Originally Posted by

**HallsofIvy** I presume you mean $\displaystyle \rho$ rather than "e"!

Certainly you should have no trouble putting $\displaystyle 5(x^2+ y^2+z^2)$ and $\displaystyle x^2+ y^2+ z^2= a^2$ into spherical coordinates ($\displaystyle x^2+ y^2+ z^2= a^2$ **is** a sphere with radius a. So I assume the problem is $\displaystyle 3z^2= x^2+ y^2$. It's not really hard to put the formulas for x, y, and z, in terms of $\displaystyle \rho$, $\displaystyle \theta$, and $\displaystyle \phi$ and reduce but I think I would be inclined to add a $\displaystyle z^2$ to each side to get $\displaystyle 4z^2= x^2+ y^2+ z^2$. Now the left side is $\displaystyle 4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi)$ and the right side is, as I said before, very easy.