Question

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

Multivariable functions

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 $$z=5$$

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 $$F=25$$