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Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.
If $f\left(x,y,z\right)$ is integrable over a solid bounded region $E$ with positive volume $V\left(E\right),$ then the average value of the function is
Note that the volume is $V\left(E\right)={\displaystyle \underset{E}{\iiint}1dV}.$
The temperature at a point $\left(x,y,z\right)$ of a solid $E$ bounded by the coordinate planes and the plane $x+y+z=1$ is $T(x,y,z)=(xy+8z+20)\text{\xb0}\text{C}\text{.}$ Find the average temperature over the solid.
Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane $x+y+z=1$ has intercepts $\left(1,0,0\right),\left(0,1,0\right),$ and $\left(0,0,1\right).$ The region $E$ looks like
Hence the triple integral of the temperature is
The volume evaluation is $V\left(E\right)={\displaystyle \underset{E}{\iiint}1}dV={\displaystyle \underset{x=0}{\overset{x=1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{y=0}{\overset{y=1-x}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{z=0}{\overset{z=1-x-y}{\int}}1dz\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{6}}}}.$
Hence the average value is ${T}_{\text{ave}}=\frac{147\text{/}40}{1\text{/}6}=\frac{6\left(147\right)}{40}=\frac{441}{20}$ degrees Celsius.
Find the average value of the function $f\left(x,y,z\right)=xyz$ over the cube with sides of length $4$ units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.
${f}_{\text{ave}}=8$
In the following exercises, evaluate the triple integrals over the rectangular solid box $B.$
$\underset{B}{\iiint}\left(2x+3{y}^{2}+4{z}^{3}\right)}dV,$ where $B=\left\{\left(x,y,z\right)|0\le x\le 1,0\le y\le 2,0\le z\le 3\right\}$
$192$
$\underset{B}{\iiint}\left(xy+yz+xz\right)}dV,$ where $B=\left\{\left(x,y,z\right)|1\le x\le 2,0\le y\le 2,1\le z\le 3\right\}$
$\underset{B}{\iiint}\left(x\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}y+z\right)}dV,$ where $B=\left\{\left(x,y,z\right)|0\le x\le 1,0\le y\le \pi ,\mathrm{-1}\le z\le 1\right\}$
$0$
$\underset{B}{\iiint}\left(z\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}x+{y}^{2}\right)}dV,$ where $B=\left\{\left(x,y,z\right)|0\le x\le \pi ,0\le y\le 1,\mathrm{-1}\le z\le 2\right\}$
In the following exercises, change the order of integration by integrating first with respect to $z,$ then $x,$ then $y.$
$\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{1}{\overset{2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{2}{\overset{3}{\int}}\left({x}^{2}+\text{ln}\phantom{\rule{0.2em}{0ex}}y+z\right)}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
$\underset{1}{\overset{2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{2}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1}{\int}}\left({x}^{2}+\text{ln}\phantom{\rule{0.2em}{0ex}}y+z\right)}}}dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy=\frac{35}{6}+2\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}2$
$\underset{0}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\mathrm{-1}}{\overset{1}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{3}{\int}}\left(z{e}^{x}+2y\right)}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
$\underset{\mathrm{-1}}{\overset{2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{4}{\int}}\left({x}^{2}z+\frac{1}{y}\right)}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
$\underset{1}{\overset{3}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{4}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\mathrm{-1}}{\overset{2}{\int}}\left({x}^{2}z+\frac{1}{y}\right)}}}dz\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy=64+12\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}3$
$\underset{1}{\overset{2}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{\mathrm{-2}}{\overset{\mathrm{-1}}{\int}}\phantom{\rule{0.2em}{0ex}}{\displaystyle \underset{0}{\overset{1}{\int}}\frac{x+y}{z}}}}dx\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dz$
Let $F,G,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}H$ be continuous 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
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