# Section13.2Double Integration and Volume¶ permalink

The definite integral of $f$ over $[a,b]\text{,}$ $\int_a^b f(x)\ dx\text{,}$ was introduced as “the signed area under the curve.” We approximated the value of this area by first subdividing $[a,b]$ into $n$ subintervals, where the $i^\text{ th }$ subinterval has length $\dx_i\text{,}$ and letting $c_i$ be any value in the $i^\text{ th }$ subinterval. We formed rectangles that approximated part of the region under the curve with width $\dx_i\text{,}$ height $f(c_i)\text{,}$ and hence with area $f(c_i)\dx_i\text{.}$ Summing all the rectangle's areas gave an approximation of the definite integral, and Theorem 5.3.20 stated that \begin{equation*} \int_a^bf(x)\ dx = \lim_{\\abs{\Delta x\}\to 0}\sum f(c_i)\dx_i, \end{equation*} connecting the area under the curve with sums of the areas of rectangles.

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

Let $R$ be a closed, bounded region in the $x$-$y$ plane and let $z=f(x,y)$ be a continuous function defined on $R\text{.}$ We wish to find the signed volume under the surface of $f$ over $R\text{.}$ (We use the term “signed volume” to denote that space above the $x$-$y$ plane, under $f\text{,}$ will have a positive volume; space above $f$ and under the $x$-$y$ 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 13.2.1(a). For simplicity's sake, we let all widths be $\dx$ and all heights be $\dy\text{.}$ 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 $i^\text{ th }$ subregion.

The volume of the rectangular solid whose base is the $i^\text{ th }$ subregion and whose height is $f(x_i,y_i)$ is $V_i=f(x_i,y_i)\dx\dy\text{.}$ Such a solid is shown in Figure 13.2.1(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.

For each subregion $R_i$ used to approximate $R\text{,}$ create the rectangular solid with base area $\dx\dy$ and height $f(x_i,y_i)\text{.}$ The sum of all rectangular solids is \begin{equation*} \ds \sum_{i=1}^n f(x_i,y_i)\dx\dy. \end{equation*}

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 $\dx_j$ and height $\dy_k\text{,}$ giving the $i^\text{ th }$ rectangle an area $\Delta A_i = \dx_j\dy_k$ and the $i^\text{ th }$ rectangular solid a volume of $f(x_i,y_i)\Delta A_i\text{.}$ Let $\norm{\Delta A}$ denote the length of the longest diagonal of all rectangles in the subdivision of $R\text{;}$ $\norm{\Delta A}\to 0$ means each rectangle's width and height are both approaching 0. If $f$ is a continuous function, as $\norm{\Delta A}$ shrinks (and hence $n\to\infty$) the summation $\ds \sum_{i=1}^n f(x_i,y_i)\Delta A_i$ approximates the signed volume better and better. This leads to a definition.

Recall that the integration symbol “$\int$” is an “elongated S,” representing the word “sum.” We interpreted $\int_a^bf(x)\ dx$ as “take the sum of the areas of rectangles over the interval $[a,b]\text{.}$” The double integral uses two integration symbols to represent a “double sum.” When adding up the volumes of rectangular solids over a partition of a region $R\text{,}$ as done in Figure 13.2.1, one could first add up the volumes across each row (one type of sum), then add these totals together (another sum), as in \begin{equation*} \sum_{j=1}^n\sum_{i=1}^mf(x_i,y_j)\Delta x_i\Delta y_j. \end{equation*}

One can rewrite this as \begin{equation*} \sum_{j=1}^n\left(\sum_{i=1}^mf(x_i,y_j)\Delta x_i\right)\Delta y_j. \end{equation*}

The summation inside the parenthesis indicates the sum of heights × widths, which gives an area; multiplying these areas by the thickness $\Delta y_j$ gives a volume. The illustration in Figure 13.2.6 relates to this understanding.

##### Definition13.2.4Double Integral, Signed Volume

Let $z=f(x,y)$ be a continuous function defined over a closed region $R$ in the $x$-$y$ plane. The signed volume $V$ under $f$ over $R$ is denoted by the double integral \begin{equation*} V = \iint_R f(x,y)\ dA. \end{equation*}

Alternate notations for the double integral are \begin{equation*} \iint_R f(x,y)\ dA=\iint_R f(x,y)\ dx\ dy=\iint_R f(x,y)\ dy\ dx. \end{equation*}

The definition above 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 area, though. Our experience has shown that evaluating the limits of sums can be tedious. We seek a more direct method.

Recall Theorem 7.2.2 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 $[a,b]\text{.}$

Consider Figure 13.2.6, where a surface $z=f(x,y)$ is drawn over a region $R\text{.}$ Fixing a particular $x$ value, we can consider the area under $f$ over $R$ where $x$ has that fixed value. That area can be found with a definite integral, namely \begin{equation*} A(x)=\int_{g_1(x)}^{g_2(x)} f(x,y)\ dy. \end{equation*}

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 $x\text{.}$

As $A(x)$ is a cross-sectional area function, we can find the signed volume $V$ under $f$ by integrating it: \begin{equation*} V = \int_a^b A(x)\ dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\ dy\right)dx = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\ dy\ dx. \end{equation*}

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 a iterated integral with the order of integration $dx\ dy\text{.}$ 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.

##### Example13.2.8Evaluating a double integral

Let $f(x,y) = xy+e^y\text{.}$ Find the signed volume under $f$ on the region $R\text{,}$ which is the rectangle with corners $(3,1)$ and $(4,2)$ pictured in Figure 13.2.9, using Fubini's Theorem and both orders of integration.

Solution
##### Example13.2.10Evaluating a double integral

Evaluate $\iint_R \big(3xy-x^2-y^2+6\big)\ dA\text{,}$ where $R$ is the triangle bounded by $x=0\text{,}$ $y=0$ and $x/2+y=1\text{,}$ as shown in Figure 13.2.11.

Solution

Note how in these two 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.

##### Key Idea13.2.12Double Integration Bounds

When evaluating $\iint_Rf(x,y)\ dA$ using an iterated integral, the bounds of integration depend only on $R\text{.}$ The surface $f$ does not determine the bounds of integration.

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.

##### Example13.2.15Evaluating a double integral

Let $f(x,y) = \sin(x) \cos(y)$ and $R$ be the triangle with vertices $(-1,0)\text{,}$ $(1,0)$ and $(0,1)$ (see Figure 13.2.16). Evaluate the double integral $\iint_Rf(x,y)\ dA\text{.}$

Solution
##### Example13.2.17Evaluating a double integral

Evaluate $\iint_R (4-y)\ dA\text{,}$ where $R$ is the region bounded by the parabolas $y^2=4x$ and $x^2=4y\text{,}$ graphed in Figure 13.2.18.

Solution

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.

##### Example13.2.19Changing the order of integration

Rewrite the iterated integral $\ds \int_0^3\int_y^3 e^{-x^2}\ dx\ dy$ with the order $dy\ dx\text{.}$ Comment on the feasibility to evaluate each integral.

Solution

Definition 5.4.26 defines the average value of a single–variable function $f(x)$ on the interval $[a,b]$ as \begin{equation*} \text{ average value of $f(x)$ on $[a,b]$ } = \frac1{b-a}\int_a^b f(x)\ dx; \end{equation*} that is, it is the “area under $f$ over an interval divided by the length of the interval.” We make an analogous statement here: the average value of $z=f(x,y)$ over a region $R$ is the volume under $f$ over $R$ divided by the area of $R\text{.}$

##### Definition13.2.22The Average Value of $f$ on $R$

Let $z=f(x,y)$ be a continuous function defined over a closed region $R$ in the $x$-$y$ plane. The average value of $f$ on $R$ is \begin{equation*} \text{ average value of $f$ on $R$ } = \frac{\ds \iint_R f(x,y)\ dA}{\ds\iint_R \ dA}. \end{equation*}

##### Example13.2.23Finding average value of a function over a region $R$

Find the average value of $f(x,y) = 4-y$ over the region $R\text{,}$ which is bounded by the parabolas $y^2=4x$ and $x^2=4y\text{.}$ Note: this is the same function and region as used in Example 13.2.17.

Solution

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 $\iint_R 1\ dA\text{;}$ again, our understanding at the time was that we were finding the area of $R\text{.}$ However, we can now view the function $z=1$ as a surface, a flat surface with constant $z$-value of 1. The double integral $\iint_R 1\ dA$ finds the volume, under $z=1\text{,}$ over $R\text{,}$ as shown in Figure 13.2.25. 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.

# Subsection13.2.1Exercises

Terms and Concepts

In the following exercises,

1. Evaluate the given iterated integral, and

2. rewrite the integral using the other order of integration.

In the following exercises:

1. Sketch the region $R$ given by the problem.

2. Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region $R\text{.}$

3. Evaluate one of the iterated integrals to find the signed volume under the surface $z=f(x,y)$ over the region $R\text{.}$