Q: Spherical Coordinates Let Q be the region above by the plane $8z=4-x-y$ and below by the cone $64z^2=x^2+y^2$. How would I setup the triple integral to find volume of Q, using spherical coordinates? I just need help with the setup part, I will do the rest.

A: This is not going to be a nice integral, but if you insist :-), here's how to set it up:
With
$$
\begin{align}
x&=r\sin\theta\cos\phi\;,\\
y&=r\sin\theta\sin\phi\;,\\
z&=r\cos\theta\;,
\end{align}
$$
the equation of the cone becomes
$$
64r^2\cos^2\theta=r^2\sin^2\theta\,(\cos^2\phi+\sin^2\phi)
$$
and thus
$$\theta=\arccos\frac1{\sqrt{65}}\;.$$
The equation of the plane becomes
$$
8r\cos\theta=4-r\sin\theta\cos\phi-r\sin\theta\sin\phi\;,
$$
and solving for $r$ yields
$$
r=\frac4{8\cos\theta+\sin\theta(\cos\phi+\sin\phi)}\;.
$$
Thus the volume is given by the integral
$$
\int_0^{2\pi}\mathrm d\phi\int_0^{\arccos1/\sqrt{65}}\mathrm d\theta\sin\theta\int_0^{4/(8\cos\theta+\sin\theta(\cos\phi+\sin\phi))}\mathrm drr^2\;.
$$
