Tuesday, March 21

Welcome, spring!

Happy birthday, Johann Sebastian Bach!
March 21, 1685 -- July 28, 1750.

Office hour shift this week
Tuesday office hours: from 3:30 to 4:30 or so.
Thursday office hours: from 9:30 to 11:30 or so.

A region in space
I began this last time. Please see here to get a start. 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=0z=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=0y=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=+/-sqrt(z-y). These will be the limits on the dx integral.

z=0z=3y=0y=zx=-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.

x=-sqrt(3)x=sqrt(3)z=x2z=3y=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=0y=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=0y=3x=-sqrt(3-y)x=+sqrt(3-y)z=x2+yz=3SQUIRREL dz dx dy.

First, this is a classroom example. The solid region is actually not very complicated. It is a convex region with boundaries given by low-degree polynomials. 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 limts seems to be difficult, and hard to describe -- I don't know yet of a computer program which can do it reliably.
z=0z=3y=0y=zx=-sqrt(z-y)x=+sqrt(z-y) SQUIRREL dx dy dz
x=-sqrt(3)x=sqrt(3)z=x2z=3y=0y=z-x2 SQUIRREL dy dz dx.
y=0y=3x=-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 in all three ways we found. The answers are shown. They are such large and silly numbers, and they all agree exactly. I am 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 have returned 0 for all of the setups, and this answer would not be very helpful. I could have specified W=1 and would have gotten (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 ...

QotD
Choose one of the other three orders and write the triple iterated integral for that order. You can't integrate SQUIRREL without more specificity, so all you can do, and what I wanted, was write the precise bounds. I think the answers are (after reading many student answers):
z=0z=3x=-sqrt(z)x=sqrt(z)y=0y=z-x2 SQUIRREL dy dx dz
x=-sqrt(3)x=sqrt(3)y=0y=3-x2z=x2+yz=3 SQUIRREL dz dy dx
y=0y=3z=yz=3x=-sqrt(z-y)x=sqrt(z-y) SQUIRREL dx dz dy

(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. The mass of this volume is C dV. Suppose Q is the distance of 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 intertia of the whole cone we need to add up the pieces of the moment of intertia. So we need The whole coneQ2C 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 coneQ2C dV becomes the triply iterated integral z=0z=H((sqrt(x2+y2))2C dAxy)dz. I wrote dAxy to remind myself that the double integral is in the xy-plane.

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

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 dAxy=r dr d. The limits on for a whole circle are 0 and 2Pi. 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 =0=2Pir=0r=RAD(r2)C r dr d.

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
=0=2Pir=0r=(R/H)zC r3 dr d.

Computation
r=0r=(R/H)zC r3 dr=(C/4)r4]r=0r=(R/H)z=(C/4)R4z4/H4.
=0=2Pi(C/4)R4z4/H4d=[(Pi C)/2]R4z4/H4. (Easy: no in the integrand so just multiply by 2Pi.)
z=0z=H[(Pi C)/2]R4z4/H4dz=[(Pi C)/10]R4z5/H4]z=0z=H=[(Pi 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 (Pi and 10)? Here is help from an engineering student. Mr. Doig reports via e-mail:
 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 (Pi/3)R2H. Mr. Doig's "a" is our R. So the formula 3/10 * m * a^2 becomes (3/10)C[(Pi/3)R2H]R2 which is indeed [(Pi C)/10]R4H. We have confirmation by high authority: "my statics text".

Cylindrical coordinates
This is a coordinate system that augments the r and 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, , from an initial ray (usually the positive x-axis) and a distance, r, from the origin.
Here are some basic axially symmetric figures:
Cylinder Example: r=5.
Cone Example: z=7r.
Elliptical paraboloid Example: z=3r2.

The location of Piscataway
Where are we? I got several answers. But I enlightened students with these facts:
Pistcataway's latitude is 40.499oN and Piscataway's longitude is 74.399oW.
Or, in more antique fashion, the latitude is 40o32´39´´ and the longitude is 74o28´28´´.
Do not be as confused as I am. This is not about stalagtites (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 Piscataway 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 Piscataway's longitude to Piscataway, 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). Piscataway 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, phi, is the angle from the positive z-axis to the radius vector. The second angle, , 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'll work a bit with them next time.

HOMEWORK

Friday, March 10

Spring is coming!
This morning while preparing for class (7 AM) I saw in my backyard an unmistakable Fox Sparrow (Passerella iliaca).
One source states:
 Birders in the eastern United States and in many parts of the West are likely to see Fox Sparrows only during migration or winter, for this large sparrow nests in northern and western Canada and in mountainous areas in the western states.
As for the name, the same source declares
 Both the Fox Sparrow's common name and the Latin-specific epithet "iliaca" refer to the "foxy" rufous red color of eastern and northern Fox Sparrows.

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 to +d. The picture displays this area, dA, magnified a lot. As mentioned, dA is an area and has dimensions length2. If d 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 d. 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 d: if the angle at the origin is doubled, the length of the intercepted arc is also doubled. Werll, "directly proportional means that there is some constant, uhhh ..., let's call it K, so that the length is K r d. 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 2Pi r. Here the d is 2Pi. So apparantly 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 d=K(360)r to be 2Pi r, you would need K=2Pi/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 never knew that before today) and therefore the constant K, if we used grads in calculus, would be 2Pi/400 which is approximately .01571, also obnoxious. Yes, things would be better if Pi 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 shwon to the right (in the first quadrant, with the curves arcs of circles centered at the origin).
How does one recogize 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,  to x, y      From x, y to r,
-------------------    -------------------
x=r cos               r2=x2+y2
y=r sin               tan =y/x```
I've given the formulas the way I most often use them. In particular, the formula for getting from x and y needs to be "adjusted" (by adding Pi) if the point whose coordinates are (x,y) is in the appropriate left half of the plane.

I recognize (primarily from the picture, but I can also use the formulas) that R is described by 0<=<=Pi/2 and by 2<=r<=4. We can convert the integral into polar coordinates:
Rx2+y2dA=0Pi/224r2 r dr d=0Pi/224r3dr d=0Pi/2(1/4)r4]r=2r=4d=60]0Pi/2=30Pi.
Of course the computation is easy. It was arranged to 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 mathematics behind 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 byu 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. Here's a quote from another site:
 ... this constant [G] was worked out by Henry Cavendish in 1798. He achieved this by measuring the gravitational attraction between two 1kg lead balls at a distance of 1 metre. I would guess that you would assume that this experiment must have found the attraction to be very small, in fact it was so small it was a stunning achievement to measure the tiny acceleration caused and to therefore calculate the gravitational attractive force (using Newton's second law). Cavendish found the force to be 6.67 x 10-11 Nm2/kg2 (or approximately 0.0000000000667 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 with 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, p (should be rho, but html doesn't have Greek letters). A small chunk of the plate ("dA") located at the point (x,y) will have mass equal to p 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 Gmp 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 hypoteneuse 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)). The square root and the square cancel. The magnitude of the force is Gmp dA/(D2+x2+y2). Several students noticed a surprizing 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, phi, between the vertical line and the line connecting the external object to dA. But cos phi 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 GmpD dA/(D2+x2+y2)3/2. Since the plate is infinite, we want R2Gmp dA/(D2+x2+y2)3/2.

Computing the integral
Many of the letters are constants: G and m and p and D. We can pull them out of the integral. 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 d, and r2=x2+y2, and all we need are the limits on the integral. For the whole plane, r should go from 0 to infinity, and should go from 0 to 2Pi. The appearance of infinity forces me to finally acknowledge that this is an improper integral.
02Pi0infinityr dr d/(D2+r2)3/2.

The inner (improper) integral
I will be careful, since I am supposed to be teaching a math course.
0infinityr dr d/(D2+x2+y2)3/2=limB-->infinity0Br dr d/(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).
0Br dr d/(D2+r2)3/2=-1/sqrt(D2+r2)]r=0r=B= -1/sqrt(D2+B2)-{-1/sqrt(D2+02)}
As B-->infinity, 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. But we need to multiply by the factors we pulled out. The answer is:
GmpD(1/D)=Gmp.

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 table. Therefore ... therefore ... the earth is flat. (Similar supporting logic: wouldn't people who wanted to lose weight climb Mt. Everest, because they would lose weight when ...).

Discussion of the claim

• The logic is flawed. If the earth were flat, then the weight would be the same. But other reasons could cause the force of gravity to be (or seem to be!) the same. A statement and its converse need not both be true!
• In fact, if the earth is spherical (which it is, essentially) the gravitational attraction of the earth acts as if the mass of the earth was concentrated at the center of the earth. The radius of the earth is reported to be about 4,000 miles, which is 21,120,000 feet. Let's say that a table is 2.5 feet high. The lecturer would need to be sensitive to a 1.4·10-14 change in weight (I squared the ratio, because of the inverse square law). The instructor is not that sensitive!

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 in 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.

The ocean
People who study the ocean (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 things 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```
Considerations which might influence this "experiment" include the following:
• The temperature measurements should be well distributed: we won't get an acceptable average temperature if the places we measure are all clumped up in one chunk of the box (the statistical phrase is that the measurements should be uniformly distributed).
• Ideally, we would like more measurements rather than fewer: with many measurements we'd have a chance of coming up with more reliable information.

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```
I will multiply the top and bottom of this fraction by (b-a)(d-c)(f-e), so we would have:
```SUM of all of the temperature measurements    (b-a)(d-c)(f-e)
------------------------------------------- · ----------------
n3                        (b-a)(d-c)(f-e)```
Just consider part of this, the fraction (b-a)(d-c)(f-e)/n3. This is the same as [(b-a)/n]·[(d-c)/n]·[(f-e)/n]. If n is large, this is the same as splitting up each of the edges of the box into n equal pieces, and what we have is a very small box of the ocean. Now if we also want the points we measure to be well-distributed, then we might expect that most of the boxes will contain exactly one sample point. We can think of f(that sample point)·[(b-a)/n]·[(d-c)/n]·[(f-e)/n] as f(that sample point)dx dy dz or as f(that sample point)dV where dV is this very small box inside the huge box of ocean. When we take the SUM we actually have an approximating Riemann sum to box of oceanT(x,y,z) dV, which is a triple integral. Whew! The limit of such approximating sums is the triple integral, but I won't go into detail because this all parallels a similar discussion for double integrals.I don't want to forget anything: there is a factor of (b-a)(d-c)(f-e) remaining on the bottom, and this is the volume of the box.
All of this is supposed to support the following definition:
The average value of the temperature in the box is
```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:
020503 x2+7yz dy dz dx
Let's compute, from the inside out: 03 x2+7yz dy dz dx=yx2+(7/2)y2z]y=0y=3=3x2+(63/2)z.
053x2+(63/2)z dz=3x2z+(63/4)z2]z=0z=5=15x2+(63/4)(25).
0215x2+(63/4)(25) dx=5x3+(63/4)(25)x]x=0x=2=40+(63/2)(25).

If this were the late 20th 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```
Incidentally, I checked and 40+(63/2)(25) is the same as (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 evaluated 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.

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 ordered for the iterated integral). Let me "integrate" (convert to iterated integrals) the function Bear 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-pane 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 region 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
02x=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.

I also tried to experiment with the other problems I did in the last class. For example, we discussed a textbook problem, #11 of section 15.3. The resulting double integral was 12x=yx=y3ex/ydx dy
So I tried the Maple command plot3d(exp(x/y),x=y..y^3,y=1..2). The first view I got is shown below to the left. Notice, please, that the scales on the three axes are all distinct and very different. The vertical axis is the most squeezed. It goes from 0 to about 50. If you want to look at the picture with all axes equal, the result is almost invisible (it becomes tall and very, very thin). The second picture is the result of my looking at the image unconstrained, but from directly above the xy-axis, so that we can't see the z-axis at all. The result should be the domain of integration. By the way, I had to magnify this picture several times to get what is shown. You might want to compare the picture shown on the right below to my picture of the domain of integration. They should match, and they do, mostly.

You can try the command plot3d(x^6*y^7,y=x^2..3*x,x=0..3) which corresponds to another problem we discussed last time. The picture, although most unreasonable, does illustrate the problem. The horizontal axes (x and y) go from from 0 to 9 and 0 to 3 respectively. The vertical axis, though, must display up to 35·108 (because 3697 is about 3.5·10^9). So this picture shows a thin strip which turns rapidly into a thin "spike" going up.

HOMEWORK
You have a workshop writeup to hand in. Please read about triple integrals and do the suggested homework problems. Practice is most important here: it will certainly improve the vague thing called intuition. I will find some iterated integrals for the solid described above and the QotD will be for students do write another iterated integral. You could also read ahead, and look at cylindrical and spherical coordinates.

Tuesday, March 7

Volume of a tetrahedron
A tetrahedron is a flat-sided object with four corners. I would like 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). 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. There's one 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. And I got this equation totally by guessing.

Here we'll use a double integral to find the volume of the tetrahedron here. 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 I got from the equation for the titlted face, just by solving it for z.

As a double integral So the volume is BaseHeight dA, and this is The triangle3-3x-(3/2)y dA. I'll convert this to an iterated integral to compute it.

47 second break for theory
Last time I was very careful to define the double integral as a limit of Riemann sums, and the domain we integrated over was a rectangle. Yet 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. The 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 peices -->0.

Converting to iterated integrals
Let's write The triangle3-3x-(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 yHighest 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:
02x=0x=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=0x=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=0x=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.
I remarked in class that, 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 triangle3-3x-(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 xRightmost 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:
01y=0y=2-2x3-3x-(3/2)y dy dx.

Computing the iterated integral
I'll first compute the inner integral:
y=0y=2-2x3-3x-(3/2)y dy= (antidifferentiate with respect to y, so x is a constant here!) 3y-3xy-(3/4)y2]y=0y=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:
img src="gifstuff/is11.gif" width=8>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:
03y=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:
09x=(1/3)yx=sqrt(y)x6y7dx dy
The computation begins with the inner integral.
x=(1/3)yx=sqrt(y)x6y7dx=(1/7)x7y8]x=(1/3)yx=sqrt(y)= (1/7){sqrt(y)}7y7-(1/7){(1/3)y}7y7=(1/7)y7/2y7-(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]) (1/7)(1/16)325-(1/7)(2/(25))325= (1/7)325{(1/16)-2/(25))

Theorem

``` 1     / 1      1  \     1    / 1      2  \
--- · | ---- - ---- | = --- ·| ---- - ---- |
8     \ 15     23 /     7    \ 15     23 /```
The proof consists of observing that the dx dy and dy dx values of the double integral must be equal by the Fubini result, and then dividing both values by 323. (The student may, of course, verify this statement using the tools of third grade arithmetic, but the prestige of double integrals is ... [priceless?].)

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

Pi? The Bible??
Some students objected when I tried to write Pi instead of 3 as I was assembling information for the computation above. I cited the Bible as a reference. Here is information about the "history of Pi" including specific biblical citations. Since we are at a secular university, I give this not for its religious value, but for ... uhhhh ... historical context. Also, maybe ... uhhhh ... maybe the value of Pi has changed with time. Uhhhh ... also everyone should know a few digits of Pi. This reference discusses how the computation of the first trillion or so decimal digits was done. You can search the first four billion bits (binary digits) of Pi for patterns here.

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 alway 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
This wasn't a random choice. The dx dy order introduces an additional kind of complexity. Consider these limits:
Bottom of yTop of yx=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:
01x=Left(y)x=Right(y)Frog dx dy +14x=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:
01x=-sqrt(y)x=sqrt(y)Frog dx dy +14x=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:
-12y=x2y=x+2Frog dy dx.
Almost surely, unless circumstances were very strange, I would set up the iterated integral this way and no in the dx dy way.

The New Jersey Chorus Frog, Pseudacris feriarum kalmi, is an endangered species in some of its range. Mr. R. Giordano made a very nice origami frog during class and gave it to me.

In my office ...
A student visited me and we did some more Frog-like problems (one with Toad and one with Lizard. Let me show you the Lizard problem. The region we looked at was bounded by y=2-2x2 and y=x4-x2. A Maple graph with these two functions is shown to the right. Just as in the previous (Frog) example, writing the dy dx integral is quite direct. My student visitor and I wrote iterated integrals for the other order:dx dy. You can try this problem. First, though, think a bit and see how many iterated integrals will be necessary.

Hint
I think three pieces are needed. Notice that y=x4-x2 is the same as x4-x2-y=0 and this is (x2)2-x2-y=0, a quadratic in x2. So you can "solve" for x2 using the quadratic formula, and then take square roots of the resulting answers to get 4 possible values of x for each value of y. Certainly this is a mess, but this is possible to do without extravagantly advanced methods.

Problem #11 of section 15.3 in the 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 Dex/ydA, where D={(x,y)|1<=y<-2, y<x<=y3}.
As to why this problem introduces a new kind of complexity, I invite you to ask Maple to integrate ex/y both dx and dy. That is, what are the respective antiderivatives? The x antiderivative is yex/y but ... there is no y antiderivative in terms of familiar functions. So if we want to get an answer to the textbook's problem, we'd better first do dx and then leave dy until later. The dx dy double integral is easy enough to write, because the region D is described suitably:
12x=yx=y3ex/ydx dy
The inner antidifferentiation gives x=yx=y3ex/ydx=yex/y]x=yx=y3=yey3/y-ye1=yey2-ey and then 12yey2-ey dy is just (1/2)ey2-(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.

Yet a different way ...
What is Rx2+y2dA, if R is the region shwon to the right (in the first quadrant, with the curves arcs of circles centered at the origin)?

But this should ring a bell: Polar Coordinates! Why?

• The integrand, x2+y2, is just r2 in polar coordinates.
• The region of integration, a quarter annulus, has boundary curves which can be described easily in polar coordinates. They are r=2 and r=4 (for the circles) and =0 and =Pi/2 (for the first quadrant). In fact this region is actually a polar rectangle.
The only part which may be difficult to understand is dA. Although we changed dA to dx dy or dy dx, dA is not just the product of d and dr. One way to see that is to realize that dA is an area, so its dimensions are length2. dr is a length, but d, an angle, is dimensionless. We'll see next time (or look in the text) that dA=r dr d.

HOMEWORK
Please read about polar coordinates (15.4) and triple integrals (15.7). I will deviate from the syllabus and cover the material listed there for lecture 15 (15.5 and 15.9) later in the course.

Friday, March 3

Just one more Lagrange multiplier problem ...
I'll find the extreme values (max and min) of g(x,y,z)=xy+3yz on the unit sphere, x2+y2+z2=1.

The initial setup is routine:
f=<2x,2y,2z>; g=<y,x+3z,3y>.
The resulting system of four equations in four variables follows.
1  2x=y
2  2y=(x+3z)
3  2z=(3y)
C  x2+y2+z2=1

Solving these equations can be difficult. The aim is to find values the variables which lead to the max and min of the objective function. The analysis of the systems can be very idiosyncratic (differing greatly for different systems). I generally try to solve for in the equations 1 and 2 and 3 and compare the results. Start with 1:
=2x/y.
Already there is some difficulty. What if y is 0? You can't ignore such alternatives, because the extreme values may be hiding there. So if y=0 then 1 forces x to be 0 and 3 shows that z must be 0 also. But (0,0,0) does not satisfy the constraint, C. So y can't be 0. Now 2 gives
=2y/(x+3z).
If x+3z=0 then y is 0 and that's not possible. And 3 gives
=2z/(3y)

We can compare the various quantities that are equal to . So 2x/y=2z/3y and therefore 3x=z. And 2x/y=2y/(x+3z) so 2x2+6zx=2y2 but since 3x=z this is just 2x2+18x2=2y2 and y2=10x2

Now C can be written only with x's, and it becomes x2+10x2+(3x)2=1, and x=+/-1/sqrt(20). We know 3x=z, so z=+/-3/sqrt(20). And y2=10x2, so y2=10/20=1/2. y must be +/-1/sqrt(2). Certainly it isn't vital in this simple classroom example, but the +/- choices of x and z must be the same, and the +/- choice of y is unlinked. We have identified 4 points in R3 as candidates for finding extreme values:
(1/sqrt(20),1/sqrt(2),3/sqrt(20)), (-1/sqrt(20),1/sqrt(2),-3/sqrt(20)), (1/sqrt(20),-1/sqrt(2),3/sqrt(20)), (-1/sqrt(20),-1/sqrt(2),-3/sqrt(20)).

Although there are methods of identifying more specific max/min behavior in Lagrange multipliers, I have never used them. I've just "plugged" the values of the candidates into the objective function. For example, I can look at xy+3yz and tell you that the largest value will come from using (1/sqrt(20),1/sqrt(2),3/sqrt(20)). And that value is 4/sqrt(40)=2/sqrt(20). So the maximum value is 2/sqrt(20). I bet the minimum will be -2/sqrt(20). The pciture displayed to the right shows the constrant (the unit sphere) and the level surface of the objective function for the maximum value. Isn't it pretty? Well, if not pretty, the tangent behavior should again confirm the geometric nature of the method.

I mentioned in class the work of Professor Wilma Olson, Department of Chemistry and Chemical Biology, who studies the structure of such molcules as DNA with mathematical and computational tools. Since "nature" likes minimum energy arrangements, much of her work consists of investigating minimization problems in many variables. One of her collaborators is Professor Bernard Coleman of the Department of Mechanics and Materials Science, who has also done further work of his own. Similar problems arise for mechanical engineers: the energy of a spring has kinetic and potential pieces, so even a simple study of a construction with a half-dozen springs immediately leads to energy minimization in 12 variables (I am really simplifying here).

Quick review of the definite integral in calc 1
Suppose we have a nice function f(x) of one variable defined on and interval, [a,b]. We might want to find the area under y=f(x) on that interval, althogh that is more of an excuse to define the definite integral than a real ambition. Here is one approach to the definition. Take a large positive integer n and divide [a,b] into n equal parts each of width x=(b-a)/n. In each subinterval choose a sample point, say qj in the jth subinterval. Compute the sum SUMj=1nf(qj)x. This is a Riemann sum approximating the definite integral. As n-->infinity, any sequence of sums is supposed to approach a unique limit, and that limit is the definite integral. But there are many choices of sample points, and maybe it isn't clear that the choices don't influence the limit. So what if we take another point pj in the jth subinterval as a sample point? Then since the width is x, certainly |qj-pj|<=x. The Mean Value Theorem (let me assume that f is differentiable) then says there's some constant, C, so that |f(qj)-f(pj)|<=Cx. (I'm not too interested in the details here -- we're only skimming!) But we can estimate the difference when different choices of sample points are made:
|SUMj=1nf(qj)x - SUMj=1nf(qj)x|<=nCxx.
The "n" comes because there are n pieces in the sum. One of the x's occurs because that's a common factor in both sums. The other comes from the MVT estimate we just stated. But now, since x=(b-a)/n, we see that the estimate of the difference is C(b-a)2/n. Two of the n's cancel. We are left with an n on the bottom, and this means that the Riemann sums do get closer as n-->infinity, no matter what choice of sample point is made.

Theorem For any choices of sample points, as n-->infinity, the Riemann sums-->a unique limit, the definite integral of f from a to b, and written abf(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 is:

FTC If F´=f, then abf(x) dx=F(b)-F(a).

Defining the double integral
In fact 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 bouindaries 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, 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.
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
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=kRf(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 x by y, with both 's very small. In addition to choosing this special chopping strategy we could also decide to add up the contributions (the f(sample point)xy) in an orderly manner. So, for example, we could add up the contributions from the lowest "row" first. In that row, since y 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" 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=cdabf(x,y)dx dy=abcdf(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". Technically and precisely these 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=2514x3y7dx 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=1x=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 try a random function: f(x,y)=sqrt(3x+8y). Well, this isn't so random (as you'll see!). I just wish I had thought of it in class. The rectangle I have in mind has these boundary lines: x=0 and x=3 and y=0 and y=2. In the previous example we converted the double integral into a dx dy iterated integral. Let me try a dy dx order here. I think that either order, again, is about the same amount of work.

So let's try:
Rsqrt(3x+8y)dA=0302sqrt(3x+8y) dy dx.

The inner integral is 02x3sqrt(3x+8y) dy. We need a "dy" antiderivative of sqrt(3x+8y). Here we can really get confused! To me writing the function as (3x+8y)1/2 makes the problem easier. I guess that the antiderivative will be something close to (3x+8y)3/2. Well, but I need to multiply by stuff to get rid of the various constants. For example, I need to multiply by (2/3) because of the power. And I need to multiply by (1/8) because of the coefficient of y that the chain rule will push out. So the answer is (2/3)(1/8)(3x+8y)3/2, and we must substitute:
(2/3)(1/8)(3x+8y)3/2]y=0y=2= (2/3)(1/8)(3x+16)3/2-(2/3)(1/8)(3x+0)3/2, and this is (1/12)(3x+16)3/2-(1/12)(3x)3/2.

And now the outer integral:
03(1/12)(3x+16)3/2-(1/12)(3x)3/2dx=(1/12)(1/3)(2/5)(3x+16)5/2-(1/12)(1/3)(2/5)(3x)5/2]03= {(1/90)(3·3+16)5/2-(1/90)(3·3)5/2} -{(1/90)(3·0+16)5/2-(1/90)(3·0)5/2}= (1/90)(255/2-95/2-165/2+0)= (1/90)(55-35-45)= (1/90)(3125-243-1024)=(1858/90)=(929/45).
Well, y'see, everything was chosen so that the final answer would have no square roots. Isn't that wonderful!

Or, in the 21st century ...

```> int(int(sqrt(3*x+8*y),x=0..3),y=0..2);
929
---
45```

QotD
Compute -2112x2y-3xy2dy dx.

The inner integral:
12x2y-3xy2dy=(1/2)x2y2-xy3]y=1y=2={(1/2)x2·22-x·23}-{(1/2)x212-x·13}=(3/2)x2-7x.
The outer integral:
-21(3/2)x2-7x dx=(1/2)x3-(7/2)x2]x=-2x=1={(1/2)-(7/2)}-{(1/2)(-2)3-(7/2)(-2)2} -3-{-4-14}=18-3=15.