$V={\displaystyle \sum _{i=1}^2{\displaystyle \sum _{j=1}^{2}f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A}}=0$

a. 26 b. Answers may vary.

$-\frac{1340}{3}$

$\frac{4-\text{ln}5}{\text{ln}5}$

$\frac{\pi }{2}$

Answers to both parts a. and b. may vary.

Type I and Type II are expressed as $\left\{\left(x,y\right)|0\le x\le 2,x^2\le y\le 2x\right\}$ and $\left\{\left(x,y\right)|0\le y\le 4,\frac{1}{2}y\le x\le \sqrt{y}\right\},$ respectively.

$\pi \text{/}4$

$\{(x,y)|0\le y\le \ln 2,1\le x\le e^y\}\cup \{(x,y)|\ln 2\le y\le e,1\le x\le 2\}\cup \{(x,y)|e\le y\le e^2,\text{ln}y\le x\le 2\}$

Same as in the example shown.

$\frac{216}{35}$

$\frac{e^2}{4}+10e-\frac{49}{4}$ cubic units

$\frac{81}{4}$ square units

$\frac{3}{4}$

$\frac{\pi }{4}$

$\frac{11}{39}\approx 0.282$

$\frac{14}{3}$

$8\pi$

$\pi \text{/}4$

$V={\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{2\sqrt{2}}{\int }}\left(16-2r^2\right)}}rdrd\theta =64\pi$ cubic units

$A=2{\displaystyle \underset{\text{−}\pi \text{/}2}{\overset{\pi \text{/}6}{\int }}{\displaystyle \underset{1+\text{sin}\theta }{\overset{3-3\text{sin}\theta }{\int }}rdrd\theta }}=8\pi +9\sqrt{3}$

$\frac{\pi }{4}$

${\displaystyle \underset{B}{\iiint }z\text{sin}x\text{cos}y}dV=8$

${\displaystyle \underset{E}{\iiint }1}dV={\displaystyle {\int }_{x=\mathrm{-3}}^{x=3}{\displaystyle {\int }_{y=\text{−}\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}{\displaystyle {\int }_{z=\text{−}\sqrt{9-x^2-y^2}}^{z=\sqrt{9-x^2-y^2}}1dzdydx=36\pi .}}}$

(i) ${\displaystyle \underset{z=0}{\overset{z=4}{\int }}{\displaystyle \underset{x=0}{\overset{x=\sqrt{4-z}}{\int }}{\displaystyle \underset{y=x^2}{\overset{y=4-z}{\int }}f\left(x,y,z\right)}}}dydxdz,$ (ii) ${\displaystyle \underset{y=0}{\overset{y=4}{\int }}{\displaystyle \underset{z=0}{\overset{z=4-y}{\int }}{\displaystyle \underset{x=0}{\overset{x=\sqrt{y}}{\int }}f\left(x,y,z\right)}}}dxdzdy,$ (iii) ${\displaystyle \underset{y=0}{\overset{y=4}{\int }}{\displaystyle \underset{x=0}{\overset{x=\sqrt{y}}{\int }}{\displaystyle \underset{z=0}{\overset{z=4-y}{\int }}f\left(x,y,z\right)}}}dzdxdy,$ (iv) ${\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=x^2}{\overset{y=4}{\int }}{\displaystyle \underset{z=0}{\overset{z=4-y}{\int }}f\left(x,y,z\right)}}}dzdydx,$ (v) ${\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{z=0}{\overset{z=4-x^2}{\int }}{\displaystyle \underset{y=x^2}{\overset{y=4-z}{\int }}f\left(x,y,z\right)}}}dydzdx$

$f_{\text{ave}}=8$

$\frac{16}{3}$

${\displaystyle \underset{E}{\iiint }f\left(r,\theta ,z\right)}rdzdrd\theta ={\displaystyle \underset{\theta =0}{\overset{\theta =\pi }{\int }}{\displaystyle \underset{r=0}{\overset{r=2\text{sin}\theta }{\int }}{\displaystyle \underset{z=0}{\overset{z=4-r\text{sin}\theta }{\int }}f\left(r,\theta ,z\right)rdzdrd\theta }}}.$

$E=\left\{\left(r,\theta ,z\right)|0\le \theta \le 2\pi ,0\le z\le 1,z\le r\le 2-z^2\right\}$ and $V={\displaystyle \underset{r=0}{\overset{r=1}{\int }}{\displaystyle \underset{z=r}{\overset{z=2-r^2}{\int }}{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}rd\theta dzdr}}}.$

$E_2=\left\{\left(r,\theta ,z\right)|0\le \theta \le 2\pi ,0\le r\le 1,r\le z\le \sqrt{4-r^2}\right\}$ and $V={\displaystyle \underset{r=0}{\overset{r=1}{\int }}{\displaystyle \underset{z=r}{\overset{z=\sqrt{4-r^2}}{\int }}{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}rd\theta dzdr}}}.$

$V\left(E\right)={\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\varphi =0}{\overset{\phi =\pi \text{/}3}{\int }}{\displaystyle \underset{\rho =0}{\overset{\rho =2}{\int }}{\rho }^2\text{sin}\phi d\rho d\phi d\theta }}}$

Rectangular: ${\displaystyle \underset{x=\mathrm{-2}}{\overset{x=2}{\int }}{\displaystyle \underset{y=\text{−}\sqrt{4-x^2}}{\overset{y=\sqrt{4-x^2}}{\int }}{\displaystyle \underset{z=\text{−}\sqrt{4-x^2-y^2}}{\overset{z=\sqrt{4-x^2-y^2}}{\int }}dzdydx-}}}{\displaystyle \underset{x=\mathrm{-1}}{\overset{x=1}{\int }}{\displaystyle \underset{y=\text{−}\sqrt{1-x^2}}{\overset{y=\sqrt{1-x^2}}{\int }}{\displaystyle \underset{z=\text{−}\sqrt{4-x^2-y^2}}{\overset{z=\sqrt{4-x^2-y^2}}{\int }}dzdydx}}}.$
Cylindrical: ${\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{r=1}{\overset{r=2}{\int }}{\displaystyle \underset{z=\text{−}\sqrt{4-r^2}}{\overset{z=\sqrt{4-r^2}}{\int }}rdzdrd\theta }}}.$
Spherical: ${\displaystyle \underset{\phi =\pi \text{/}6}{\overset{\phi =5\pi \text{/}6}{\int }}{\displaystyle \underset{\theta =0}{\overset{\theta =2\pi }{\int }}{\displaystyle \underset{\rho =\text{csc}\phi }{\overset{\rho =2}{\int }}{\rho }^2\text{sin}\phi d\rho d\theta d\phi }}}.$

$\frac{9\pi }{8}\text{kg}$

$M_x=\frac{81\pi }{64}$ and $M_y=\frac{81\pi }{64}$

$\stackrel{\text{−}}{x}=\frac{M_y}{m}=\frac{81\pi \text{/}64}{9\pi \text{/}8}=\frac{9}{8}$ and $\stackrel{\text{−}}{y}=\frac{M_x}{m}=\frac{81\pi \text{/}64}{9\pi \text{/}8}=\frac{9}{8}.$

$\stackrel{\text{−}}{x}=\frac{M_y}{m}=\frac{1\text{/}20}{1\text{/}12}=\frac{3}{5}$ and $\stackrel{\text{−}}{y}=\frac{M_x}{m}=\frac{1\text{/}24}{1\text{/}12}=\frac{1}{2}$

$x_c=\frac{M_y}{m}=\frac{1\text{/}15}{1\text{/}6}=\frac{2}{5}\text{and}{y}_c=\frac{M_x}{m}=\frac{1\text{/}12}{1\text{/}6}=\frac{1}{2}$

$I_x={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=0}{\overset{y=x}{\int }}{y}^2\sqrt{xy}dydx}}=\frac{64}{35}$ and $I_y={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=0}{\overset{y=x}{\int }}{x}^2\sqrt{xy}dydx}}=\frac{64}{35}.$ Also, $I_0={\displaystyle \underset{x=0}{\overset{x=2}{\int }}{\displaystyle \underset{y=0}{\overset{y=x}{\int }}\left(x^2+y^2\right)\sqrt{xy}dydx}}=\frac{128}{35}.$

$R_x=\frac{6\sqrt{35}}{35},$ $R_y=\frac{6\sqrt{35}}{35},$ and $R_0=\frac{6\sqrt{70}}{35}.$

$\frac{54}{35}=1.543$

$\left(\frac{3}{2},\frac{9}{8},\frac{1}{2}\right)$

The moments of inertia of the tetrahedron $Q$ about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}$ are $99\text{/}35,36\text{/}7,\text{and}243\text{/}35,$ respectively.

$T^{\mathrm{-1}}(x,y)=(u,v)$ where $u=\frac{3x-y}{3}$ and $v=\frac{y}{3}$

$J\left(u,v\right)=\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}=\left|\begin{array}{lll}\frac{\partial x}{\partial u}\hfill & & \frac{\partial x}{\partial v}\hfill \\ \frac{\partial y}{\partial u}\hfill & & \frac{\partial y}{\partial v}\hfill \end{array}\right|=\left|\begin{array}{lll}1\hfill & & 1\hfill \\ 0\hfill & & 2\hfill \end{array}\right|=2$

${\displaystyle \underset{0}{\overset{\pi \text{/}2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{r}^{3}drd\theta }}$

$x=\frac{1}{2}\left(v+u\right)$ and $y=\frac{1}{2}\left(v-u\right)$ and ${\displaystyle \underset{2}{\overset{4}{\int }}{\displaystyle \underset{\mathrm{-u}}{\overset{u}{\int }}\frac{4}{u^2}\left(\frac{1}{2}\right)}}dvdu.$

$\frac{1}{2}\left(\text{sin}2-2\right)$

${\displaystyle \underset{0}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{1}{\overset{2}{\int }}\left(\frac{v}{3}+\frac{vw}{3u}\right)}}}dudvdw=2+\text{ln}8$

27.

0.

21.3.

a. 28 ${\text{ft}}^3$ b. 1.75 ft.

a. $0.112$ b. $f_{\text{ave}}\simeq 0.175;$ here $f(0.4,0.2)\simeq 0.1,$ $f(0.2,0.6)\simeq \mathrm{-0.2},$ $f(0.8,0.2)\simeq 0.6,$ and $f(0.8,0.6)\simeq 0.2.$

$2\pi .$

40.

$\frac{81}{2}+39\sqrt[3]{2}.$

$e-1.$

$15-\frac{10\sqrt{2}}{9}.$

0.

$(e-1)(1+\text{sin}1-\text{cos}1).$

$\frac{3}{4}\text{ln}\left(\frac{5}{3}\right)+2{\text{ln}}^{2}2-\text{ln}2.$

$\frac{1}{8}\left[\left(2\sqrt{3}-3\right)\pi +6\text{ln}2\right].$

$\frac{1}{4}{e}^4\left(e^4-1\right).$

$4(e-1)\left(2-\sqrt{e}\right).$

$-\frac{\pi }{4}+\text{ln}\left(\frac{5}{4}\right)-\frac{1}{2}\text{ln}2+\text{arctan}2.$

$\frac{1}{2}.$

$\frac{1}{2}\left(2\text{cosh}1+\text{cosh}2-3\right).$

a. $f(x,y)=\frac{1}{2}xy\left(x^2+y^2\right)$ b. $V={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}f(x,y)dxdy}}=\frac{1}{8}$ c. $f_{\text{ave}}=\frac{1}{8};$
d.

a. For $m=n=2,$ $I=4e^{\mathrm{-0.5}}\approx 2.43$ b. $f_{\text{ave}}=e^{\mathrm{-0.5}}\simeq 0.61;$
c.

a. $\frac{2}{n+1}+\frac{1}{4}$ b. $\frac{1}{4}$

$56.5\text{°}$ F; here $f(x_1^{*},y_1^{*})=71,$ $f(x_2^{*},y_1^{*})=72,$ $f(x_1^{*},y_2^{*})=40,$ $f(x_2^{*},y_2^{*})=43,$ where $x_i^{*}$ and $y_j^{*}$ are the midpoints of the subintervals of the partitions of $[a,b]$ and $[c,d],$ respectively.

$\frac{27}{20}$

Type I but not Type II

$\frac{\pi }{2}$

$\frac{1}{6}\left(8+3\pi \right)$

$\frac{1000}{3}$

Type I and Type II

The region $D$ is not of Type I: it does not lie between two vertical lines and the graphs of two continuous functions $g_1(x)$ and $g_2(x).$ The region $D$ is not of Type II: it does not lie between two horizontal lines and the graphs of two continuous functions $h_1(y)$ and $h_2(y).$

$\frac{\pi }{2}$

$0$

$\frac{2}{3}$

$\frac{41}{20}$

$\mathrm{-63}$

$\pi$

a. Answers may vary; b. $\frac{2}{3}$

a. Answers may vary; b. $\frac{7}{3}$

$\frac{8\pi }{3}$

$e-\frac{3}{2}$

$\frac{1}{3}$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{x-1}{\overset{1-x}{\int }}xdydx}}={\displaystyle \underset{\mathrm{-1}}{\overset{0}{\int }}{\displaystyle \underset{0}{\overset{y+1}{\int }}xdxdy}}+{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1-y}{\int }}xdxd}}y=\frac{1}{3}$

$\underset{\text{−1}}{\overset{1}{\int }}\underset{\text{−}\sqrt{1–y^2}}{\overset{\sqrt{1–y^2}}{\int }}ydxdy=\underset{\text{−1}}{\overset{1}{\int }}\underset{\text{−}\sqrt{1–x^2}}{\overset{\sqrt{1–x^2}}{\int }}ydydx=0$

${\displaystyle \underset{D}{\iint }\left(x^2-y^2\right)dA}={\displaystyle \underset{\mathrm{-1}}{\overset{1}{\int }}{\displaystyle \underset{y^4-1}{\overset{1-y^4}{\int }}\left(x^2-y^2\right)dxdy}}=\frac{464}{4095}$

$\frac{4}{5}$

$\frac{5\pi }{32}$

$1$

$2$

a. $\frac{1}{3};$ b. $\frac{1}{6};$ c. $\frac{1}{6}$

a. $\frac{4}{3};$ b. $2\pi ;$ c. $\frac{6\pi -4}{3}$

$0\text{and}0.865474;$ $A\left(D\right)=0.621135$

$P\left[X+Y\le 6\right]=1+\frac{3}{2e^2}-\frac{5}{e^{6\text{/}5}}\approx 0.45;$ there is a $45\text{\%}$ chance that a customer will spend $6$ minutes in the drive-thru line.

$D=\left\{\left(r,\theta \right)|4\le r\le 5,\frac{\pi }{2}\le \theta \le \pi \right\}$

$D=\left\{\left(r,\theta \right)|0\le r\le \sqrt{2},0\le \theta \le \pi \right\}$

$D=\left\{\left(r,\theta \right)|0\le r\le 4\text{sin}\theta ,0\le \theta \le \pi \right\}$

$D=\left\{\left(r,\theta \right)|3\le r\le 5,\frac{\pi }{4}\le \theta \le \frac{\pi }{2}\right\}$

$D=\left\{\left(r,\theta \right)|3\le r\le 5,\frac{3\pi }{4}\le \theta \le \frac{5\pi }{4}\right\}$

$D=\left\{\left(r,\theta \right)|0\le r\le \text{tan}\theta \text{sec}\theta ,0\le \theta \le \frac{\pi }{4}\right\}$

$0$

$\frac{63\pi }{16}$

$\frac{3367\pi }{18}$

$\frac{35{\pi }^2}{576}$

$\frac{7{\mathrm{\pi }}^2}{576}\left(21-e^2+e^4\right)$

$\frac{5}{2}\text{ln}\left(1+\sqrt{2}\right)$