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

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:
₀²₀⁵₀³ 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 late 20^th 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 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
₀²_x=0^x=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 ₁²_x=y^x=y3e^x/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.

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 triangle}3-3x-(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 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.
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 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:
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:
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]) (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 /

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

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 y}^{Top 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-2^x=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 _De^x/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 e^x/y both dx and dy. That is, what are the respective antiderivatives? The x antiderivative is ye^x/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:
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.

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.
  

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.

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 q_j in the j^th subinterval. Compute the sum SUM_j=1ⁿf(q_j)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 p_j in the j^th subinterval as a sample point? Then since the width is x, certainly |q_j-p_j|<=x. The Mean Value Theorem (let me assume that f is differentiable) then says there's some constant, C, so that |f(q_j)-f(p_j)|<=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:
|SUM_j=1ⁿf(q_j)x - SUM_j=1ⁿf(q_j)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 _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 is:

FTC If F´=f, then _a^bf(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=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 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=_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". 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=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 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=0^y=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)(25^5/2-9^5/2-16^5/2+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 21^st 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=1^y=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=-2^x=1={(1/2)-(7/2)}-{(1/2)(-2)³-(7/2)(-2)²} -3-{-4-14}=18-3=15.
As I sadly admitted in class, somehow I got the answer 6 for this. Oh well.

Course workload
I hope to have one more workshop problem (so that I can grade each student's efforts at least once). There should also be two more Maple labs, but I'd like them to be interesting and useful. So we'll see.

HOMEWORK
Please read the first sections of chapter 15. I'll discuss 15.3 and probably 15.4 on Tuesday. Please hand in these problems at recitation on Wednesday: 14.8: 6 15.2: 7, 15 15.3: 4, 11

Note I will submit formal warning grades to students who got a D or an F on the first exam, or who are registered for the course and did not take the exam.

Maintained by greenfie@math.rutgers.edu and last modified 3/23/2006.