula. What maybe could be done is tiny Samples taken at various parts of the plate, according to some method (maybe depending on accessibility or expense or ... anything). Then these samples could have their density measured, and maybe then, after dividing the plate (thoughtwise!) into pieces, the sample densities could be multiplied by the areas of the pieces. The sum of these products would then be an estimate for the mass in the plate. (It is a Riemann sum.) If a better (more accurate?) estimate was wanted, maybe then use more sample points, smaller areas, etc. The process is exactly the same mathematics as the definition of the double integral. Reality? Well, you might work for Schlumberger and your sample points might cost three to five million dollars each as you try to investigate the oil or gas quantities of some region. You'd then really think a bit about the whole process. And that's what they do.

Some very simple examples of double integrals
I did some examples similar to these.
Example A The rectangle R is defined by x=3 and x=7 and y=5 and y=8. The function is f(x,y)=700. Then ∫∫_Rf(x,y)dA is 700(7–3)(8–5). Of course, I am using the fact that the volume described by the double integral is the volume of a rectangular solid with edge dimensions 700 and 7–3 and 8–5.
Example B Here I took f(x,y) to be 5–x²+y², definitely a more complicated function than the previous example's. The rectangle I took was defined by x=–3, x=3, y=–3, and y=3. Let's temporarily discard the 5 and concentrate on –x²+y². If I interchange x and y the sign of the function's value changes. But the rectangular domain is symmetric about (0,0), so the net value (+'s and –'s cancelling!) of the double integral of –x²+y² over the rectangle is 0 (!). Now I "integrate" the 5, and the result is 5(3–(–3))(3–(–3)). So we did have a more complicated function but the choice of domain made the hard part of the function drop out.

Basic properties
The basic properties of the double integral are exactly like those of the 1-dimensional version.

• Linearity If f and g are both nice functions defined on the rectangle R, then ∫∫_Rf(x,y)+g(x,y)DA=∫∫_Rf(x,y)dA+∫∫_Rg(x,y)dA; if k is a constant, then ∫∫_Rkf(x,y)dA=k∫∫_Rf(x,y)dA.
• Order If f(x,y)≤g(x,y) for all points (x,y) in the rectangle, then ∫∫_Rf(x,y)dA≤∫∫_Rg(x,y)dA.

How to compute?
Maybe all this theory is very nice, but let me show you the way most double integrals are computed. One method of chopping up a rectangle is to use vertical and horizontal lines, parallel to the sides. So we get a grid of subrectangles, each Δx by Δy, with both Δ's very small. In addition to choosing this special chopping strategy we could also decide to add up the contributions (the f(sample point)ΔxΔy) in an orderly manner. So, for example, we could add up the contributions from the lowest "row" first. In that row, since Δy is small, y hardly varies at all. The sum of a row sure looks like the definite integral with respect to x only for "that" value of y. The same can be done for each row. When the row sums are done, we now have the y's to worry about. But this is a y integral. Of course, a completely symmetric procedure can be used in the other order: first dy, with x held constant, and then dx. Again what's here is not a "proof" (this is not too difficult, actually) but I hope the discussion supports the following result.

Fubini's Theorem
Suppose f(x,y) is a continuous function in a rectangle R defined by x=a, x=b, y=c, and y=d. Then
∫∫_Rf(x,y)dA=∫_c^d∫_a^bf(x,y)dx dy=∫_a^b∫_c^df(x,y)dy dx.
Comments The first "creature" in the equation above is called a double integral. The two others are officially called iterated integrals: iterated means "repeated" (you m ight think that these things should be called "partial integrals" in analogy with partial derivatives, but ... they are not). Technically and precisely these two kinds of integrals are different creatures. Also please note that the outside integration limits go with the outside d-variable -- sometimes this can be confusing. When it is, I write things like "x=a" instead of just "a" so I don't confuse myself.

Example 1
Let me try to compute ∫∫_R x³y⁷dA where R is the rectangle defined by x=1, x=4, y=2 and y=5. The Fubini Theorem allows me to "trade in" the double integral for an iterated integral, either dx dy or dy dx. There are some occasions where one order or the other might be preferable (we'll see this later) but here I don't think that happens. So:
∫∫_Rx³y⁷dA=∫₂⁵∫₁⁴x³y⁷dx dy.

I'll begin by computing the inner integral:
∫₁⁴x³y⁷dx=(1/4)x⁴y⁷]₁⁴. In this antidifferentiation, the y⁷ is a constant (this is, not surprizingly, exactly the inverse of partial differentiation). Then we evaluate and get (1/4)4⁴y⁷–(1/4)1⁴y⁷=(255/4)y⁷. I remarked in class that I sometimes lose my way in these computations, and need to write (1/4)x⁴y⁷]_x=1^x=4 to insure that I remember to substitute for the correct variable.

Now ∫₂⁵(255/4)y⁷dy=(255/4)(1/8)y⁸]₂⁵=(255/4)(1/8)5⁸–(255/4)(1/8)2⁸.

My silicon buddy ... A report from Maple: > int(int(x^3*y^7,x=1..4),y=2..5); 99544095 -------- 32

Example 2
Let's compute ∫∫_Rf(x,y) dA where R is the rectangle shown to the right and determined by these inequalities: 0≤x≤Π/4 and –Π/2≤y≤Π, and where f(x,y)=cos(2x+3y).

One order was done in class, and the other was, how wonderful, the QotD!

Let me try this dy dx. So:
    ∫_x=0^x=Π∫_y=–Π/2^y=Πcos(2x+3y)dy dx     

• FTC in y (the inner integral):
  ∫_y=–Π/2^y=Πcos(2x+3y)dy=(1/3)sin(2x+3y)]_y=–Π/2^y=Π=(1/3)sin(2x+3Π)–(1/3)sin(2x–3Π/2).
• FTC in x (the outer integral):
  ∫_x=0^x=Π/4(1/3)sin(2x+3Π)–(1/3)sin(2x–3Π/2)dx=–(1/6)cos(2x+3Π)+(1/6)cos(2x–3Π/2)]_x=0^x=Π/4=
  (–(1/6)cos(7Π/2)+(1/6)cos(–2Π/2))–(–(1/6)cos(3Π)+(1/6)cos(–3Π/2))=
  (–(1/6)0+(1/6)(–1))–(–(1/6)(–1)+(1/6)0)=–1/3.

Now, just for fun (?), I'll do this dx dy. So:
    ∫_y=–Π/2^y=Π∫_x=0^x=Πcos(2x+3y)dx dy     

• FTC in x (the inner integral):
  ∫_x=0^x=Π/4cos(2x+3y)dx=(1/2)sin(2x+3y)]_x=0^x=Π/4=(1/2)sin(Π+3y)–sin(0+3y).
• FTC in y (the outer integral):
  ∫_y=–Π/2^y=Π(1/2)sin(Π+3y)–(1/2)sin(3y)dy= –(1/6)cos(Π+3y)+(1/6)cos(3y)]_y=–Π/2^y=Π=
  (–(1/6)cos(4Π)+(1/6)cos(3Π))–(–(1/6)cos(–Π/2)+(1/6)cos(–3Π/2))=
  (–(1/6)1+(1/6)(–1)))–(–(1/6)0+(1/6)0)=–1/3.

Or, of course > int(int(cos(2*x+3*y),y=-Pi/2..Pi),x=0..(1/4)*Pi); -1/3 > int(int(cos(2*x+3*y),x=0..(1/4)*Pi),y=-Pi/2..Pi); -1/3

APOLOGY to sections 15, 16, & 17     So I computed the double integral on Friday, October 22, on the region defined by 0≤x≤Π/4 (that's fine) but also with the y restriction –Π/4≤x≤Π (so the lower y bound is different from what I wanted). The answer is (sqrt(2)/6)–1/6 as we got in class. Certainly this is not as simple and "nice" as the –1/3 answer I anticipated. I'm sorry that I made this mistake.

More max/min "review" from 1 variable calculus In one variable calculus, the study of max/min included both local max and min and also max and min in an interval. The theoretical result which supports this is the following: A continuous function on a closed bounded interval always attains its maximum and its minimum. The theoretical justification of this statement is rarely mentioned in calc 1, but usually some examples are presented to show that the assumptions are needed. So: in an open interval, not containing its endpoints, a continuous function need not attain its min or its max (consider tan(x) on (–Π/2,Π/2) for example). And a function which is NOT continuous on a closed bounded interval need not attain its min or its max (consider the piecewise function f(x)=1/x for x≠0 and f(x)=0 on the interval [–1,1], for example).

When we hunted for extrema (max/min) on a closed bounded interval, then we had to check both the interior critical points and the end points. The theory in >1 variable A similar theoretical result is true in more than one dimension. That result is: a continuous function on a closed bounded set must attain its maximum and its minimum. Here "bounded" means the whole set sits inside some (perhaps very big!) ball. And "closed" means that all of the limits of sequences in the set are in the set: the edges of the set are included. Let me do an example.

Example on a square
Consider the polynomial f(x,y)=x²+y²+2+y and let's try to find its max and min values on the square with sides determined by x=±1 and y=±1. Inside the square, we can look for peaks or pits by checking for critical points. So let's consider:
f_x=2x and f_y=2y+1. The only critical point is (0,–1/2). It would be difficult for f's value at (0,–1/2) (which is 1.75) to be both the max and the min of f on the square. But f can have extreme values on the boundary of the square without these values occurring at critical points.

The boundary of the square is fairly simple to investigate. For example, one side has x=1 and –1≤y≤1. On that side f(1,y)=1²+y²+2+y=y²+y+3. Hey: the (1 dim) c.p. for this is at 2y+1=0 so y=–1/2. And the max/min for just that side will be either at –1/2 or –1 or 1 (the last two values of y are the endpoints of the interval, of course). Now f(1,–1)=1²+(–1)²+2–1=3 and f(1,–1/2)=1²+(–1/2)²+2–(1/2)=2.75 and f(1,1)=1²+1²+2+1=5. Whew!

We could check the side with y=–1. Then –1≤x≤1 and f(x,–1)=x²+(–1)²+2+(–1)=x²+2. There's a (1 dim) c.p. at x=0, and the value at that point and the endpoints should be considered: f(0,–1)=0²+(–1)²+2+(–1)=2 and f(–1,–1)=(–1)²+(–1)²+2+(–1)=3 and f(1,–1)=1²+(–1)²+2+(–1)=3.

I think I won't look at the other sides. The minimum value TURNS OUT TO BE 1.75 and the maximum value, 5. A picture of the situation is shown to the right.

I haven't told you the real problem: there are far more "closed and bounded" objects than square or rectangles, and, not like the 1 dimensional case, the boundaries of the objects can be rather complicated. Interior max and mins can be found by searching for critical points, using the techniques we previously discussed. But ... for strange boundaries, ideas we haven't seen before are used. Let me show you an example.

A rather simple example Find the maximum and minimum values of 3x–4y+5z on the unit sphere x2+y2+z2=1. Here is perhaps a more complicated picture, with the constraint (the unit sphere) and five planes representing where f(x,y,z)=3x–4y+5z=–8 and –3 and 1 and 5 and 9. The picture is supposed to help you understand that max/min occur where the planes will be tangent to the sphere. The system of Lagrange multiplier equations (three of them here, since we are in R3) together with the constraint follows. 2x=3(λ) 2y=–4(λ) 2z=5(λ) x2+y2+z2=1 The left-hand sides are the components of ∇(x2+y2+z2) and the right-hand sides are λ multiplying the components of ∇(3x–4y+5z). You can solve for x and y and z in terms of λ, and substitute these values in the constraint equation, getting λ=±(2/sqrt(50)). Then 3x–4y+5z turns out to be (for the two choices of λ, generating two candidates for where extreme values take place) sqrt(50) and –sqrt(50). Here is a final picture of the constraint and the two planes given by 3x–4y+5z=±sqrt(50).

Proofs, etc.: the dual (?) nature of math I learned and "liked" Lagrange multipliers in a several variable calculus course, just as I hope you might. The justification for the method was more or less what I have shown you. So I knew it was "true". But I never saw a "proof" of the Lagrange multiplier method until my second year of grad school. Sigh. It really isn't that difficult to prove. Maybe I didn't (even as an apprentice professional mathematician!) feel the need to prove such a lovely idea.

A more complicated dimension 2 example
Find the point(s) on y=1/x² which is(are) closest to the origin and find the distance implied.

Here we are trying to minimize sqrt(x²+y²) when (x,y) satisfies the constraint y–1/x²=0. I want to practice using Lagrange multipliers (I acknowledge that other tricks can be used).

So the objective function is sqrt(x²+y²): wait -- I want to be a little bit nice to myself. Minimizing this function, which is always non-negative is the same as minimizing its square.

This is an idea which is frequently very useful in practice (and it also occurs, for example, in the workshop problem whose solution is to be submitted tomorrow!): distance and energy and other "natura;" quantities may have algebraic descriptions with square roots. The minimum or the maximum of a quantity described as SQRT(FROG) can be obtained by finding the max or min of FROG itself, and then taking the square root. This is because SQRT is a strictly increasing function. Getting rid of SQRT is nice -- derivatives are easier, etc.

What are the Lagrange multiplier equations? The gradient of x²+y² is <2x,2y> and the gradient of y–1/x² is <2/x³,1> and therefore we have the following "system" of three equations in three unknowns to solve:

2x=(λ)(2/x3)
2y=(λ)(1)
y–1/x2=0

The accompanying Maple graph shows a parts of the graphs of y=1/x² and x²+y²=2^1/3+2^–2/3 in the first quadrant. They are tangent, as the previous discussion suggests.

A heated spherical object (?!)
As a last example I considered the following problem: suppose a solid object occupies the space specified by x²+y²+z²≤1. Suppose at the point (x,y,z), the temperature of the object is given by T(x,y,z)=xy²z³ (I am not asserting that this is a physically realistic problem!). What are the maximum and minimum temperatures in the object?

Well, the max/min are either inside the ball or on the surface. let me analyze these separately.

Inside the ball
If the max/min occur inside the ball, then they must happen at critical points. Well, ∇T=<y²z³,2xyz³,3xy²z²>. For which x, y, and z are all of the components equal to 0? This seems almost silly. Look at the first component: if y²z³=0 then either y or z must be 0. But if one of y or z is 0, then the temperature, xy²z³, must be 0. This is rather annoying: there are many, many critical points, but none of them are maxes or mins because the temperature can easily be positive or negative!

On the surface of the ball
So we need to find the max/min of T(x,y,z) subject to the constraint of being on the surface of the ball: f(x,y,z)=x²+y²+z²=1. Well, we've got the vector equation λ∇T=∇f. This works out to the ...

QotD
One vector equation is here three scalar equations. We also need the constraint. So we have:

(1)  λy2z3=2x
(2)  λ2xyz3=2y
(3)  λ3xy2z2=2z
(4)  x2+y2+z2=1

Solving the Lagrange multiplier equations This collection of equations is sufficiently complicated that I've got to think for a while. Note first that all of x and y and z should be not equal to 0 since that would make T=0 and we covered that when we did the interior critical point analysis. This means I can divide without thinking too much (λ also can't be 0 because then that would force one of the x, y, and z to be 0). Divide equation (1) by equation (2). The result is y/(2x)=x/y so that y2=2x2. Divide equation (1) by equation (3) and the result is z/(3x)=x/z so that z2=3x2. These two equations allow me to plug into (4), the constraint equation, and the equation x2+y2+z2=1 becomes 6x2=1 so x=±1/sqrt(6). The signs of x and y and z are not connected since they are related by (THING)2=(OTHER THING)2. Thus y=±sqrt(2)/sqrt(6) and z=±sqrt(3)/sqrt(6). There are EIGHT candidate points on the boundary corresponding to all the possible sign choices of the coordinates. The product of the signs determines whether one of the 8 points gives a max (there are four of those, and the sign product is positive, and the value of temperature is sqrt(3)/36) or gives a min (where the sign product is nagative, at four points where the temperature is –sqrt(3)/36). To the right is a picture of the sphere, x2+y2+z2=1, in blue. That should be easy to recognize. In green is a portion of the level surface, T(x,y,z)=xy2z3=sqrt(3)/36. That's a sort of strange surface. It seems to have 4 pieces, and the pieces are all tangent to the sphere, exactly as the Lagrange multiplier method predicts. But the picture is complicated. Things are intricate. Higher dimensional problems may have lots of special points to consider.

Return of the exam
I returned the exam. Here is a discussion of the grades and grading. Please check your exam grading (and arithmetic!) with the partial credit guidelines before any discussion with me. I remark, as I did in class, that the letter grades are only a guideline for students. To me, there is not much difference between 54 and 55, even though the letter assignments are distinct. I had sent the following message to students who got D's or F's on the first exam, and students with low C's should also read and think about the message.

The first exam has been graded. This message is going to students who got D's or F's on that exam. The exam was graded too late for me to use the official university warning system. I want to signal these students that their chance of success in the course is low.

I certainly can't address particular situations but something should be done to learn more and better, because what you are doing so far is not working. Please decide if what I write applies to you. You may also wish to discuss your particular situation with either of the recitation instructors or with me.

1. You should reconsider your time allocation. Perhaps decrease the time you spend working on jobs or in organized extracurricular activities. Reconsider how you spend your study time: plan seriously to increase the efficiency and usefulness of this time.

2. Consider your overall academic load for the semester. Students who do very poorly in this course's first exam rarely finish with a grade of C or better. A grade of D doesn't guarantee you know very much, even enough to use the ideas and methods of the course in applications as many of you will need. You may want to drop this course or another course. Withdrawal with a W is possible before October 27. If you can't do what's in the previous paragraph, consider dropping this course. That might be the more mature action.

I hope this message is useful. Exams should be returned at the end of class Tuesday.

Sincerely,
S. Greenfield

Workshops
I tried to indicate some of my reasons for asking students to do "workshop" problems, including working in small groups and then writing up responses. This is all related to useful acdademic and vocational skills, and I know this by direct evidience. The probability that I will read your solution is about one-third so please try to do a good job. Thank you.

Second derivative test for two variables
Suppose f(x,y) is a differentiable function, and (x₀,y₀) is a critical point. That is, f_x(x₀,y₀)=0 and f_y(x₀,y₀)=0. Then compute D=f_xx(x₀,y₀)f_yy(x₀,y₀) –(f_xy(x₀,y₀))². Here we go:

If D>0	and if fxx(x0,y0)>0 then f has a local minimum at (x0,y0).
	and if fxx(x0,y0)<0 then f has a local maximum at (x0,y0).
If D<0 then f has a saddle point at (x0,y0).
If D=0 then this second derivative test supplies no information.

Please note that when D>0, the signs of f_xx and f_yy must agree (as I said in class, this is not obvious!) so that you can check either of these numbers. The textbook calls D the discriminant. It is also called the Hessian in some places (for example, Maple and Mathematica). Also I mentioned that I think of D as the determinant of this:

( fxx fxy ) 
( fyx fyy )

Why is this "test" valid?
The test essentially results from computing the second directional derivative at the critical point and seeing how to ensure that this result is always positive (or always negative or ...). That together with results from one variable calculus (on concavity) will insure some kinds of local behavior near the critical point. The textbook has such a discussion. Another way of understanding and verifying this result follows from a two-dimensional version of Taylor's Theorem. Taylor's Theorem also allows extensions of this derivative "test" to more variables and higher derivatives.
In this course, we concentrate on stating the result and then giving some examples. That is enough of a task! I hope you saw the test in your recitations Wednesday. First, let me give an example from a textbook. A textbook example - YAWN ... - well, not just any textbook (or even ours!) but from a different, slightly older, textbook.

Euler's example
Leonhard Euler (1707–1783) was a great and very prolific mathematician. He published Institutiones Calculi Differentialis (In English, Methods of the Differential Calculus) in 1755. It was an influential text, one of the earliest calculus texts, and was the first source of criteria for discovering local extrema of functions of several variables. In it Euler investigated the following specific example: V=x³+y²–3xy +(3/2)x. He asserted that V has a minimum at both (1,3/2) and (1/2,3/4). Was Euler correct? (My source for this information is A History of Mathematics by Victor J. Katz, Harper Collins, 1993, p.517.)

Let's compute. V_x=3x²–3y+{3/2} and V_y=2y–3x. Critical points occur where both V_x and V_y are 0. There 2y=3x or y= {3/2}x, so the V_x condition becomes: 3x²–{9/2}x+{3/2}=0 or 6x²–9x+3=0 which factors (even then textbook problems were predictable!) into (2x–1)(3x–3)=0. The critical points are as Euler asserted: (1,{3/2}) and ({1/2},{3/4}). Logically just checking that Euler's points are critical points is not enough -- we should check that he found all critical points, which we did.

Now to test the type of the critical points: V_xx=6x, V_xy=–3, V_yx=–3, and V_yy=2. So the discriminant is the determinant of

(6x   –3) 
(–3    2)

Two comments about what I did

1. Solving non-linear equations can really be quite difficult, and each collection of such equations can present different "challenges". Any method that gets a solution is a good method, and there's likely to be more than one good method!
2. Since I sometimes make mistakes computing derivatives, I usually compute both the xy and the yx derivatives and quickly see that they're equal. This provides a small check in this computation.

One strange example
The function f(x,y)=–(x²–1)²–(x²y–x–1)² is given. This is not the world's most horrible function. It is "only" a polynomial of degree 6. Let me find the critical points.

Well, f_x=–2(x²–1)2x–2(x²y–x–1)(2xy–1) and f_y=–2(x²y–x–1)x².

Consider the equation f_y=0 first. Well, maybe x=0. Then f_y=0 and f_x=0–2(–1)(–1) is not 0. So this doesn't get me any critical points.

Now to get f_y=0 we can ask that x²y–x–1=0. Then f_x=0 becomes (x²–1)2x=0 since the other piece becomes 0. Then either x=0 or x=1 or x=–1. Whew! If x=0, x²y–x–1=0 becomes –1=0: false. If x=1, x²y–x–1=0 becomes y–2=0 so y=2. Therefore (1,2) is a critical point. If x=–1, x²y–x–1=0 becomes y=0, and (–1,0) is a critical point.

If you don't like this logical torture, try the following:

> f:=-(x^2-1)^2-(x^2*y-x-1)^2;
                               2     2     2           2
                       f := -(x  - 1)  - (x  y - x - 1)
> solve({diff(f,x),diff(f,y)});
                        {x = 1, y = 2}, {x = -1, y = 0}

The pictures certainly shouldn't be convincing evidence, but they do seem to support the assertion that the function has local maximums at both critical points!

Let me try the second derivative at, say, (–1,0). Sometimes computations are good for the soul (sometimes?).
f_x=–2(x²–1)2x–2(x²y–x–1)(2xy–1) so
f_xx=–2(2x)(2x)–2(x²–1)2–2(2xy–1)(2xy–1)–2(x²y–x–1)(2y)  and   f_xy=–2(x²)(2xy–1)–2(x²y–x–1)(2x).

f_y=–2(x²y–x–1)x² so
f_yx=–2(2xy–1)x²–2(x²y–x–1)2x  and  f_yy=–2(x²)x².

At (–1,0):
    f_xx=–2(–2)(–2)–2((–1)²–1)2–2(–1)(–1)=–10
    f_xy=–2((–1)²)(–1)–2(–(–1)–1)(2(–1))=2
    f_yx=–2(–1)(–1)²–2(–(–1)–1)2(–1)=2
    f_yy=–2((–1)²)(–1)²=–2
So D=(–10)(–2)–2(2)=16>0 and f_xx<0: this is a local max.

The other critical point also is a local max. I am getting too tired to try this one. As I mentioned in class, I needed three tries to do the first one correctly when preparing this example by myself!

Why do I find this disconcerting? Well, imagine we walk from one peak to another (shown to the right, the blue "trail"). Shouldn't we somehow pass through a saddle? Well, in fact, no, we don't need to: maybe the lowest point on the blue trail is not a critical point -- the tangent plane to the surface at that point may be tilted. In this example, the tangent plane is always tilted at every point except the two peaks.

I tried to generate a good Maple graph of f, but everything I tried didn't help me visualize things better. Maybe someone else can come up with a good picture.

The situation in 1 variable calculus is considerably different. If I have two local maxes (and, yeah, if the function is continuous, differentiable, etc.: nice) then there must be a local min between them.

Another strange example
Here f(x,y)=3xe^y–x³–e^3y. Then we compute f_x=3e^y–3x² and f_y=3xe^y–3e^3y. If f_y=0, then e^y3(x–e^2y)=0 so x=e^2y (the other factor, e^y, is a value of the exponential function and is never 0). Then f_x=3e^y–3x²=0 leads to 3e^y–3(e^2y)²=0 or e^y–e^4y=0 or e^y(1–e^3y)=0. Since exp is never 0, we need 1=e^3y and this occurs only when y=0. Therefore this function has only one critical point, at (1,0). My friend does this, by the way:

> f:=3*x*exp(y)-x^3-exp(3*y);
                                           3
                        f := 3 x exp(y) - x  - exp(3 y)
> solve({diff(f,x),diff(f,y)});
                               2                                 2
 {x = 1, y = 0}, {x = RootOf(_Z  + _Z + 1), y = ln(-1 - RootOf(_Z  + _Z + 1))}

The left graph below is a local picture of the critical point. This seems to convincingly support the assertion that (1,0) is a local strict maximum of the function. But I'll compute.

Since f_x=3e^y–3x² then f_xx=–6x  and  f_xy=3e^y;
Since f_y=3xe^y–3e^3y then f_yx=3e^y  and  f_yy=3xe^y–9e^3y. We evaluate and need the determinant of

( –6  3 )
( 3  3–9)

In the graph on the right, x goes from –5 to 5 and y varies just between –.05 and .05: therefore y is just about 0, and 3xe^y–x³–e^3y is just about 3x–x³–1. Certainly this shows that the function has no absolute max or min.

The situation in 1 variable calculus is considerably different. Suppose we have a function defined on all of the real numbers (o.k., a differentiable function) which has exactly one critical point and that critical point is a local maximum. Then that local maximum is a global, absolute maximum for the function. What the heck is happening in several variables? I just don't understand, that is, really understand beyond superficially.

Consider the ellipse x²+5y²=1. Find the rectangle of largest area inscribed in this ellipse with sides parallel to the coordinate axes. Of course this turns into: maximize 4xy (the objective function) subject to x²+5y²=1 (the constraint). Consideration of the geometry (varying rectangles) suggests that there is indeed a "biggest" rectangle, somewhere.

How the "heck" does a calc 1 student solve this problem since the function to be maximized, 4xy, has two variables? Here are two methods. Reduction of dimension, simple version This is probably the orthodox way for students in CALC 1 to solve the problem. Since x2+5y2=1, we know that y=sqrt((1–x2)/5). Then the area is F(x)=4x·sqrt((1–x2)/5). The domain for this function is 0≤x≤1. General theory from one variable calculus states that max/min are obtained at end points or critical points. But F(0)=F(1)=1. So the max is gotten where F'(x)=0. We computed this. Of course, in a random situation, it may be very difficult to solve (effectively!) for one of the variables in terms of the other. Reduction of dimension by parameterization In calc 2 students might be aware of parameterization of the curve. This is another way to "lower dimension". We could make an inspired "guess": try x=cos(θ) and y=sin(θ)/sqrt(5). Then the pair (x,y) is on the ellipse, and since the max is obtained somewhere in the first quadrant, we are left with maximizing 4xy=(4/sqrt(5))cos(θ)sin(θ)=(2/sqrt(5))sin(2θ) for θ between 0 and Π/2. This can be solved almost "by inspection": just take θ to be Π/4. The max value is then 4/sqrt(5). Of course, in a "random" situation it may be very difficult to get nice parameterizations.

A weird 2-dimensional way to do the problem

I had Maple sketch some level curves of 4xy, the objective function, and compare them with the constraint curve x2+5y2=1. Here is the result of these Maple commands. A:=contourplot(x*y,x=-1.1..1.1,y=-1.1..1.1,color=red, thickness=2,scaling=constrained,grid=[50,50], contours=[.02,.05,.08,.2,.3,.5,-.02,-.05,-.08,-.2,-.3,-.5]): B:=implicitplot(x^2+5*y^2=1, x=-4..4,y=-4..4,color=blue, thickness=2, scaling=constrained, grid=[80,80]): display({A,B}); The picture is shown to the right.
A close-up view Suppose you consider a level curve of the objective function that crosses the constraint curve, as shown. One math word which applies to this situation is that the two curves are transversal. So we have 4xy=C crossing x2+5y2=1. What happens if we "wiggle" C a little bit, so we consider 4xy=C+ε and 4xy=C–ε (here ε is supposed to be a very small number). Now it seems reasonable (4xy is certainly continuous, so its values don't hop around or break or anything) that these level curves are close to 4xy=C. These level curves must also cross the constraint curve. That means the function 4xy has values C+ε and C–ε on the constraint curve. (The level curves are exactly where that function takes on its values!) Since there are both larger and smaller values of 4xy on the constraint curve, C can't be an extreme value (either max or min) for 4xy on x2+5y2=1.	Local picture near a level curve corresponding to a non-extreme value
Another close-up view This seems to imply, if you examine the picture closely, that the largest (and the smallest) values of 4xy will be at points on the ellipse where the ellipse will be tangent to level curves of the constraint, x2+5y2=1. If the level curves of the objective function are not tangent, then we will be able to vary the values of the constant generating that contour and get bigger and smaller values of the objective function on the constraint curve. If the level curves are tangent then the normal vectors of the constraint curve (∇f at that point) and the objective function (∇g) at that point will both be perpendicular to the same line (in three dimensions it would be a tangent plane). These gradient vectors may not be exactly the same vector, but one of them must be a scalar multiple of the other.	Local picture near a level curve corresponding to an extreme value

Now the algebraic side
If f(x,y)=4xy and g(x,y)=x²+5y²=1, then at such points (extreme values of the objective function on the constraint curve), there is some real number λ so that ∇g=λ ∇f (everyone uses this Greek letter) because the tangent lines are the same, and therefore the normal vectors must be parallel: one must be a scalar multiple of the other. This one vector equation in R² gives two scalar equations, one for each component of the vectors:

2x=(λ)4y
10y=(λ)4x

Here is a detailed solution of the system of equations. The first and second equations tell me that λ=2x/4y=x/2y and λ=10y/4x=5y/2x so that x/2y=5y/2x and 2x²=10y² and therefore x²=5y². Now combine this with the constraint equation, x²+5y²=1. We see that 2x² must be 1, so that x should be ±1/sqrt{2} and y=±1/sqrt{10}. For the maximum value of 4xy, take either both plus signs or both minus signs. Then 4xy will be 4/sqrt{20}=2/sqrt{5}.
To the right is the ellipse, the constraint, and a level curve of the objective function for the supposed maximum value: so it is 4xy=2/sqrt{5}. And, just as predicted, these curves are tangent.

Fan mail for the Lagrange multiplier method
I think it is wonderful that a relatively small amount of algebraic effort can produce such a lovely geometric result (the specific hyperbola which is tangent to the ellipse). This reassures me that things algebraic and geometric both reflect the same reality.

• On Sunday, October 10, Mr. Vidit Nanda will have a review session from from 3:00 PM to 4:30 PM in SEC 202 (on Busch Campus)
• On Monday, October 11, Mr. Gabe Bouch will have a review session from from 7:00 PM to 9:00 PM in LSH B269 (on Livingston -- nothing was available on Busch)

Next in the course is an introduction to max/min in several variables.

Review of 1 variable
I try not to work hard, so I thought maybe a quick review of extreme value material from 1 variable calculus would be useful. The names of ideas to recall include these:
critical point, maximum, minimum, absolute maximum, absolute minimum, local maximum, local minimum.

Fermat's fact
What I called "Fermat's fact" was the following wonderful observation in one-variable calculus:
If f is differentiable at x₀ and if f´(x₀) is not 0, then f does not have an extreme value at x₀.

The picture shows a "proof" (well, I hope fairly convincing to a picture person). If there is a tilt in the tangent line, then there are both higher and lower values near x₀. If x₀ is either kind of extreme value (max/min), then we see that f´(x₀) cannot be 0.

Critical number
Therefore the following definition was created.
x₀ is a critical point of the function f if either f is not differentiable at x₀ or f´(x₀)=0.
For simplicity in this discussion I'll assume that f is defined in some interval that has x₀ inside it (in the interior). Here are some pictures of critical points in 1 variable.

What the zoo shows in the first two pictures are functions which are not differentiable at a point. The first such does not have an extreme value, but the second has a local min. Such nondifferentiable behavior occurs frequently in a number of applications. It is just considered bad taste to show pictures like this in calc 1 but really there are areas of applications (industrial engineering, operations research) where such pictures are typical, not exceptional. The other pictures are differentiable at the critical point. At the first picture (locally like x³, say), the function doesn't have an extreme value (max or min). At the other two, there is a local min and a local max, respectively.

Identifying ("classifying") the type of critical point
Taylor's Theorem sort of helps us understand the 1 variable case, at least where the point is "critical" because the derivative is 0. Here we have f(x)=f(x₀)+f´(x₀)(x–x₀)+f´´(x₀)/2(x–x₀)²+H.O.T. where the H.O.T.'s are at least 3^rd order, and →0 faster than the other terms. If the first derivative is 0 at x₀, this becomes
f(x)=(Value at x₀)+0+(Some number)(x–x₀)²+H.O.T.
The second derivative test in one variable is the recognition that if Some number is positive then locally the graph of the function looks like a parabola opening up, so the function must have a local min. If Some number is negative the parabola opens down, and the function must have a local max. If Some number is actually 0, then the H.O.T.'s get involved, and the situation can't be deduced from the second derivative: not enough information. Examples supplied by students (x³ and x⁴ and –x⁴) showed that if f´(x₀)=0 and also f´´(x₀)=0 then the function could have no local max/min and a local min or a local max at x₀. Things are complicated.

And now let's look at more than 1 dimension: maximum, minimum, absolute maximum, absolute minimum, local maximum, local minimum. These words and phrases mean more or less then same, but max/min in more than one variable is much more complicated. Some examples will be useful. We deal now with f(x,y).

(x₀,y₀) is a local minimum of f if f((x₀,y₀)≤f(x,y) for (x,y)'s close to (x₀,y₀).
(x₀,y₀) is a local maximum of f if f((x₀,y₀)≥f(x,y) for (x,y)'s close to (x₀,y₀).
If instead of local we want absolute then "close to" gets exchanged for "all (x,y) in the domain of f". If we want strict local max or strict local min then we should ask that ≤ becomes < for (x,y)'s not equal to (x₀,y₀). There's a similar modification for strict local min. (The peak of a mountain is usually a strict local max, while a ridgeline in the hills may not be strict -- other points may have equal height.
(x₀,y₀) is a critical point of f if either at least one of ∂f/∂x(x₀,y₀) or ∂f/∂y(x₀,y₀) does not exist
OR
∂f/∂x(x₀,y₀)=0 and ∂f/∂y(x₀,y₀)=0.

The simple pictures with simple formulas
Here are some pictures and some formulas.

Discussion and formulas	The pictures
A cone I wanted to give an example of a function which might make you think a bit. Consider f(x,y)=sqrt(x2+y2). The contour curves (z=constant) are all circles centered at the origin because they are x2+y2= constant2. But a "trace" with y=0 in the xz-plane has z=sqrt(x2). This is not a straight line (square root is a function with domain non-negative reals and range also non-negative reals!). It is z=|x|. So the circles pack themselves to come to a point at (0,0,0). This is certainly a local min, and it is certainly also a critical point but no partial derivatives exist at x=0 and y=0.
Min A function defined on all of R2 with a local (and absolute) minimum is f(x,y)=x2+y2. The graph of this function is a surface called a paraboloid. It is a nice, smooth "cup" opening up. Vertical slices through (0,0) are all parabolas opening up and the contour lines are circles. The red dot is the critical point and the brown plane is the tangent plane at that point (the xy-plane).
Min The simplest local and absolute strict maximum is, of course, just the reflection of the previous example, done with minus signs algebraically. So here f(x,y)=–x2–y2, and (0,0) provides a strict maximum. The graph is a paraboloid whose axis of symmetry is again the z-axis. This graph opens "down".
A saddle The function f(x,y)=–x2+y2 gives a nice example of a saddle point. The xz-slice (where y=0) shows the curve z=–x2 and the yz-slice (where x=0) shows z=y2. Each has a (strict) extreme point at 0. One is a max and one is a min. Such behavior is called a saddle point. Perhaps the behavior most similar in one variable calculus would be that of the function x3 (an inflection point). But in 2 and more variables the local picture can be much more complicated. Here the surface is more complicated, and my picture is certainly not so good. But the tangent plane and critical point are the same. The tangent plane cuts through the surface (similar to the way a tangent line at an inflection point in 1 variable calculus cuts through the graph of a curve).

Using Fermat's fact here
If a point is a local extreme point of some function f in several variables, and if that function is differentiable at that point, then all of the first partial derivatives of the function must be 0 at that point. If that's not true, just "slice" the function at that point in the direction of the derivative which is not 0. The one variable Fermat fact implies that the function does not have an extreme value (max or min) at the point in one variable, and therefore the function in several variables has both higher and lower values near the point. Therefore (whew!):
    An extreme point must be a critical point.
Our functions will almost always be differentiable (not like the graph of the cone above), so our functions will have their extreme values where ∇f=0. This doesn't mean that non-differentiable functions (functions with jumps or corners) are not important or interesting in mathematics and its applications (again: linear optimization which is a major tool in some areas of industrial engineering, shock waves in physical phenomena). Just learning to use the tools for higher dimensional analysis of differentiable functions is a big enough task.

Another instructor's final exam question Suppose f(x,y,z,w)=3x6+8y12+55z8+9w64. What are the critical points of f and what type (max, min, saddle) are they? This problem seems a bit forbidding. But fx=3·6x5. The only way this is 0 is for x to be 0. Similar remarks for fy, fz, and fw imply that the only critical point for this function is (0,0,0,0). What type of critical point is (0,0,0,0)? Well, f(0,0,0,0)=0. And if any of the coordinates in (x,y,z,w) is not 0, then (since we have only even powers!) f(x,y,z,w)>0. Therefore this critical point is an absolute minimum.

Suppose z=f(x,y), and f is differentiable. What is the geometric meaning of "(x₀,y₀) is a critical point of f"? Since ∇f(x₀,y₀)=0, both of the first partial derivatives are 0. Therefore z=f(x₀,y₀) (that is, z=a constant) is the tangent plane to z=f(x,y) at the point (x₀,y₀,f(x₀,y₀)). The "flat" plane through the point, parallel to the xy-coordinate plane, is tangent to the surface. This can be difficult to "see" in a graph, though.

Monkey saddle
The examples already shown are the standard critical points for functions of two variables. But there are other kinds of critical points. The graph z=x³–3xy² shows one of them. Again the origin, (0,0), is the only critical point. This is because z_x=3x²–3y² and z_y=–6xy. For z_y to be equal to 0, either x or y must be 0. But then use z_x=0 to conclude that the other variable must be 0 also. For this function, the xy-plane is the tangent plane at the origin because (0,0,0) is on the graph. This critical point's local behavior is up/down repeated three times (at equally spaced 120^o angular intervals) if you walk around the surface in a small circle centered at the origin. The critical point is called a monkey saddle because, presumably, a monkey could sit on it with spaces for two legs and a tail to hang down.

Critical points in more than one variable can have very different local pictures, and there has been a great deal of effort expended trying to understand them.

Our next task will be to learn and use a two variable version of the second derivative test,

Maintained by greenfie@math.rutgers.edu and last modified 10/8/2010.