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???.
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???
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???.