Average value for triple integrals

Formula for average value over an object

To find the average value of a function over some object $E$, we’ll use the formula

$f_{avg}=\frac{1}{V(E)}\int\int\int_Ef(x,y,z)\ dV$

where $V(E)$ is the volume of the object $E$.

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In order to use the formula, we’ll have find the volume of the object, plus the domain of $x$, $y$, and $z$ so that we can set limits of integration, turn the triple integral into an iterated integral, and replace $dV$ with $dz\ dy\ dx$.

How to calculate average value using a triple integral

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Average value over a cube

Example

Find the average value of the function over a cube with side length $2$, lying in the first octant with one corner at the origin $(0,0,0)$ and three sides lying in the coordinate planes.

$f(x,y,z)=3xyz^2$

We’ll start by finding the volume of the cube. Since we’re dealing with a cube with side length $2$, the volume will be

$V(E)=(2)(2)(2)$

$V(E)=8$

To find the limits of integration, we have to look at the object we’ve been given. In this case, it’s a cube whose corner is sitting at $(0,0,0)$ on the origin. Since the cube has side length $2$, the limits of integration are $x=[0,2]$, $y=[0,2]$ and $z=[0,2]$.

Plugging everything we’ve found into the triple integral formula for average value, including the function itself, we get

$f_{avg}=\frac18\int^2_0\int^2_0\int^2_03xyz^2\ dz\ dy\ dx$

$f_{avg}=\frac38\int^2_0\int^2_0\int^2_0xyz^2\ dz\ dy\ dx$

In order to use the formula, we’ll have find the volume of the object, plus the domain of x, y, and z.

Integrating with respect to $z$, we get

$f_{avg}=\frac38\int^2_0\int^2_0\frac13xyz^3\Big|_{z=0}^{z=2}\ dy\ dx$

$f_{avg}=\frac18\int^2_0\int^2_0xyz^3\Big|_{z=0}^{z=2}\ dy\ dx$

$f_{avg}=\frac18\int^2_0\int^2_0xy(2)^3-xy(0)^3\ dy\ dx$

$f_{avg}=\frac18\int^2_0\int^2_08xy\ dy\ dx$

Now we’ll integrate with respect to $y$.

$f_{avg}=\frac18\int^2_08\left(\frac12\right)xy^2\Big|_{y=0}^{y=2}\ dx$

$f_{avg}=\frac12\int^2_0xy^2\Big|_{y=0}^{y=2}\ dx$

$f_{avg}=\frac12\int^2_0x(2)^2-x(0)^2\ dx$

$f_{avg}=\frac12\int^2_04x\ dx$

Finally we’ll integrate with respect to $x$.

$f_{avg}=\frac12\left(\frac42x^2\right)\Big|_0^2$

$f_{avg}=x^2\Big|_0^2$

$f_{avg}=(2)^2-(0)^2$

$f_{avg}=4$

The average value of the function $f(x,y,z)=3xyz^2$ over the cube $E$ is $4$.

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