Give an example of a region that cannot be expressed by either of the forms shown in Figure 34. (One example is the disk with a hole in the middle between the graphs of $x^{2}+y^{2}=1$ and $x^{2}+y^{2}=2$ in Figure $10 . )$

The idea of the average value of a function, discussed earlier for functions of the form $y=f(x)$ , can be extended to functions of more than one independent variable. For a function $z=f(x, y)$ ,

the average value of $f$ over a region $R$ is defined as

$\frac{1}{A} \iint_{R} f(x, y) d x d y$

where $A$ is the area of the region $R .$ Find the average value foreach function over the regions $R$ having the given boundaries.

## Discussion

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## Recommended Questions

In the following exercises, the function $f$ is given in terms of double integrals.

a. Determine the explicit form of the function $f$

b. Find the volume of the solid under the surface $z=f(x, y)$ and above the region $R$ .

c. Find the average value of the function $f$ on $R$ .

d. Use a computer algebra system (CAS) to plot $z=f(x, y)$ and $z=f_{\text ave }$ in the same system of coordinates.

$f(x, y)=\int_{0}^{y} \int_{0}^{x}(x s+y t) d s d t$ where $(x, y) \in R=[0,1] \times[0,1]$

In the following exercises, the function $f$ is given in terms of double integrals.

a. Determine the explicit form of the function $f$

b. Find the volume of the solid under the surface $z=f(x, y)$ and above the region $R$ .

c. Find the average value of the function $f$ on $R$ .

d. Use a computer algebra system (CAS) to plot $z=f(x, y)$ and $z=f_{\text ave }$ in the same system of coordinates.

$f(x, y)=\int_{0}^{x} \int_{0}^{y}[\cos (s)+\cos (t)] d t d s$ where $(x, y) \in R=[0,3] \times[0,3]$

Evaluate the double integral by first identifying it as the volume of a solid.

$\iint_{R}(4-2 y) d A, \quad R=[0,1] \times[0,1]$

Evaluate the double integral by first identifying it as the volume of a solid.

$ \iint_R (4 - 2y)\ dA $, $ R = [0, 1] \times [0, 1] $

Interpret the volume integral in the Divergence Theorem.

Draw a solid whose volume is given by the double integral $\int_{0}^{6} \int_{1}^{2} 10 d y d x .$ Then evaluate the integral using geometry.

Evaluate the double integral by first identifying it as the volume of a solid.

$ \iint_R (2x + 1)\ dA $, $ R = \{(x, y) \mid 0 \le x \le 2, 0 \le y \le 4 \} $

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The prism in the first octant bounded by $z=2-4 x$ and $y=8$.

Volume enclosed by paraboloids $\quad$ Let $D$ be the region bounded by the paraboloids $z=8-x^{2}-y^{2}$ and $z=x^{2}+y^{2} .$ Write six different triple iterated integrals for the volume of $D .$ Evaluate one of the integrals.

Finding the Volume of a Solid Explain how to

find the volume of a solid with a known cross section.

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by the cylinder $y=9-x^{2}$ and the paraboloid $y=2 x^{2}+3 z^{2}$.

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded below by the cone $z=\sqrt{x^{2}+y^{2}}$ and bounded above by the sphere $x^{2}+y^{2}+z^{2}=8$.

Sketch the region of integration and the solid whose volume is given by the double integral.

\begin{equation}

\int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} d x d y\end{equation}

Evaluate the double integral by first identifying it as the volume of a solid.

$ \iint_R \sqrt{2}\ dA $, $ R = \{(x, y) \mid 2 \le x \le 6, -1 \le y \le 5 \} $

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by $x=0, x=2, y=z, y=z+1, z=0$ and $z=4$

Find the surface area of the volume generated when the following curves revolve around the $x$ -axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

$$

y=\sqrt{4-x^{2}} \text { from } x=0 \text { to } x=2

$$

Use a double integral to derive the formula for the volume of

a sphere of radius $a$

Find the surface area of the volume generated when the following curves revolve around the $y$ -axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

$$

y=x^{2} \text { from } x=0 \text { to } x=2

$$

Write an integral for the average value of $f(x, y, z)=x y z$ over the region bounded by the paraboloid $z=9-x^{2}-y^{2}$ and the $x y$ -plane (assuming the volume of the region is known).

Rotation of the region in Figure 12 about the $y$ -axis produces a

solid with two types of different cross sections. Compute the volume as

a sum of two integrals, one for $-12 \leq y \leq 4$ and one for $4 \leq y \leq 12$ .

sketch the region of integration and the solid whose volume is given by the double integral.

$$\int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} d x d y$

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid in the first octant bounded by the plane $2 x+3 y+6 z=12$ and the coordinate planes.

Draw the solid region whose volume is given by the following double integrals. Then find the volume of the solid.

$$\int_{0}^{6} \int_{1}^{2} 10 d y d x$$

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by the surfaces $z=e^{y}$ and $z=1$ over the rectangle $\{(x, y): 0 \leq x \leq 1$ $0 \leq y \leq \ln 2\}$

(a) Estimate the volume of the solid that lies below the surface $z=x+2 y^{2}$ and above the rectangle $R=[0,2] \times[0,4] .$ Use a Riemann sum with $m=n=2$ and choose the sample points to be lower

right corners.

(b) Use the Midpoint Rule to estimate the volume in part (a).

(c) Evaluate the double integral and compare your answer with the estimates in parts (a) and (b).

Volume of solid

the volume of the region in the first octant enclosed by the cylinder $x^{2}+z^{2}=4$ and the plane $y=3 .$ Evaluate one of the

integrals.

Volume A region in the first quadrant is bounded above by the

curve $y=\cosh x,$ below by the curve $y=\sinh x,$ and on the left

and right by the $y$ -axis and the line $x=2,$ respectively. Find the

volume of the solid generated by revolving the region about the

$x$ -axis.

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by $x=0, x=2, y=0, y=e^{-z}, z=0,$ and $z=1$.

Volume of solid Write six different iterated triple integrals

for the volume of the region in the first octant enclosed by the

cylinder $x^{2}+z^{2}=4$ and the plane $y=3 .$ Evaluate one of the

integrals.

Volume of solid Write six different iterated triple integrals for the volume of the region in the first octant enclosed by the cylinder $x^{2}+z^{2}=4$ and the plane $y=3 .$ Evaluate one of the integrals.

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by $x=0, y=z^{2}, z=0,$ and $z=2-x-y$

In Section $6.2,$ Exercise $6,$ we revolved about the $y$ -axis the region between the curve $y=9 x / \sqrt{x^{3}+9}$ and the $x$ -axis from $x=0$

to $x=3$ to generate a solid of volume $36 \pi .$ What volume do you get if you revolve the region about the $x$ -axis instead?

Find the surface area of the volume generated when the following curves revolve around the $x$ -axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

$$

y=x^{3} \text { from } x=0 \text { to } x=1

$$

Find the surface area of the volume generated when the following curves revolve around the $y$ -axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

$$

y=\sqrt[3]{x} \text { from } x=1 \text { to } x=27

$$

Use a triple integral to find the volume of the following solids. (FIGURE CAN'T COPY)

The solid bounded by $x=0, x=1-z^{2}, y=0, z=0,$ and $z=1-y$.

Evaluate the double integral by first identifying it as the volume of a solid.

$\iint_{R} 3 d A, \quad R=\{(x, y) |-2 \leqslant x \leqslant 2,1 \leqslant y \leqslant 6\}$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $x$ -axis. If you cannot evaluate the integral exactly, use your calculator to approximate it.

$$y=\sqrt{4-x^{2}} \text { from } x=0 \text { to } x=2$$