INTEGRATION in several variables, DOUBLE INTEGRALS

     The fundamental ideas involved in defining, interpreting and evaluating the integral

$\displaystyle{\int\int_D \, f(x,\, y)\, dx dy}$

 of a function $z = f(x, y)$ of two variables over a region $D$ in the $xy$-plane are a lot like the ones for the integral

$\displaystyle{\int_a^b\, f(x)\,dx}$

of a function $y = f(x)$ of a function of one variable over an interval $[a,\, b]$ in the $x$-axis. In the one variable case we assumed at the beginning that $f(x) > 0$ on $[a,\, b]$ so that the graph of $f$ lay above the $x$-axis, and then we tried to make sense of the notion of the area under the graph of $f$ and above $[a,\, b]$. So let's try the same approach in two variables, starting with a simple function, say a plane above a rectangle $D$ in the $xy$-plane having sides parallel to the axes. The region below the plane and above $D$ will now be a solid like the one shown to the right below.

   Problem 1: find the volume of the solid when $z = f(x, y)$ is the plane  $z = A-Bx$ and $D = [a,\,b]\times[c,\, d]$.

This is simply a problem in geometry, not calculus! But let's think about it the same way we did for partial derivatives: take a vertical slice $y = Y$ of the solid for fixed $Y$. The trace of the solid on the vertical plane is always a trapezoid congruent to the face $ABQP,$ independently of $Y$. On the other hand,
$A = (a,\, c), \,\,\,\, B = (b,\,c), \,\,\,\, C = (b,\,d)$
and
$P = (a,\, c,\,f(a,\,c)),\,\,\,\, Q = (b,\,c,\, f(b,\,c)),$
so the side $\overline{AP}$ of the trapezoid has length $f(a,\,c)$, while side $\overline{BQ}$ has length $f(b,\,c)$ and the distance between these parallel sides is $(b-a)$. In this case,
 $\displaystyle{area(ABQP)  =  \frac{1}{2}\left(f(a,\,c) + f(b,\, c)\right)(b-a)}$;
also, because $f(x,\, y)$ does not depend on $y$,
 $\displaystyle{area(trace) = \frac{1}{2}\left(f(a,c)+f(b,\,c)\right)(b-a)  =  \int_a^b\, f(x, Y)\, dx}$,
for each slice $y = Y$. So the solid has
$\displaystyle{volume = area(ABQP)(d-c)  =  \int_c^d\left(\int_a^b\, f(x, y)\, dx\right) dy},
 by geometry and the fact that $\int_a^b f(x, y)\, dx$ doesn't depend on $y$!!
 

 
     Next is an animated example showing a solid which rotates first to provide an overhead view of the region $D$ before being sliced by a vertical plane into two pieces. Then one half retreats to reveal the trace, which sure looks like the standard 'area under a curve' graphic! That area is the area of cross-section. The key question is how to get the volume of the solid knowing such areas of cross-section. 


   Problem 2: find the volume of the solid under the paraboloid
$z = f(x,\, y)  =  4 - x^2 - y^2$
and over $D  =  [0,\,1]\times[0,\,1]$.     
   The graph of $f(x, y)$ is shown to the right with the cross-section assumed to be the trace on a vertical plane for fixed $x$. This time
$\displaystyle{A(x)  = area(trace)  =  \int_0^1\, f(x,\, y)\, dy  = \frac{11}{3} - x^2}$,
defines a function $A(x)$ (do the calculation!). It's the area of cross-section at $x$, which both the graphic and the integral show varies with $x$. There's nothing to prevent us from integrating again, however, now with respect to $x$.
   But is there any reason why integrating $A(x)$ should give volume? It did before because of the simple geometry in problem 1.  We really need to decide exactly what volume means!


     In everyday terms, think of the solid in problem 2 as a 'loaf of bread' whose volume we estimate by taking the sum of the volume of successively thin slices of bread, the trapezoid of Problem 1 being like one such slice. This gives us a Riemann sum estimate of the volume whose accuracy depends on how thin the slices are; but in the limit, we'll get an exact value for the volume. Let's see how this works for the same function
$z = f(x,\, y)  =  4 - x^2 - y^2$
over $D  =  [0,\,1]\times[0,\,1]$ as in problem 2. And as before define the area of cross-section $A(x)$ by
$\displaystyle{A(x)   =   \int_0^1\, f(x,\,y)\, dy  =  \frac{11}{3} - x^2, \,\,\,\,\,\, 0 \le x \le 1.$   

       To the right is the graph of $A(x)$ with the interval $[0,\,1]$ subdivided into $m$ equal sub-intervals of length $\Delta x$ and for each sub-interval a rectangle drawn whose height is $A(x_j)$, choosing a point $x_j$ in that sub-interval. In one variable, we'd next look at the Riemann sum
$\displaystyle{\sum_{j \,=\,1}^m\, A(x_j)\Delta x}$;
giving us an estimate for the area under the graph of $A(x)$, while
$\displaystyle{\int_0^1\, A(x)\, dx  =  \lim_{m\,\to\,\infty}\,\left(\sum_{j\,=\,1}^m\, A(x_j) \Delta x\right)}$
 would be the exact area.
     But $A(x)$ itself is an area, so what do the sum and limit say about $f(x,\,y)$? Well, take a slice of thickness $\Delta x$ through the solid in Problem 2. Just as in Problem 1, the volume of this slice is $A(x_j^{*})\Delta x$.   So by stacking up a whole bunch of these slices we approximate the solid in Problem 2, and the Riemann sum is the volume of this approximating solid. 


The limit
$\displaystyle{\lim_{m\,\to\,\infty}\left(\sum_{j\,=\,1}^m\, A(x_j) \Delta x\right) =  \int_0^1\,A(x)\, dx  =  \int_0^1\left(\int_0^1\, f(x,\,y)dy \right)dx}$
of the Riemann sum thus expresses the volume of the particular solid in Problem 2 as a Repeated Integral or Iterated Integral,  integrating first with respect to $y$ and then with respect to $x$.


If we were being more careful, however, we'd start out with a general $D$ in the $xy$-plane like the one to the right, approximate $D$ with a mesh of rectangles, all of the same area $\Delta x \Delta y$, and on each rectangle like the blue one construct a rectangular solid having volume $f(x_j, y_k) \Delta x \Delta y$ by choosing a point $(x_j, y_k)$ in the rectangle. Adding up all these volumes we get a double Riemann sum
$\displaystyle{\sum_{j\,=\,1}^{m}\,\sum_{k\,=\,1}^n\, f(x_j,  y_k) \Delta x \Delta y}.$
The Double Integral of $f(x, y)$ over $D$ is then defined as
$\displaystyle{\int}\displaystyle{ \int_D\, f(x, y)\, dx dy   =  \lim_{m, n\, \to\, \infty}\,\left( \sum_{j\,=\,1}^{m}\,\sum_{k\,=\,1}^{n}\,f(x_j, y_k) \Delta x \Delta y\right)}$.
 
Ah, now I think I begin to see how the double Riemann sum for $f(x, y)$ gives us the earlier single Riemann sum for $A(x)$: take $D$ to be the rectangle $[0, 1]\times [0, 1]$, then
$\displaystyle{A(x_j)  =  \int_0^1\, f(x_j, y)\, dy  =  \lim_{n\,\to\, \infty}\, \left(\sum_{k\,=\,1}^n\, f(x_j, y_k) \Delta y\right)}$,
in which case
$\displaystyle{\int_0^1\left(\int_0^1\, f(x, y)\, dy\right) dx  =  \lim_{m\, \to \, \infty} \left(\sum_{j\, =\, 1}^m\, A(x_j) \Delta x\right)} =  \displaystyle{\lim_{m, n \,\to \, \infty}\,\left( \sum_{j\,=\,1}^m\,\sum_{k\,=\,1}^n \,f(x_j, y_k) \Delta x \Delta y \right)}  =  \displaystyle{\int} \displaystyle{\int_D \, f(x, y )\,dx dy. }$
A famous theorem, called Fubini's theorem, tells us that this works for all rectangles $D$ and for pretty much all functions $f(x, y)$. In fact, we can even interchange the order of integration by integrating first with respect to $x$ and then with respect to $y$. Let's summarize because we've done a lot!

        the double integral $\displaystyle{\int}\displaystyle{ \int_D\, f(x, y)\, dx dy}$ over a region $D$ in the $xy$-plane is defined as the limit of a double Riemann sum of volumes of rectangular solids and is interpreted as volume when $f$ is positive;

        by fixing either $x$ or $y$, the double integral can be thought of as integrating areas of cross-sections, i.e., by building up from the one variable case, which in turn tells us how to evaluate double integrals one variable at a time;

       a pattern has been set for integrals of functions of any number of variables - a triple integral will be built up by integrating volumes of cross-sections, for instance. 

To complete Problem 2: the solid under the paraboloid and over $D = [0,\,1] \times [0, 1]$ has volume

 =  $\displaystyle{\int}\displaystyle{\int_D}\,(4-x^2 - y^2)\, dx dy  =  \displaystyle{\int_0^1}\left(\displaystyle{\int_0^1}\, (4-x^2-y^2)\, dy\right)dx  =  \displaystyle{\int_0^1}\, (\frac{11}{3} - x^2)\, dx =  \displaystyle{\frac{10}{3}}.$

     It's probably clear by now that the domain of integration causes the main headaches in dealing with double integrals, because once we've know the limits of integration, we're back to the integrals of one variable calculus. Texts usually make a big fuss, disinguishing between regions of type I and II, or Vertically simple and Horizontally simple as they often call them. But really all one has to do is draw the region D and work from there! Let's look at two examples:


Problem 3: evaluate the integral
$I  =  \displaystyle{\int}\displaystyle{\int}_D \,(x+y)\,dx dy $
when $D$ consists of all $(x, y)$ such that  $0 \le y \le \sqrt{9-x^2},\,\,\, 0 \le x \le 3$.
Since $y^2 = 9 - x^2$ is a circle centered at the origin of radius 3, the region $D$ is the shaded quadrant of the circle of radius 3 shown to the right. First fix $x$, integrate with respect to $y$ along the black line, and finally let $x$ vary. Then as a repeated integral:
$I  =  \displaystyle{\int_0^3}\left(\displaystyle{\int}_0^{\sqrt{9 - x^2}}\, (x+y)\, dy \right)dx$.
But
$\displaystyle{\int}_0^{\sqrt{9-x^2}}\, (x+y)\, dy  =  \left[ xy +\displaystyle{\frac{1}{2}} y^2 \right]_0^{\sqrt{9-x^2}}  =  x\sqrt{9-x^2} + \displaystyle{\frac{1}{2}}(9-x^2),$
so that $I$ becomes $$\int_0^3\, (x\sqrt{9-x^2}+\frac{1}{2}(9-x^2))\,dx  =  I_1+I_2.$$ We do these separately. $$I_2  =  \left[\,\frac{9}{2}x -\frac{1}{6}x^3\,\right]_0^3  =  9.$$
To evaluate $I_1$, set $u = 9-x^2$. Then $$I_1  =  \frac{1}{2}\int_0^9\, \sqrt{u}\, du  =  9,$$
and so $I = 18$.