# 5.4 Triple integrals  (Page 6/8)

 Page 6 / 8

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 Show that

$\underset{a}{\overset{b}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{c}{\overset{d}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{e}{\overset{f}{\int }}{F}^{\prime }\left(x\right){G}^{\prime }\left(y\right){H}^{\prime }\left(z\right)dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\left[F\left(b\right)-F\left(a\right)\right]\phantom{\rule{0.2em}{0ex}}\left[G\left(d\right)-G\left(c\right)\right]\phantom{\rule{0.2em}{0ex}}\left[H\left(f\right)-H\left(e\right)\right].$

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 $E=\left\{\left(x,y,z\right)|0\le x\le \frac{\pi }{2},\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}x\le y\le \text{cos}\phantom{\rule{0.2em}{0ex}}x,-1\le z\le 1\right\}$

$2$

$\underset{E}{\iiint }\left(xy+yz+xz\right)dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 1,\text{−}{x}^{2}\le y\le {x}^{2},0\le z\le 1\right\}$

In the following exercises, evaluate the triple integrals over the indicated bounded region $E.$

$\underset{E}{\iiint }\left(x+2yz\right)dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 1,0\le y\le x,0\le z\le 5-x-y\right\}$

$\frac{439}{120}$

$\underset{E}{\iiint }\left({x}^{3}+{y}^{3}+{z}^{3}\right)dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 2,0\le y\le 2x,0\le z\le 4-x-y\right\}$

$\underset{E}{\iiint }y\phantom{\rule{0.2em}{0ex}}dV,$ where $E=\left\{\left(x,y,z\right)|-1\le x\le 1,\text{−}\sqrt{1-{x}^{2}}\le y\le \sqrt{1-{x}^{2}},0\le z\le 1-{x}^{2}-{y}^{2}\right\}$

$0$

$\underset{E}{\iiint }x\phantom{\rule{0.2em}{0ex}}dV,$ where $E=\left\{\left(x,y,z\right)|-2\le x\le 2,-4\sqrt{1-{x}^{2}}\le y\le \sqrt{4-{x}^{2}},0\le z\le 4-{x}^{2}-{y}^{2}\right\}$

In the following exercises, evaluate the triple integrals over the bounded region $E$ of the form $E=\left\{\left(x,y,z\right)|{g}_{1}\left(y\right)\le x\le {g}_{2}\left(y\right),c\le y\le d,e\le z\le f\right\}.$

$\underset{E}{\iiint }{x}^{2}dV,$ where $E=\left\{\left(x,y,z\right)|1-{y}^{2}\le x\le {y}^{2}-1,-1\le y\le 1,1\le z\le 2\right\}$

$-\frac{64}{105}$

$\underset{E}{\iiint }\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x+y\right)dV,$ where $E=\left\{\left(x,y,z\right)|-{y}^{4}\le x\le {y}^{4},0\le y\le 2,0\le z\le 4\right\}$

$\underset{E}{\iiint }\left(x-yz\right)dV,$ where $E=\left\{\left(x,y,z\right)|-{y}^{6}\le x\le \sqrt{y},0\le y\le 1x,-1\le z\le 1\right\}$

$\frac{11}{26}$

$\underset{E}{\iiint }zdV,$ where $E=\left\{\left(x,y,z\right)|2-2y\le x\le 2+\sqrt{y},0\le y\le 1x,2\le z\le 3\right\}$

In the following exercises, evaluate the triple integrals over the bounded region

$E=\left\{\left(x,y,z\right)|{g}_{1}\left(y\right)\le x\le {g}_{2}\left(y\right),c\le y\le d,{u}_{1}\left(x,y\right)\le z\le {u}_{2}\left(x,y\right)\right\}.$

$\underset{E}{\iiint }zdV,$ where $E=\left\{\left(x,y,z\right)|-y\le x\le y,0\le y\le 1,0\le z\le 1-{x}^{4}-{y}^{4}\right\}$

$\frac{113}{450}$

$\underset{E}{\iiint }\left(xz+1\right)dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le \sqrt{y},0\le y\le 2,0\le z\le 1-{x}^{2}-{y}^{2}\right\}$

$\underset{E}{\iiint }\left(x-z\right)dV,$ where $E=\left\{\left(x,y,z\right)|-\sqrt{1-{y}^{2}}\le x\le y,0\le y\le \frac{1}{2}x,0\le z\le 1-{x}^{2}-{y}^{2}\right\}$

$\frac{1}{160}\left(6\sqrt{3}-41\right)$

$\underset{E}{\iiint }\left(x+y\right)dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le \sqrt{1-{y}^{2}},0\le y\le 1x,0\le z\le 1-x\right\}$

In the following exercises, evaluate the triple integrals over the bounded region

$E=\left\{\left(x,y,z\right)|\left(x,y\right)\in D,{u}_{1}\left(x,y\right)x\le z\le {u}_{2}\left(x,y\right)\right\},$ where $D$ is the projection of $E$ onto the $xy$ -plane.

$\underset{D}{\iint }\left(\underset{1}{\overset{2}{\int }}\left(x+z\right)dz\right)dA,$ where $D=\left\{\left(x,y\right)|{x}^{2}+{y}^{2}\le 1\right\}$

$\frac{3\pi }{2}$

$\underset{D}{\iint }\left(\underset{1}{\overset{3}{\int }}x\left(z+1\right)dz\right)dA,$ where $D=\left\{\left(x,y\right)|{x}^{2}-{y}^{2}\ge 1,x\le \sqrt{5}\right\}$

$\underset{D}{\iint }\left(\underset{0}{\overset{10-x-y}{\int }}\left(x+2z\right)dz\right)dA,$ where $D=\left\{\left(x,y\right)|y\ge 0,x\ge 0,x+y\le 10\right\}$

$1250$

$\underset{D}{\iint }\left(\underset{0}{\overset{4{x}^{2}+4{y}^{2}}{\int }}y\phantom{\rule{0.2em}{0ex}}dz\right)dA,$ where $D=\left\{\left(x,y\right)|{x}^{2}+{y}^{2}\le 4,y\ge 1,x\ge 0\right\}$

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 $y,\phantom{\rule{0.2em}{0ex}}\text{and then}\phantom{\rule{0.2em}{0ex}}x.$

$\underset{0}{\overset{5}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{-3}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\sqrt{9-{y}^{2}}}{\int }}z\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=90$

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 $\underset{-2}{\overset{0}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{x}{\overset{0}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{{x}^{2}+{y}^{2}}{\int }}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx.$ 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 $\underset{-1}{\overset{0}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{\text{−}{x}^{2}}{\overset{0}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{1+\sqrt{{x}^{2}+{y}^{2}}}{\int }}dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx.$ 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: $y,z,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}z,x,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y.$

$\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{2}{\overset{4}{\int }}\left({x}^{2}{z}^{2}+1\right)dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$

$\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{2}{\overset{4}{\int }}\left({y}^{2}{z}^{2}+1\right)dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy;$ $\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{1}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{2}{\overset{4}{\int }}\left({x}^{2}{y}^{2}+1\right)dy\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dx$

$\underset{1}{\overset{3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\text{−}x+1}{\int }}\left(2x+5y+7z\right)dy\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz$

$\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{\text{−}y}{\overset{y}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{1-{x}^{4}-{y}^{4}}{\int }}\text{ln}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy$

$\underset{-1}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{\text{−}{y}^{6}}{\overset{\sqrt{y}}{\int }}\left(x+yz\right)dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$

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 $a>0.$

$V=\underset{\text{−}a}{\overset{a}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{\text{−}\sqrt{{a}^{2}-{z}^{2}}}{\overset{\sqrt{{a}^{2}-{z}^{2}}}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{\sqrt{{x}^{2}+{z}^{2}}}{\overset{{a}^{2}}{\int }}dy\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dz$

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