Using spherical coordinates to find volume of a region Use spherical coordinates to find the volume of the region lying above $z = \sqrt{3x^2+3y^2}$ and within the $x^2+y^2+z^2=2az$, $a>0$.
So far I know that the first graph is a cone and the second one is some kind of sphere. I have completed the square so that the new equation is: $$x^2+y^2+(z-a)^2=a^2$$
I know how to convert to spherical coordinates but the $a$ is throwing me a bit.
 A: So its nice that you have changed one equation to simpler coordinates but you haven't changed the other. So let's change them both by
$$\begin{array}{ccc} \tilde{x} = x & \tilde{y}=y & \tilde{z}=z-a \end{array}$$
This yields the bounds $$\begin{array}{cc}\tilde{x}^2 + \tilde{y}^2 + \tilde{z}^2 = a^2 & \tilde{z} + a = \sqrt{3\tilde{x}^2 + 3 \tilde{y}^2} \end{array}$$
Proceed from here by using cylindrical coordinates making the equations,
$$\begin{array}{cc} r^2 + \tilde{z}^2 = a^2 & \tilde{z} + a = \sqrt{3}r\end{array}$$
Recall the Jacbian for cylindrical coordinates is: $\text{d}V = r \ \text{d}r \text{d}\theta \text{d}\tilde{z}$
A: First use cylindrical coordinates to understand what you are dealing with. You have $z>\sqrt 3 |r|$ and $r^2+(z-a)^2<a^2$ I have drawn this on a paper:

You need to know a little geometry. We see that the volume we need is a sphere cap and a cone. If you have difficulty finding the heights ($3/2a$ and $a/2$) or the radius of the sphere cap ($a\sqrt 3/2$) you can ask about it.
Another picture:

First find the volume of the sphere "slice" (starting from the center and including the sphere cap) (blue+green in the second image). $\phi$ integral is omitted since clindrical symmetry is present. It is multiplied by $2\pi$
$$2\pi \int ^{\pi/3}_0 \int ^a_0 dr\,d\theta$$
Now we need to subtract the little cone between the sphere cap and the center. Then add the big cone connecting the bottom of the sphere and the sphere cap. (red+green - green) I am using the ordinary cone volume formula.
The answer should come up to be (after quite a bit of simplification):
$$\frac{7 \pi  a^3}{12}$$
If you have trouble with one part in particular I can explain it.
