APEX Calculus

We use a similar approach in this section to find volume under a surface.

Let $R$ be a closed, bounded region in the $xy$-plane and let $z=f(x,y)$ be a continuous function defined on $R\text{.}$ We wish to find the signed volume under the graph of $f$ over $R\text{.}$ (We use the term “signed volume” to denote that space above the $xy$-plane, under $f\text{,}$ will have a positive volume; space above $f$ and under the $xy$-plane will have a “negative” volume, similar to the notion of signed area used before.)

We start by partitioning $R$ into $n$ rectangular subregions as shown in Figure 14.2.2.(a). For simplicity’s sake, we let all widths be $\mathrm{\Delta }x$ and all heights be $\mathrm{\Delta }y\text{.}$ Note that the sum of the areas of the rectangles is not equal to the area of $R\text{,}$ but rather is a close approximation. Arbitrarily number the rectangles 1 through $n\text{,}$ and pick a point $(x_i,y_i)$ in the $i$th subregion.

The volume of the rectangular solid whose base is the $i$th subregion and whose height is $f(x_i,y_i)$ is $V_i=f(x_i,y_i)\mathrm{\Delta }x\mathrm{\Delta }y\text{.}$ Such a solid is shown in Figure 14.2.2.(b). Note how this rectangular solid only approximates the true volume under the surface; part of the solid is above the surface and part is below.

This approximates the signed volume under $f$ over $R\text{.}$ As we have done before, to get a better approximation we can use more rectangles to approximate the region $R\text{.}$

In general, each rectangle could have a different width $\mathrm{\Delta }{x}_j$ and height $\mathrm{\Delta }{y}_k\text{,}$ giving the $i$th rectangle an area $\mathrm{\Delta }{A}_i=\mathrm{\Delta }{x}_j\mathrm{\Delta }{y}_k$ and the $i$th rectangular solid a volume of $f(x_i,y_i)\mathrm{\Delta }{A}_i\text{.}$ Let $\Vert \mathrm{\Delta }A\Vert$ denote the length of the longest diagonal of all rectangles in the subdivision of $R\text{;}$ $\Vert \mathrm{\Delta }A\Vert \to 0$ means each rectangle’s width and height are both approaching 0. If $f$ is a continuous function, as $\Vert \mathrm{\Delta }A\Vert$ shrinks (and hence $n\to \mathrm{\infty }$) the summation ${\displaystyle \sum _{i=1}^{n}f(x_i,y_i)\mathrm{\Delta }{A}_i}$ approximates the signed volume better and better. This leads to a definition.

Definition 14.2.3 does not state how to find the signed volume, though the notation offers a hint. We need the next two theorems to evaluate double integrals to find volume.

This theorem states that we can find the exact signed volume using a limit of sums. The partition of the region $R$ is not specified, so any partitioning where the diagonal of each rectangle shrinks to 0 results in the same answer.

This does not offer a very satisfying way of computing volume, though. Our experience has shown that evaluating the limits of sums can be tedious. We seek a more direct method.

Recall Theorem 7.2.3 in Section 7.2. This stated that if $A(x)$ gives the cross-sectional area of a solid at $x\text{,}$ then ${\int }_a^{b}A(x)dx$ gave the volume of that solid over $[a,b]\text{.}$

Remember that though the integrand contains $x\text{,}$ we are viewing $x$ as fixed. Also note that the bounds of integration are functions of $x\text{:}$ the bounds depend on the value of $x\text{.}$

This gives a concrete method for finding signed volume under a surface. We could do a similar procedure where we started with $y$ fixed, resulting in an iterated integral with the order of integration $dxdy\text{.}$ The following theorem states that both methods give the same result, which is the value of the double integral. It is such an important theorem it has a name associated with it.

Note that once again the bounds of integration follow the “curve to curve, point to point” pattern discussed in the previous section. In fact, one of the main points of the previous section is developing the skill of describing a region $R$ with the bounds of an iterated integral. Once this skill is developed, we can use double integrals to compute many quantities, not just signed volume under a surface.

Note how in these examples that the bounds of integration depend only on $R\text{;}$ the bounds of integration have nothing to do with $f(x,y)\text{.}$ This is an important concept, so we include it as a Key Idea.

Before doing another example, we give some properties of double integrals. Each should make sense if we view them in the context of finding signed volume under a surface, over a region.

In the previous section we practiced changing the order of integration of a given iterated integral, where the region $R$ was not explicitly given. Changing the bounds of an integral is more than just an test of understanding. Rather, there are cases where integrating in one order is really hard, if not impossible, whereas integrating with the other order is feasible.

The previous section introduced the iterated integral in the context of finding the area of plane regions. This section has extended our understanding of iterated integrals; now we see they can be used to find the signed volume under a surface.

This new understanding allows us to revisit what we did in the previous section. Given a region $R$ in the plane, we computed ${\iint }_{R}1dA\text{;}$ again, our understanding at the time was that we were finding the area of $R\text{.}$ However, we can now view the graph $z=1$ as a surface, a flat surface with constant $z$-value of 1. The double integral ${\iint }_{R}1dA$ finds the volume, under $z=1\text{,}$ over $R\text{,}$ as shown in Figure 14.2.31. Basic geometry tells us that if the base of a general right cylinder has area $A\text{,}$ its volume is $A\cdot h\text{,}$ where $h$ is the height. In our case, the height is 1. We were “actually” computing the volume of a solid, though we interpreted the number as an area.

The next section extends our abilities to find “volumes under surfaces.” Currently, some integrals are hard to compute because either the region $R$ we are integrating over is hard to define with rectangular curves, or the integrand itself is hard to deal with. Some of these problems can be solved by converting everything into polar coordinates.