Question

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

Multivariable functions
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asked 2021-01-13

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

Answers (1)

2021-01-14

Average value of \(\displaystyle{F}=\frac{{1}}{{v}}\int\int\int_{{E}}{F}{\left({x},{y},{z}\right)}{d}{V}\)
Where, \(\displaystyle{d}{V}={\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}{\quad\text{and}\quad}{V}\) be the volume over the given region.
Now, \(x=[0,5],\) \(y=[0,5], z=[0,5]\)
Then find the Volume
\(\displaystyle{V}={\int_{{{z}={0}}}^{{5}}}{\int_{{{y}={0}}}^{{5}}}{\int_{{{x}={0}}}^{{5}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
\(\displaystyle{V}={\left({\int_{{{z}={0}}}^{{5}}}{\left.{d}{x}\right.}\right)}{\left({\int_{{{y}={0}}}^{{5}}}{\left.{d}{y}\right.}\right)}{\left({\int_{{{x}={0}}}^{{5}}}{\left.{d}{z}\right.}\right)}\)
\(\displaystyle{V}={{\left({x}\right)}_{{0}}^{{5}}}{{\left({y}\right)}_{{0}}^{{5}}}{{\left({z}\right)}_{{0}}^{{5}}}\)
\(\displaystyle{V}={5}\times{5}\times{5}\)
\(V=125\)
Now Average value of \(\displaystyle{F}=\frac{{1}}{{V}}\int\int\int_{{E}}{F}{\left({x},{y},{z}\right)}{d}{V}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{\int_{{{y}={0}}}^{{5}}}{\int_{{{x}={0}}}^{{5}}}{\left({x}^{{2}}+{y}^{{2}}+{z}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{\int_{{{y}={0}}}^{{5}}}{{\left(\frac{{x}^{{3}}}{{3}}+{y}^{{2}}{x}+{z}^{{2}}{x}\right)}_{{{x}={0}}}^{{5}}}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{\int_{{{y}={0}}}^{{5}}}{\left(\frac{{125}}{{3}}+{5}{y}^{{2}}+{5}{z}^{{2}}\right)}{\left.{d}{y}\right.}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{{\left(\frac{{125}}{{3}}{y}+\frac{{{5}{y}^{{3}}}}{{3}}+{5}{z}^{{2}}{y}\right)}_{{{x}={0}}}^{{5}}}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{\left(\frac{{125}}{{3}}\times{5}+\frac{{{5}\times{5}^{{3}}}}{{3}}+{25}{z}^{{2}}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\int_{{{z}={0}}}^{{5}}}{\left(\frac{{1250}}{{3}}+{25}{z}^{{2}}\right)}{\left.{d}{z}\right.}\)
\(\displaystyle=\frac{{1}}{{125}}{\left(\frac{{1250}}{{3}}{z}+\frac{{{25}{z}^{{3}}}}{{3}}\right)}\)
\(\displaystyle\frac{{1}}{{125}}{\left(\frac{{6250}}{{3}}+\frac{{3125}}{{3}}\right)}\)
\(\displaystyle=\frac{{3125}}{{125}}\)
\(=25\)
Average value of \(F=25\)

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