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Let $F,G,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}H$ be differential functions on $\left[a,b\right],\left[c,d\right],$ and $\left[e,f\right],$ respectively, where $a,b,c,d,e,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}f$ are real numbers such that $a<b,c<d,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}e<f.$ Show that
In the following exercises, evaluate the triple integrals over the bounded region $E=\left\{\left(x,y,z\right)|a\le x\le b,{h}_{1}\left(x\right)\le y\le {h}_{2}\left(x\right),e\le z\le f\right\}.$
$\underset{E}{\iiint}\left(2x+5y+7z\right)}dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 1,0\le y\le -x+1,1\le z\le 2\right\}$
$\frac{77}{12}$
$\underset{E}{\iiint}\left(y\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x+z\right)}dV,$ where $E=\left\{\left(x,y,z\right)|1\le x\le e,0\le y\le \text{ln}\phantom{\rule{0.2em}{0ex}}x,0\le z\le 1\right\}$
$\underset{E}{\iiint}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x+\text{sin}\phantom{\rule{0.2em}{0ex}}y\right)}dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le \frac{\pi}{2},\text{\u2212}\text{cos}\phantom{\rule{0.2em}{0ex}}x\le y\le \text{cos}\phantom{\rule{0.2em}{0ex}}x,\mathrm{-1}\le z\le 1\right\}$
$2$
$\underset{E}{\iiint}\left(xy+yz+xz\right)}dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 1,\text{\u2212}{x}^{2}\le y\le {x}^{2},0\le z\le 1\right\}$
In the following exercises, evaluate the triple integrals over the indicated bounded region $E.$
$\underset{E}{\iiint}\left(x+2yz\right)dV},$ where $E=\left\{(x,y,z)|0\le x\le 1,0\le y\le x,0\le z\le 5-x-y\right\}$
$\frac{439}{120}$
$\underset{E}{\iiint}\left({x}^{3}+{y}^{3}+{z}^{3}\right)}dV,$ where $E=\left\{(x,y,z)|0\le x\le 2,0\le y\le 2x,0\le z\le 4-x-y\right\}$
$\underset{E}{\iiint}y\phantom{\rule{0.2em}{0ex}}dV},$ where $E=\left\{(x,y,z)|-1\le x\le 1,\text{\u2212}\sqrt{1-{x}^{2}}\le y\le \sqrt{1-{x}^{2}},0\le z\le 1-{x}^{2}-{y}^{2}\right\}$
$0$
$\underset{E}{\iiint}x\phantom{\rule{0.2em}{0ex}}dV},$ where $E=\left\{(x,y,z)|-2\le x\le 2,\mathrm{-4}\sqrt{1-{x}^{2}}\le y\le \sqrt{4-{x}^{2}},0\le z\le 4-{x}^{2}-{y}^{2}\right\}$
In the following exercises, evaluate the triple integrals over the bounded region $E$ of the form $E=\left\{(x,y,z)|{g}_{1}\left(y\right)\le x\le {g}_{2}\left(y\right),c\le y\le d,e\le z\le f\right\}.$
$\underset{E}{\iiint}{x}^{2}dV},$ where $E=\left\{(x,y,z)|1-{y}^{2}\le x\le {y}^{2}-1,\mathrm{-1}\le y\le 1,1\le z\le 2\right\}$
$-\frac{64}{105}$
$\underset{E}{\iiint}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x+y\right)}dV,$ where $E=\left\{(x,y,z)|-{y}^{4}\le x\le {y}^{4},0\le y\le 2,0\le z\le 4\right\}$
$\underset{E}{\iiint}\left(x-yz\right)}dV,$ where $E=\left\{(x,y,z)|-{y}^{6}\le x\le \sqrt{y},0\le y\le 1x,\mathrm{-1}\le z\le 1\right\}$
$\frac{11}{26}$
$\underset{E}{\iiint}z}dV,$ where $E=\left\{(x,y,z)|2-2y\le x\le 2+\sqrt{y},0\le y\le 1x,2\le z\le 3\right\}$
In the following exercises, evaluate the triple integrals over the bounded region
$\underset{E}{\iiint}z}dV,$ where $E=\left\{(x,y,z)|-y\le x\le y,0\le y\le 1,0\le z\le 1-{x}^{4}-{y}^{4}\right\}$
$\frac{113}{450}$
$\underset{E}{\iiint}\left(xz+1\right)}dV,$ where $E=\left\{(x,y,z)|0\le x\le \sqrt{y},0\le y\le 2,0\le z\le 1-{x}^{2}-{y}^{2}\right\}$
$\underset{E}{\iiint}\left(x-z\right)}dV,$ where $E=\left\{(x,y,z)|-\sqrt{1-{y}^{2}}\le x\le y,0\le y\le \frac{1}{2}x,0\le z\le 1-{x}^{2}-{y}^{2}\right\}$
$\frac{1}{160}\left(6\sqrt{3}-41\right)$
$\underset{E}{\iiint}\left(x+y\right)}dV,$ where $E=\left\{(x,y,z)|0\le x\le \sqrt{1-{y}^{2}},0\le y\le 1x,0\le z\le 1-x\right\}$
In the following exercises, evaluate the triple integrals over the bounded region
$E=\left\{(x,y,z)|\left(x,y\right)\in D,{u}_{1}\left(x,y\right)x\le z\le {u}_{2}\left(x,y\right)\right\},$ where $D$ is the projection of $E$ onto the $xy$ -plane.
$\underset{D}{\iint}\left({\displaystyle \underset{1}{\overset{2}{\int}}\left(x+z\right)dz}\right)}dA,$ where $D=\left\{(x,y)|{x}^{2}+{y}^{2}\le 1\right\}$
$\frac{3\pi}{2}$
$\underset{D}{\iint}\left({\displaystyle \underset{1}{\overset{3}{\int}}x\left(z+1\right)dz}\right)}dA,$ where $D=\left\{(x,y)|{x}^{2}-{y}^{2}\ge 1,x\le \sqrt{5}\right\}$
$\underset{D}{\iint}\left({\displaystyle \underset{0}{\overset{10-x-y}{\int}}\left(x+2z\right)dz}\right)}dA,$ where $D=\left\{(x,y)|y\ge 0,x\ge 0,x+y\le 10\right\}$
$1250$
$\underset{D}{\iint}\left({\displaystyle \underset{0}{\overset{4{x}^{2}+4{y}^{2}}{\int}}y\phantom{\rule{0.2em}{0ex}}dz}\right)}dA,$ where $D=\left\{(x,y)|{x}^{2}+{y}^{2}\le 4,y\ge 1,x\ge 0\right\}$
The solid $E$ bounded by ${y}^{2}+{z}^{2}=9,z=0,$ and $x=5$ is shown in the following figure. Evaluate the integral $\underset{E}{\iiint}z\phantom{\rule{0.2em}{0ex}}dV$ by integrating first with respect to $z,$ then $y,\phantom{\rule{0.2em}{0ex}}\text{and then}\phantom{\rule{0.2em}{0ex}}x.$
$\underset{0}{\overset{5}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\mathrm{-3}}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{\sqrt{9-{y}^{2}}}{\int}}z\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}}=90$
The solid $E$ bounded by $y=\sqrt{x},$ $x=4,$ $y=0,$ and $z=1$ is given in the following figure. Evaluate the integral $\underset{E}{\iiint}xyz\phantom{\rule{0.2em}{0ex}}dV$ by integrating first with respect to $x,$ then $y,$ and then $z.$
[T] The volume of a solid $E$ is given by the integral $\underset{\mathrm{-2}}{\overset{0}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{x}{\overset{0}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{{x}^{2}+{y}^{2}}{\int}}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}}.$ Use a computer algebra system (CAS) to graph $E$ and find its volume. Round your answer to two decimal places.
$V=5.33$
[T] The volume of a solid $E$ is given by the integral $\underset{\mathrm{-1}}{\overset{0}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\text{\u2212}{x}^{2}}{\overset{0}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1+\sqrt{{x}^{2}+{y}^{2}}}{\int}}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}}}.$ Use a CAS to graph $E$ and find its volume $V.$ Round your answer to two decimal places.
In the following exercises, use two circular permutations of the variables $x,y,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z$ to write new integrals whose values equal the value of the original integral. A circular permutation of $x,y,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}z$ is the arrangement of the numbers in one of the following orders: $y,z,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}z,x,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y.$
$\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{2}{\overset{4}{\int}}\left({x}^{2}{z}^{2}+1\right)}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
$\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{2}{\overset{4}{\int}}\left({y}^{2}{z}^{2}+1\right)}}}dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy;$ $\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{2}{\overset{4}{\int}}\left({x}^{2}{y}^{2}+1\right)}}}dy\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dx$
$\underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{\text{\u2212}x+1}{\int}}\left(2x+5y+7z\right)}}}dy\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz$
$\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\text{\u2212}y}{\overset{y}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1-{x}^{4}-{y}^{4}}{\int}}\text{ln}\phantom{\rule{0.2em}{0ex}}x}}}\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy$
$\underset{\mathrm{-1}}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\text{\u2212}{y}^{6}}{\overset{\sqrt{y}}{\int}}\left(x+yz\right)}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
Set up the integral that gives the volume of the solid $E$ bounded by ${y}^{2}={x}^{2}+{z}^{2}$ and $y={a}^{2},$ where $a>0.$
$V={\displaystyle \underset{\text{\u2212}a}{\overset{a}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\text{\u2212}\sqrt{{a}^{2}-{z}^{2}}}{\overset{\sqrt{{a}^{2}-{z}^{2}}}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\sqrt{{x}^{2}+{z}^{2}}}{\overset{{a}^{2}}{\int}}dy\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz}}}$
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