# Average value over a multivariable function using triple integrals. Find the average value of

Falak Kinney 2021-01-13 Answered

Average value over a multivariable function using triple integrals. Find the average value of $F\left(x,y,z\right)={x}^{2}+{y}^{2}+{z}^{2}$ over the cube in the first octant bounded bt the coordinate planes and the planes $x=5$, $y=5$, and $z=5$

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## Expert Answer

Arnold Odonnell
Answered 2021-01-14 Author has 109 answers

Average value of $F=\frac{1}{v}\int \int {\int }_{E}F\left(x,y,z\right)dV$
Where, $dV=dxdydz\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}V$ be the volume over the given region.
Now, $x=\left[0,5\right],$ $y=\left[0,5\right],z=\left[0,5\right]$
Then find the Volume
$V={\int }_{z=0}^{5}{\int }_{y=0}^{5}{\int }_{x=0}^{5}dxdydz$
$V=\left({\int }_{z=0}^{5}dx\right)\left({\int }_{y=0}^{5}dy\right)\left({\int }_{x=0}^{5}dz\right)$
$V={\left(x\right)}_{0}^{5}{\left(y\right)}_{0}^{5}{\left(z\right)}_{0}^{5}$
$V=5×5×5$
$V=125$
Now Average value of $F=\frac{1}{V}\int \int {\int }_{E}F\left(x,y,z\right)dV$
$=\frac{1}{125}{\int }_{z=0}^{5}{\int }_{y=0}^{5}{\int }_{x=0}^{5}\left({x}^{2}+{y}^{2}+{z}^{2}\right)dxdydz$
$=\frac{1}{125}{\int }_{z=0}^{5}{\int }_{y=0}^{5}{\left(\frac{{x}^{3}}{3}+{y}^{2}x+{z}^{2}x\right)}_{x=0}^{5}dydz$
$=\frac{1}{125}{\int }_{z=0}^{5}{\int }_{y=0}^{5}\left(\frac{125}{3}+5{y}^{2}+5{z}^{2}\right)dydz$