Originally Posted by

**Dwill90** Hmm. The only difference between your integral and the one I had was that I had $\displaystyle p^6$ as a factor, but you had $\displaystyle p^5$. So I tracked the ps in this manner:

$\displaystyle z^3$ becomes $\displaystyle p^3 * cos^3(phi)$;

$\displaystyle sqrt(x^2 + y^2 + z^2)$ becomes $\displaystyle p$;

and the jacobian contributes $\displaystyle p^2 * sin(phi)$.

$\displaystyle p^3 * p * p^2 = p^6$. Have I made an error?

Using the integral you've provided, I evaluated it fully out, leaving the phi integral for last, and obtained:

$\displaystyle pi/6 * integral(0, pi, cos^3(phi) * sin(phi), dphi)$

After using a u-substitution of u = cos(phi), I have the integral

$\displaystyle integral(1, -1, u^3, du)$

Which then evaluates to 0. I guess the answer really is zero...