## Section4.2Double Integrals: Volume and Average Value

Consider a surface $f(x,y)\text{.}$ In this section, we are interested in computing either the volume under $f$ or the average function value of $f$ over a certain area in the $x$-$y$-plane. You might temporarily think of this surface as representing physical topography—a hilly landscape, perhaps. What is the average height of the surface (or average altitude of the landscape) over some region?

### Subsection4.2.1Volume and Average Value over a Rectangular Region

As with most such problems, we start by thinking about how we might approximate the answer. Suppose the region is a rectangle, $[a,b]\times[c,d]\text{.}$ We can divide the rectangle into a grid, $m$ subdivisions in one direction and $n$ in the other, as indicated in the 2D graph below, where $m=6$ and $n=4\text{.}$ We pick $x$-values $x_0\text{,}$ $x_1\text{,}$…, $x_{m-1}$ in each subdivision in the $x$-direction, and similarly in the $y$-direction. At each of the points $(x_i,y_j)$ in one of the smaller rectangles in the grid, we compute the height of the surface: $f(x_i,y_j)\text{.}$ Now the average of these heights should be (depending on the fineness of the grid) close to the average height of the surface:

\begin{equation*} \ds\frac{f(x_0,y_0)+f(x_1,y_0)+\cdots+f(x_0,y_1)+f(x_1,y_1)+\cdots+ f(x_{m-1},y_{n-1})}{mn} \end{equation*}

As both $m$ and $n$ go to infinity, we expect this approximation to converge to a fixed value, the actual average height of the surface. For reasonably nice functions this does indeed happen. Using sigma notation, we can rewrite the approximation:

\begin{align*} \frac{1}{mn}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i) \amp =\frac{1}{(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\frac{b-a}{m}\frac{d-c}{n}\\ \amp =\frac{1}{(b-a)(d-c)}\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{.} \end{align*}

The two parts of this product have useful meaning: $(b-a)(d-c)$ is of course the area of the rectangle, and the double sum adds up $mn$ terms of the form $f(x_j,y_i)\Delta x\Delta y\text{,}$ which is the height of the surface at a point multiplied by the area of one of the small rectangles into which we have divided the large rectangle. In short, each term $f(x_j,y_i)\Delta x\Delta y$ is the volume of a tall, thin, rectangular box, and is approximately the volume under the surface and above one of the small rectangles; see Figure 4.1. When we add all of these up, we get an approximation to the volume under the surface and above the rectangle $R=[a,b]\times[c,d]\text{.}$ When we take the limit as $m$ and $n$ go to infinity, the double sum becomes the actual volume under the surface, which we divide by $(b-a)(d-c)$ to get the average height. (a) The surface $z$ on $[0.5,3.5]\times[0.5,2.5]\text{.}$

Double sums like this come up in many applications, so in a way it is the most important part of this example; dividing by $(b-a)(d-c)$ is a simple extra step that allows the computation of an average. This computation is independent of $f$ being positive, and so as we did in the single variable case, we introduce a special notation for the limit of such a double sum, which is referred to as the double integral of $f$ over the region $R\text{.}$

###### Definition4.11. Double Integral over a Rectangular Region.

Let $f(x,y)$ be a continuous function defined over the rectangle $R=[a,b]\times[c,d]\text{,}$ then the double integral of $f$ over $R$ is denoted by

\begin{equation*} \iint_R f(x,y)\,dA=\iint_R f(x,y)\,dx\,dy=\lim_{m\to\infty}\lim_{n\to\infty} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y\text{,} \end{equation*}

provided the limit exists.

Note: The notation $dA$ indicates a small element of area, without specifying any particular order for the variables $x$ and $y\text{;}$ it is shorter and more generic than writing $dx\,dy\text{.}$

We now capture our results from the earlier calculations using the notation of the double integral.

Note: Just as with single-variable integration, this last theorem can be stated for more general functions that can take on negative values, as long as we understand that $V$ represents a signed volume.

### Subsection4.2.2Computing Double Integrals over Rectangular Regions

The next question, of course, is: How do we compute these double integrals? You might think that we will need some two-dimensional version of the Fundamental Theorem of Calculus, but as it turns out we can get away with just the single variable version, applied twice.

Going back to the double sum, we can rewrite it to emphasize a particular order in which we want to add the terms:

\begin{equation*} \sum_{i=0}^{n-1}\left(\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\right)\Delta y\text{.} \end{equation*}

In the sum in parentheses, only the value of $x_j$ is changing; $y_i$ is temporarily constant. As $m$ goes to infinity, this sum has the right form to turn into an integral:

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x = \int_a^b f(x,y_i)\,dx\text{.} \end{equation*}

So after we take the limit as $m$ goes to infinity, the sum is

\begin{equation*} \sum_{i=0}^{n-1}\left(\int_a^b f(x,y_i)\,dx\right)\Delta y\text{.} \end{equation*}

Of course, for different values of $y_i$ this integral has different values; in other words, it is really a function applied to $y_i\text{:}$

\begin{equation*} G(y)=\int_a^b f(x,y)\,dx\text{.} \end{equation*}

If we substitute back into the sum we get

\begin{equation*} \sum_{i=0}^{n-1} G(y_i)\Delta y\text{.} \end{equation*}

This sum has a nice interpretation. The value $G(y_i)$ is the area of a cross section of the region under the surface $f(x,y)\text{,}$ namely, when $y=y_i\text{.}$ The quantity $G(y_i)\Delta y$ can be interpreted as the volume of a solid with face area $G(y_i)$ and thickness $\Delta y\text{.}$ Think of the surface $f(x,y)$ as the top of a loaf of sliced bread. Each slice has a cross-sectional area and a thickness; $G(y_i)\Delta y$ corresponds to the volume of a single slice of bread. Adding these up approximates the total volume of the loaf. (This is very similar to the technique we used to compute volumes in Section 3.3, except that there we need the cross-sections to be in some way “the same” .) Figure 4.2 shows this “sliced loaf” approximation using the same surface as shown in Figure 4.1. Nicely enough, this sum looks just like the sort of sum that turns into an integral, namely,

\begin{align*} \lim_{n\to\infty}\sum_{i=0}^{n-1} G(y_i)\Delta y\amp =\int_c^d G(y)\,dy\\ \amp =\int_c^d \int_a^b f(x,y)\,dx\,dy\text{.} \end{align*}

Let's be clear about what this means: we first will compute the inner integral, temporarily treating $y$ as a constant. We will do this by finding an antiderivative with respect to $x\text{,}$ then substituting $x=a$ and $x=b$ and subtracting, as usual. The result will be an expression with no $x$ variable but some occurrences of $y\text{.}$ Then the outer integral will be an ordinary one-variable problem, with $y$ as the variable.

###### Example4.14. Average Height.

Figure 4.1 shows the function $\sin(xy)+6/5$ on $[0.5,3.5]\times[0.5,2.5]\text{.}$ Find the average height of this surface.

Solution

The volume under this surface is

\begin{equation*} \int_{0.5}^{2.5}\int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\,dy\text{.} \end{equation*}

The inner integral is

\begin{equation*} \begin{split} \int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\amp= \left.{-\cos(xy)\over y}+{6x\over5}\right|_{0.5}^{3.5}\\ \amp= {-\cos(3.5y)\over y}+{\cos(0.5y)\over y}+{18\over5}. \end{split} \end{equation*}

Unfortunately, this gives a function for which we can't find a simple antiderivative. To complete the problem we could use Sage or similar software to approximate the integral. Doing this gives a volume of approximately $8.84\text{,}$ so the average height is approximately $8.84/6\approx 1.47\text{.}$

Because addition and multiplication are commutative and associative, we can rewrite the original double sum:

\begin{equation*} \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}f(x_j,y_i)\Delta x\Delta y=\sum_{j=0}^{m-1}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y\Delta x\text{.} \end{equation*}

Now if we repeat the development above, the inner sum turns into an integral:

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y = \int_c^d f(x_j,y)\,dy\text{,} \end{equation*}

and then the outer sum turns into an integral:

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-1}\left(\int_c^d f(x_j,y)\,dy \right)\Delta x = \int_a^b\int_c^d f(x,y)\,dy\,dx\text{.} \end{equation*}

In other words, we can compute the integrals in either order, first with respect to $x$ then $y\text{,}$ or vice versa. Thinking of the loaf of bread, this corresponds to slicing the loaf in a direction perpendicular to the first.

We summarize our findings by providing a general guideline for how the double integral over a rectangle, such as the one shown below, is computed. ###### Guideline for Computing a Double Integral over a Rectangle.

Let $f(x,y)$ be a continuous function defined over the rectangle $R=[a,b]\times[c,d]\text{.}$ Follow these steps to evaluate the double integral

\begin{equation*} \int_c^d \int_a^b f(x,y)\,dx\,dy = \int_a^b \int_c^d f(x,y)\,dy\,dx\text{.} \end{equation*}
1. Choose the double integral, $\ds{\int_c^d \int_a^b f(x,y)\,dx\,dy}$ or $\ds{\int_a^b \int_c^d f(x,y)\,dy\,dx}\text{.}$

2. Compute the inner integral:

1. If $\ds \int_a^b f(x,y)\,dx\text{,}$ then temporarily treat $y$ as a constant. Find an antiderivative with respect to $x\text{,}$ then evaluate over the integration bounds $x = a$ and $x = b\text{.}$

2. If $\ds \int_c^d f(x,y)\,dy\text{,}$ then temporarily treat $x$ as a constant. Find an antiderivative with respect to $y\text{,}$ then evaluate over the integration bounds $y=c$ and $y=d\text{.}$

3. Compute the outer integral, which will be an ordinary one-variable integral of the form

\begin{equation*} \int_c^d G(y)\, dy \text{ or } \int_a^b G(x)\,dx\text{.} \end{equation*}

We haven't really proved that the value of a double integral is equal to the value of the corresponding two single integrals in either order of integration, but provided the function is continuous, this is true. The result is called Fubini's Theorem, which we state here without proof.

We provide one example, where we compute the volume under a surface in two ways by switching the order of integration.

###### Example4.16. Compute Volume in Two Ways.

We compute $\ds\iint_R 1+(x-1)^2+4y^2\,dA\text{,}$ where $R=[0,3]\times[0,2]\text{,}$ in two ways.

Solution

First,

\begin{align*} \int_0^3\int_0^2 1+(x-1)^2+4y^2\,dy\,dx \amp =\int_0^3\left. y+(x-1)^2y+{4\over 3}y^3\right|_0^2\,dx\\ \amp =\int_0^3 2+2(x-1)^2+{32\over 3}\,dx\\ \amp =\left. 2x + {2\over 3}(x-1)^3+{32\over 3}x\right|_0^3\\ \amp =6+{2\over 3}\cdot 8 + {32\over 3}\cdot3-(0-1\cdot{2\over3}+0)\\ \amp =44\text{.} \end{align*}

In the other order:

\begin{align*} \int_0^2\int_0^3 1+(x-1)^2+4y^2\,dx\,dy \amp =\int_0^2\left. x+{(x-1)^3\over3}+4y^2x\right|_0^3\,dy\\ \amp =\int_0^2 3+{8\over3}+12y^2+{1\over3}\,dy\\ \amp =\left. 3y+{8\over3}y+4y^3+{1\over3}y\right|_0^2\\ \amp =6+{16\over3}+32+{2\over3}\\ \amp =44\text{.} \end{align*}
Note: In this example there is no particular reason to favor one direction over the other; in some cases, one direction might be much easier than the other, so it's usually worth considering the two different possibilities.

### Subsection4.2.3Computing Double Integrals over any Region

Frequently we will be interested in a region that is not simply a rectangle. Let's compute the volume under the surface $x+2y^2$ above the region described by $0\le x\le1$ and $0\le y\le x^2\text{,}$ shown below. In principle there is nothing more difficult about this problem. If we imagine the three-dimensional region under the surface and above the parabolic region as an oddly shaped loaf of bread, we can still slice it up, approximate the volume of each slice, and add these volumes up. For example, if we slice perpendicular to the $x$-axis at $x_i\text{,}$ the thickness of a slice will be $\Delta x$ and the area of the slice will be

\begin{equation*} \int_0^{x_i^2} x_i+2y^2\,dy\text{.} \end{equation*}

When we add these up and take the limit as $\Delta x$ goes to 0, we get the double integral

\begin{align*} \int_0^1 \int_0^{x^2} x+2y^2\,dy\,dx \amp =\int_0^1 \left.xy+{2\over3}y^3\right|_0^{x^2}\,dx\\ \amp =\int_0^1 x^3+{2\over3}x^6\,dx\\ \amp =\left. {x^4\over4}+{2\over21}x^7\right|_0^1\\ \amp ={1\over4}+{2\over21}={29\over84}\text{.} \end{align*}

We could just as well slice the solid perpendicular to the $y$-axis, in which case we get

\begin{align*} \int_0^1 \int_{\sqrt y}^1 x+2y^2\,dx\,dy \amp =\int_0^1 \left.{x^2\over2}+2y^2x\right|_{\sqrt y}^1 \,dy\\ \amp =\int_0^1 {1\over2}+2y^2-{y\over2}-2y^2\sqrt y\,dy\\ \amp =\left.{y\over2}+{2\over3}y^3-{y^2\over4}-{4\over7}y^{7/2}\right|_0^1\\ \amp ={1\over2}+{2\over3}-{1\over4}-{4\over7}={29\over84}\text{.} \end{align*}

What is the average height of the surface over this region? As before, it is the volume divided by the area of the base, but now we need to use integration to compute the area of the base, since it is not a simple rectangle. The area is

\begin{equation*} \int_0^1 x^2\,dx={1\over3}\text{,} \end{equation*}

so the average height is

\begin{equation*} \frac{29}{84}\div\frac{1}{3} = \frac{29}{28}\text{.} \end{equation*}

Although we have not proven that the order of integration can be switched, we nonetheless capture our results and state the general version of Fubini's Theorem without proof.

Note: Although Fubini's Theorem tells us that the order of integration does not matter in a double integral, the theorem does not tell us which of the double integrals is easier to compute. Experience through practice allows us to decide whether to choose to set up a double integral with $dx\,dy$ or $dy\,dx\text{.}$

We explore the order of integration with one more example of a double integral.

###### Example4.18. Volume of Region.

Find the volume under the surface $\ds z=\sqrt{1-x^2}$ and above the triangle formed by $y=x\text{,}$ $x=1\text{,}$ and the $x$-axis.

Solution

Let's consider the two possible ways to set this up:

Which appears easier? In the first, the first (inner) integral is easy, because we need an antiderivative with respect to $y\text{,}$ and the entire integrand $\ds\sqrt{1-x^2}$ is constant with respect to $y\text{.}$ Of course, the second integral may be more difficult. In the second, the first integral is mildly unpleasant—a trig substitution. So let's try the first one, since the first step is easy, and see where that leaves us.

\begin{equation*} \int_0^1 \int_0^x \sqrt{1-x^2}\,dy\,dx= \int_0^1 \left. y\sqrt{1-x^2}\right|_0^x\,dx= \int_0^1 x\sqrt{1-x^2}\,dx\text{.} \end{equation*}

This is quite easy, since the substitution $u=1-x^2$ works:

\begin{equation*} \int x\sqrt{1-x^2}\,dx=-{1\over 2}\int \sqrt u\,du ={1\over3}u^{3/2}=-{1\over3}(1-x^2)^{3/2}\text{.} \end{equation*}

Then

\begin{equation*} \int_0^1 x\sqrt{1-x^2}\,dx=\left. -{1\over3}(1-x^2)^{3/2}\right|_0^1 ={1\over3}\text{.} \end{equation*}

This is a good example of how the order of integration can affect the complexity of the problem. In this case it is possible to do the other order, but it is a bit messier. In some cases one order may lead to an integral that is either difficult or impossible to evaluate; it's usually worth considering both possibilities before going very far.

##### Exercises for Section 4.2.

Compute the following double integrals.

1. $\ds \int_{0}^{2}\int_{0}^{4} 1+x \,dy\,dx$

$16$
Solution
\begin{equation*} \begin{split} \int_0^2 \int_0^4 (1+x)\,dy\,dx \amp = \int_0^2 (1+x) y\big\vert_0^4\,dx \\ \amp= \int_0^2 4(1+x)\,dx = 4\left[x+\frac{x^2}{2}\right]_0^2 = 16.\end{split} \end{equation*}
2. $\ds \int_{-1}^{1}\int_{0}^{2} x+y\,dy\,dx$

$4$
Solution
\begin{equation*} \begin{split} \int_{-1}^1\int_0^2 (x+y)\,dy\,dx \amp= \int_{-1}^1 \left[xy + \frac{y^2}{2}\right]_0^2 \\ \amp= \int_{-1}^2 \left(2x+2\right)\,dx = \left[x^2+2x\right]_{-1}^1 = 4. \end{split} \end{equation*}
3. $\ds \int_{1}^{2}\int_{0}^{y} xy \,dx\,dy$

$15/8$
Solution
\begin{aligned}\int_1^2\int_0^y xy \,dx\,dy \amp = \int_1^2 y \int_0^y x \,dx\,dy \\ \amp = \int_1^2 y\left[\frac{1}{2}x^2\right]_0^y\,dy \\ \amp = \int_1^2 \frac{1}{2}y^3\,dy = \frac{1}{2}\left[\frac{1}{4}y^4\right]_1^2 = \frac{15}{8} \end{aligned}
4. $\ds \int_{0}^{1}\int_{y^2/2}^{\sqrt y} \,dx\,dy$

$1/2$
Solution
\begin{equation*} \begin{split} \int_0^1\int_{y^2/2}^{\sqrt{y}}\,dx\,dy \amp= \int_0^1 x\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp= \int_0^1 \sqrt{y}-\frac{y^2}{2}\,dy\\ \amp \left[\frac{2y^{3/2}}{3}-\frac{y^3}{6}\right]_0^1 = \frac{1}{2} \end{split} \end{equation*}
5. $\ds \int_{1}^{2}\int_{1}^{x} {x^2\over y^2}\,dy\,dx$

$5/6$
Solution
\begin{equation*} \begin{split} \int_1^2\int_1^x \frac{x^2}{y^2}\,dy\,dx \amp= \int_1^2 x^2 \left[-\frac{1}{y}\right]_1^x\,dx \\ \amp= \int_1^2 x^2 \left(-\frac{1}{x}+1\right)\,dx = \int_1^2 \left(-x + x^2\right)\,dx\\ \amp= \left[-\frac{x^2}{2} + \frac{x^3}{3}\right]_1^2 = \frac{5}{6} \end{split} \end{equation*}
6. $\ds \int_{0}^{1}\int_{0}^{x^2} y e^x\,dy\,dx$

$\frac{9e}{2} - 12$
Solution
\begin{equation*} \begin{split}\int_0^1 \int_0^{x^2} e^x y\,dy\,dx = \int_0^1 \frac{e^x y^2}{2} \big\vert_0^{x^2}\,dx = \int_0^1 \frac{e^x x^4}{2}\,dx \end{equation*}
We now use integration by parts to solve the corresponding indefinite integral, with $u=x^4\text{,}$ $dv = e^xdx\text{,}$ and so $v=e^x\text{,}$ and $du=4x^3\,dx\text{:}$
\begin{equation*} \int \frac{e^x x^4}{2}\,dx = \frac{1}{2} \left[x^4e^x - 4 \int e^x x^3\,dx \right] \end{equation*}
We now continue by applying Integration by parts twice more with $dv = e^x\,dx\text{:}$
\begin{equation*} \begin{split} \int \frac{e^x x^4}{2}\,dx \amp= \frac{1}{2} \left[x^4e^x - 4 \int e^x x^3\,dx \right] \\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12 \int e^x x^2\,dx \right]\\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24 \int e^x x\,dx \right]\\ \amp= \frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x \,dx \right] \end{split} \end{equation*}
Hence, we have
\begin{equation*} \begin{split} \int_0^1 \int_0^{x^2} \frac{y}{e^x}\,dy\,dx \amp= \int_0^1 \frac{e^x x^4}{2}\,dx \\ \amp=\frac{1}{2} \left[x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x \,dx \right]_0^1\\ \amp= \frac{9e}{2} - 12.\end{split} \end{equation*}
7. $\ds \int_{0}^{\sqrt{\pi/2}}\int_{0}^{x^2} x\cos y\,dy\,dx$

$1/2$
Solution
\begin{aligned}\int_0^1\int_{y^2/2}^{\sqrt{y}}\,dx\,dy \amp = \int_0^1 x\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp = \int_0^1 \sqrt{y}-\frac{y^2}{2}\,dy\\ \amp = \left[\frac{2y^{3/2}}{3}-\frac{y^3}{6}\right]_0^1 = \frac{1}{2} \end{aligned}
8. $\ds \int_{0}^{\pi/2}\int_{0}^{\cos\theta}r^2(\cos\theta-r) \,dr\,d\theta$

$\pi/64$
Solution

\begin{aligned}\int_0^{\pi/2}\int_0^{\cos\theta} r^2 \left(\cos\theta - r\right)\,dr\,d\theta \amp =\int_0^{\pi/2} \left[\frac{r^3}{3}\cos\theta - \frac{r^4}{4}\right]_0^{\cos\theta} \,d\theta \\ \amp =\int_0^{\pi/2} \left[\frac{\cos^4\theta}{3}-\frac{\cos^4\theta}{4}\right]\,d\theta = \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \end{aligned} Now use the identity $\displaystyle{\cos^2\theta = \frac{1}{2}(1+\cos (2\theta))}\text{:}$

\begin{equation*} \begin{split} \cos^4\theta \amp = (\cos^2\theta)^2 = \frac{1}{2}(1+\cos 2\theta)^2 \\ \amp = \frac{1}{4} \left(1 + \cos^22\theta + 2 \cos 2\theta\right) \\ \amp = \frac{1}{4}\left(1 + \frac{1}{2}(1+\cos 4\theta) + 2 \cos 2\theta\right) \\ \amp = \frac{3}{8} + \frac{1}{8} \cos 4\theta + \frac{1}{2}\cos 2\theta. \end{split} \end{equation*}

Therefore,

\begin{equation*} \begin{split} \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \amp = \frac{1}{12}\int_0^{\pi/2} \left(\frac{3}{8} + \frac{1}{8} \cos 4\theta + \frac{1}{2}\cos 2\theta\right)\\ \amp = \frac{1}{12}\left[\frac{3\theta}{8} + \frac{\sin 4\theta}{32} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2}\\ \amp = \frac{\pi}{64}. \end{split} \end{equation*}

We find that

\begin{equation*} \int_0^{\pi/2}\int_0^{\cos\theta} r^2 \left(\cos\theta - r\right)\,dr\,d\theta = \frac{\pi}{64} \text{.} \end{equation*}
9. $\ds \int_0^1\int_{\sqrt{y}}^1 \sqrt{x^3+1}\,dx\,dy$

$\frac{2}{9}\left(2\sqrt{2}-1\right)$
Solution
First, we notice that we cannot evaluate the integral
\begin{equation*} \int \sqrt{x^3+1} \,dx, \end{equation*}
and so we try switching the order of integration. The region of integration is as follows: Therefore,
\begin{equation*} \int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3+1} \,dx \,dy = \int_0^1 \int_0^{x^2} \sqrt{x^3+1} \,dy \,dx = \int_0^1 x^2 \sqrt{x^3+1} \,dx. \end{equation*}
We now carry out the integration by making the following substitution: let $u=x^3+1\text{,}$ then $du = 3x^2\,dx\text{.}$ Then as $x$ goes from $0 \to 1\text{,}$ $u$ goes from $1 \to 2\text{:}$
\begin{equation*} \int_0^1 x^2 \sqrt{x^3+1} \,dx = \int_1^2 \frac{1}{3} \sqrt{u}\,du = \frac{2}{9} u^{3/2}\big\vert_1^2 = \frac{2}{9} \left(2\sqrt{2}-1\right). \end{equation*}
Therefore, we have found that
\begin{equation*} \int_0^1 \int_{\sqrt{y}}^1 \sqrt{x^3+1} \,dx \,dy = \frac{2}{9}\left(2\sqrt{2}-1\right). \end{equation*}
10. $\ds \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2}\,dy\,dx$

$(2\sqrt2-1)/6$
Solution
First, notice that we cannot integrate
\begin{equation*} \int x\sqrt{1+y^2}\,dy. \end{equation*}
Therefore, we switch the order of integation: Therefore,
\begin{equation*} \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2} \,dy \,dx = \int_0^1 \int_0^{\sqrt{y}} x\sqrt{1+y^2} \,dx \,dy = \int_0^1 \frac{y}{2} \sqrt{1+y^2}\,dy. \end{equation*}
Now use the substitution: $u=1+y^2$ with $du = 2y\,dy\text{.}$ Then as $y$ goes from $0 \to 1\text{,}$ we have $u$ going from $1 \to 2\text{.}$ Hence,
\begin{equation*} \int_0^1 \frac{y}{2} \sqrt{1+y^2}\,dy = \int_1^2 \frac{\sqrt{u}}{4}\,du = \frac{u^{3/2}}{6}\big\vert_1^2 = \frac{1}{6} \left(2\sqrt{2}-1\right). \end{equation*}
Therefore, we have shown that
\begin{equation*} \int_0^1 \int_{x^2}^1 x\sqrt{1+y^2} \,dy \,dx = \frac{1}{6} \left(2\sqrt{2}-1\right). \end{equation*}
11. $\ds \int_0^1 \int_0^y {2\over\sqrt{1-x^2}}\,dx\,dy$

$\pi-2$
Solution
We will find that this problem is easier if we first switch the order of integration: Therefore,
\begin{equation*} \int_0^1 \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \int_0^1 \int_x^1 \frac{2}{\sqrt{1-x^2}}\,dy \,dx = \int_0^1 \frac{2(1-x)}{\sqrt{1-x^2}}\,dx. \end{equation*}
Hence, we split this integral into two:
\begin{equation*} \int_0^1 \frac{2}{\sqrt{1-x^2}}\,dx = 2\sin^{-1}(x)\big\vert_0^1 = \pi. \end{equation*}
For the second integral, we use the following substitution: $u=1-x^2$ with $du=-2x\,dx\text{.}$ Therefore,
\begin{equation*} \int_0^1 \frac{-2x}{\sqrt{1-x^2}}\,dx = \int_0^1 \frac{1}{\sqrt{u}}\,du = -2. \end{equation*}
Altogether, we find
\begin{equation*} \int_0^1 \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \pi -2. \end{equation*}
12. $\ds \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy$

$(e^9-1)/6$
Solution
We first switch the order of integration: Therefore,
\begin{equation*} \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy = \int_0^3\int_0^{x/3} e^{x^2}\,dy\,dy = \int_0^3 \frac{x}{3} e^{x^2}\,dx. \end{equation*}
This is an integral which we can solve using substitution: let $u=x^2$ with $du=2x\,dx\text{.}$ Then,
\begin{equation*} \int_0^3 \frac{x}{3} e^{x^2}\,dx = \int_0^9 \frac{1}{6} e^u\,du = \frac{1}{6} \left(e^9-1\right). \end{equation*}
We have found that
\begin{equation*} \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy = \frac{1}{6} \left(e^9-1\right). \end{equation*}
13. $\ds \int_{-1}^1\int_0^{1-x^2} x^2-\sqrt{y}\,dy\,dx$

$\ds {4\over15}-{\pi\over4}$
14. $\ds \int_{0}^{\sqrt2/2}\int_{-\sqrt{1-2x^2}}^{\sqrt{1-2x^2}} x\,dy\,dx$

$1/3$

Determine the volume of the region that is bounded as follows:

1. below $z=1-y$ and above the region $-1\le x\le 1\text{,}$ $0\le y\le 1-x^2$

$4/5$
Solution
The area which bounds the surface in the $x$-$y$-plane is shown below. Therefore, the volume is computed as follows:
\begin{equation*} \begin{split} V= \int_{-1}^1 \int_{0}^{1-x^2} (1-y)\,dy\,dx \amp= \int_{-1}^1 y-\frac{y^2}{2} \big\vert_0^{1-x^2} \,dx\\ \amp= \frac{1}{2}\int_{-1}^1 (1-x^4) \,dx = \frac{1}{2}\left[x-\frac{x^5}{5}\right]_{-1}^1 = \frac{4}{5}.\end{split} \end{equation*}
2. enclosed by $y=x^2\text{,}$ $y=4\text{,}$ $z=x^2\text{,}$ $z=0$

$128/15$
Solution
In the $x$-$y$-plane, the surface is bounded by: And so we set up the following integral to compute the volume:
\begin{equation*} \begin{split} V=\int_{-2}^2 \int_{x^2}^4 \int_0^{x^2} \,dz\,dy\,dx \amp= \int_{-2}^2 \int_{x^2}^4 x^2\,dy\,dy \\ \amp= \int_{-2}^2 -x^4+4x^2\,dx = \frac{128}{15}.\end{split} \end{equation*}
3. in the first octant bounded by $y^2=4-x$ and $y=2z$

$2$
Solution
The region is bounded by the following cross-sections: Therefore, we compute the following as follows:
\begin{equation*} V = \int_0^4 \int_0^{\sqrt{4-x}} \frac{y}{2} \,dy\,dx=2. \end{equation*}
4. in the first octant bounded by $y^2=4x\text{,}$ $2x+y=4\text{,}$ $z=y\text{,}$ and $y=0$

$\frac{11}{3}$
Solution
In the first octant of the $x$-$y$-plane, the surface is bounded as follows: Therefore, we set up the integral:
\begin{equation*} V = \int_0^1 \int_{2\sqrt{x}}^{4-2x} \int_0^y dz\,dy\,dx = \frac{11}{3}. \end{equation*}
5. in the first octant bounded by $x+y+z=9\text{,}$ $2x+3y=18\text{,}$ and $x+3y=9\text{.}$

$\frac{81}{2}$
Solution
In the $x$-$y$-plane, the region is bounded by: Therefore, we compute the volume of the region:
\begin{equation*} V = \int_0^9 \int_{(9-x)/3}^{2(9-x)/3} (9-x-y) \,dy\,dx = \frac{81}{2}. \end{equation*}
6. in the first octant bounded by $x^2+y^2=a^2$ and $z=x+y$

$\frac{2a^3}{3}$
Solution
In the $x$-$y$-plane, the region is bounded as follows: Therefore, we set up the following integral to compute the volume:
\begin{equation*} V = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} (x+y)\,dy\,dx = \frac{2}{3} a^3. \end{equation*}
7. bounded by $4x^2+y^2=4z$ and $z=2$

$4\pi$
Solution
This volume is an elliptic paraboloid. For fixed $z\text{,}$ the cross-sections of the volume in the $x$-$y$ plane are ellipses. Therefore, we consider this to be the volume of the surface $z= \frac{1}{4}\left(4x^2+y^2\right)$ below the ellipse $\frac{x^2}{2} + \frac{y^2}{8}=1\text{:}$
\begin{equation*} V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{8-4x^2}}^{\sqrt{8-4x^2}} \,dy\, dx = 4\pi. \end{equation*}
8. bounded by $z=x^2+y^2$ and $z=4$

$8\pi$
Solution
The volume bounded by the paraboloid $z=x^2+y^2$ and the plane $z=4$ is the volume above the surface $z=x^2+y^2$ below the circle $4=x^2+y^2\text{.}$ This volume can be computed by:
\begin{equation*} V = \int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (x^2+y^2) \,dx \,dy = 8\pi. \end{equation*}
9. under the surface $z=xy$ and above the triangle with vertices $(1,1,0)\text{,}$ $(4,1,0)\text{,}$ $(1,2,0)$ Answer
$31/8$
Solution

The area which bounds the region in the $x$-$y$-plane is shown below. Therefore, if we let $x$ vary from $1$ to $4\text{,}$ we see that $y$ is bounded between $1$ and the line $y=\frac{7}{3}-\frac{x}{3}\text{.}$ And so the volume under the surface $z=xy$ and above the area shown in green is calculated by

\begin{equation*} \begin{split} \int_1^4\int_1^{7/3-x/3} xy\,dy\,dx \amp = \int_1^4 x\left[\frac{1}{2}y^2\right]_1^{7/3-x/3} \,dx\\ \amp =\int_1^4 \frac{x}{18}\left(x^2-14x+40\right)\,dx \\ \amp = \left[\frac{1}{18}\left(\frac{x^4}{4}-\frac{14x^3}{3}+20x^2\right)\right]_1^4 =\frac{31}{8}. \end{split} \end{equation*}

Evaluate $\ds\iint x^2\,dA$ over the region in the first quadrant bounded by the hyperbola $xy=16$ and the lines $y=x\text{,}$ $y=0\text{,}$ and $x=8\text{.}$

$448$
Solution
The volume is bounded by the following area in the $x$-$y$-plane: Therefore, we compute the following integrals:
\begin{equation*} V = \int_0^4 \int_0^x x^2 \,dy\,dx + \int_4^8 \int_0^{16/x} x^2 \,dy \,dx = 448. \end{equation*}

A swimming pool is circular with a 40 meter diameter. The depth is constant along east-west lines and increases linearly from 2 meters at the south end to 7 meters at the north end. Find the volume of the pool.

$1800\pi$${ m}^3$
Solution
The domain of the density function is the circular area shown below: where $x$ and $y$ are measured in m. Let $z$ denote the density in metres as a function of $x$ and $y\text{.}$ The density is a plane with the following cross-sections: Therefore, the density can be expressed as
\begin{equation*} z=f(x,y) = \frac{x+y}{8}+\frac{9}{2}. \end{equation*}
Hence, the volume of the pool can be computed by the following integral:
\begin{equation*} V = \int_{-20}^{20} \int_{-\sqrt{20^2-x^2}}^{\sqrt{20^2-x^2}} \left[\frac{x+y}{8}+\frac{9}{2}\right]\,dy\,dx = 1800\pi \ \text{m}^3. \end{equation*}

Find the average value of $f(x,y)=e^y\sqrt{x+e^y}$ on the rectangle with vertices $(0,0)\text{,}$ $(4,0$), $(4,1)$ and $(0,1)\text{.}$

$\ds{(e^2+8e+16)\over15}\sqrt{e+4}-{5\sqrt5\over3}-{e^{5/2}\over15} +{1\over15}$
Solution
We wish to compute the average value of $f(x,y) = e^y\sqrt{x+e^y}$ over the rectangle shown below: The total area of the rectangle is
\begin{equation*} \int_0^4\int_0^1 \,dy\,dx = 4, \end{equation*}
and so we compute
\begin{equation*} f_{avg} = \frac{1}{4} \int_0^4\int_0^1 e^y\sqrt{x+e^y}\, dy\,dx. \end{equation*}
Evaluating this integral, we find
\begin{equation*} f_{avg} = \frac{1}{15}\left[1-25\sqrt{5}-e^{5/2}+16\sqrt{4+e} +8e\sqrt{4+e} +e^2\sqrt{4+e}\right]. \end{equation*}

Reverse the order of integration on each of the following integrals

1. $\ds\int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx \,dy$

$\ds\int_0^3\int_0^{9-x^2} f(x,y)\,dx\,dy$
Solution

The integral

\begin{equation*} \int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx\,dy \end{equation*}

gives the volume under the surface $z=f(x,y)$ and above the region in the $x$-$y$-plane shown below: So if instead we let $x$ vary from 0 to 3, we see that $y$ is bounded between $y=0$ and the curve $x=\sqrt{9-y} \implies y = 9-x^2$ (for $x \in [0,3]$). Therefore,

\begin{equation*} \int_0^9 \int_0^{\sqrt{9-y}} f(x,y)\,dx\,dy = \int_0^3\int_0^{9-x^2} f(x,y)\,dx\,dy\text{.} \end{equation*}
2. $\ds\int_1^2 \int_0^{\ln x} f(x,y) \; dy \; dx$

$\int_0^{\ln(2)} \int_{e^y}^{2} f(x,y)\,dx\,dy$
Solution
The area of integration is shown below: Therefore,
\begin{equation*} \int_1^2 \int_0^{\ln x} f(x,y)\,dy\,dx = \int_0^{\ln(2)} \int_{e^y}^{2} f(x,y)\,dx\,dy. \end{equation*}
3. $\ds\int_0^1 \int_{\arcsin y}^{\pi/2} f(x,y) \; dx \; dy$

$\int_0^{\pi/2} \int_0^{\sin x} f(x,y) \,dy\,dx$
Solution
The area of integration is shown below: Therefore, we see that
\begin{equation*} \int_0^1 \int_{\arcsin(y)}^{\pi/2} f(x,y)\,dx\,dy = \int_0^{\pi/2} \int_0^{\sin x} f(x,y) \,dy\,dx. \end{equation*}
4. $\ds\int_0^1 \int_{4x}^{4} f(x,y) \; dy \; dx$

$\int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy$
Solution
The area of integration is shown below: Therefore, we see that
\begin{equation*} \int_0^1 \int_{4x}^4 f(x,y)\,dy\,dx = \int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy. \end{equation*}
5. $\ds\int_0^3 \int_{0}^{\sqrt{9-y^2}} f(x,y) \; dx \; dy$

$\int_0^3 \int_0^{\sqrt{9-x^2}} f(x,y)\,dy\,dx$
Note that the circle centred at the origin with radius 3 has equation $x^2+y^2=9\text{,}$ and in the first quadrant can be expressed as $x=\sqrt{9-y^2}$ or $y=\sqrt{9-x^2}\text{.}$ Therefore, the area of integration is: 