# Find the limits for the average triple integral

I am trying to do the following problem: Calculate the average of $f(x,y,z) = x^2 + y^2 + z^2$ over the set of points satisfying $|x| + |y| + |z| ≤ 1$.

So,

$\frac{1}{volume of D}\iiint Fdv$ = $\frac{1}{volume of D}\iiint (x^2 + y^2 + z^2 )dz dy dx$

My problem is how to determine D. So, how do I find the limits of integration.

• @nicomezi: spherical coordinates over a prism could be quite cumbersome. – Martin Argerami Apr 21 '17 at 2:22

To simplify computations, note that the integrand does not vary when the variables change sign. Because both the integrand and the region are symmetric in the three variables, the integral will be the same when restricted to each of the eight octants. Thus we may assume that $x,y,z\geq0$, and then multiply our integral by $8$.
The inequality $x+y+z\leq1$ forces $x\leq1$. So we could let $0\leq x\leq1$. Now we have $$y+z\leq 1-x.$$ So we can let $$0\leq y \leq1-x,$$ and finally $z\leq1-x-y$, i.e. $$0\leq z\leq 1-x-y.$$ Thus \begin{align} \iiint_D(x^2+y^2+z^2)\,dV&=8\int_{0}^1\int_{0}^{1-x}\int_{0}^{1-x-y}(x^2+y^2+z^2)\,dz\,dy\,dx\\ \ \\ &=8\int_0^1\int_0^{1-x}(x^2+y^2)(1-x-y)+\frac13\,(1-x-y)^3\,dy\,dx\\ \ \\ &=8\int_0^1\frac1{12}(x-1)^2(7x^2-2x+1)+\frac1{12}(x-1)^4\,dx\\ \ \\ &=\frac8{20}=\frac25. \end{align} The volume of $D$ is $$8\,\int_0^1\int_0^{1-x} \int_0^{1-x-y} 1 \,dz\, dy\, dx=\frac86=\frac43.$$ So the average is $$\frac{\frac25}{\frac43}=\frac3{10}.$$