5.4 Triple integrals  (Page 6/8)

Let $F,G,\text{and}H$ be differential functions on $\left[a,b\right],\left[c,d\right],$ and $\left[e,f\right],$ respectively, where $a,b,c,d,e,\text{and}f$ are real numbers such that $a<b,c<d,\text{and}e<f.$ Show that

${\displaystyle \underset{E}{\iiint }\left(2x+5y+7z\right)}dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le 1,0\le y\le -x+1,1\le z\le 2\right\}$

${\displaystyle \underset{E}{\iiint }\left(y\text{ln}x+z\right)}dV,$ where $E=\left\{\left(x,y,z\right)|1\le x\le e,0\le y\le \text{ln}x,0\le z\le 1\right\}$

${\displaystyle \underset{E}{\iiint }\left(\text{sin}x+\text{sin}y\right)}dV,$ where $E=\left\{\left(x,y,z\right)|0\le x\le \frac{\pi }{2},\text{−}\text{cos}x\le y\le \text{cos}x,\mathrm{-1}\le z\le 1\right\}$

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

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

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

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

${\displaystyle \underset{E}{\iiint }xdV},$ where $E=\left\{(x,y,z)|-2\le x\le 2,\mathrm{-4}\sqrt{1-x^2}\le y\le \sqrt{4-x^2},0\le z\le 4-x^2-y^2\right\}$

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

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

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

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

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

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

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

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

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

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

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

${\displaystyle \underset{D}{\iint }\left({\displaystyle \underset{0}{\overset{4x^2+4y^2}{\int }}ydz}\right)}dA,$ 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 ${\displaystyle \underset{E}{\iiint }zdV}$ by integrating first with respect to $z,$ then $y,\text{and then}x.$

The solid $E$ bounded by $y=\sqrt{x},$ $x=4,$ $y=0,$ and $z=1$ is given in the following figure. Evaluate the integral ${\displaystyle \underset{E}{\iiint }xyzdV}$ by integrating first with respect to $x,$ then $y,$ and then $z.$

[T] The volume of a solid $E$ is given by the integral ${\displaystyle \underset{\mathrm{-2}}{\overset{0}{\int }}{\displaystyle \underset{x}{\overset{0}{\int }}{\displaystyle \underset{0}{\overset{x^2+y^2}{\int }}dzdydx}}}.$ Use a computer algebra system (CAS) to graph $E$ and find its volume. Round your answer to two decimal places.

[T] The volume of a solid $E$ is given by the integral ${\displaystyle \underset{\mathrm{-1}}{\overset{0}{\int }}{\displaystyle \underset{\text{−}{x}^2}{\overset{0}{\int }}{\displaystyle \underset{0}{\overset{1+\sqrt{x^2+y^2}}{\int }}dzdydx}}}.$ Use a CAS to graph $E$ and find its volume $V.$ Round your answer to two decimal places.

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{1}{\overset{3}{\int }}{\displaystyle \underset{2}{\overset{4}{\int }}\left(x^2{z}^2+1\right)}}}dxdydz$

${\displaystyle \underset{1}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{\text{−}x+1}{\int }}\left(2x+5y+7z\right)}}}dydxdz$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{\text{−}y}{\overset{y}{\int }}{\displaystyle \underset{0}{\overset{1-x^4-y^4}{\int }}\text{ln}x}}}dzdxdy$

${\displaystyle \underset{\mathrm{-1}}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{\text{−}{y}^6}{\overset{\sqrt{y}}{\int }}\left(x+yz\right)}}}dxdydz$

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 $a>0.$