Find the volume between the cone $y = \sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$. 
Find the volume between the cone $y = \sqrt {x^2 + z^2} $ and the sphere $x^2 + y^2 + z^2 = 49$.

I know that the volume we're interested in is the volume of the intersection between the sphere of radius $7$ and a an upside down cone in the direction of the $y$-axis, but I have no clue on how to set up the bounds of integration. I'm guessing we are supposed to do this in spherical coordinates, but how would we determine the limits of integration?
Thank you.
 A: For given input eliminate $x,z$ etc. and you are left with a circle on the sphere:
$$ x= 7/\sqrt2 ,y=  7/\sqrt2 \,\cos t, z=  7/\sqrt2 \, \sin t  $$
You can use established result Gauss Bonnet thm to advantage, since $k_g , K $ are constant as a differential geometry approach.
$$ k_g= \frac{1}{7},\, s= 2 \pi \frac{7}{\sqrt2},\, \int k_g ds = \frac{\pi}{\sqrt2} $$
Integral curvature (solid angle) 
$$ \int \int K dA = 2 \pi- \frac{ \pi}{\sqrt2}$$
and Volume is solid angle times $R^3/3$
$$= \pi(2 - \frac{ 1}{\sqrt2}) \, \frac{7^3}{3}. $$
A: If you want to do it in spherical coordinates notice $z=\sqrt{x^2+y^2}$ translates to $z=r$ and $x^2+y^2+z^2=49$ translates to $\rho=7$.
If $z=r$ then $\frac{r}{z}=\tan (\phi)=1$ so $y=r$ translates to $\phi=\frac{\pi}{4}$.
Hence what we have is,
$$V=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{7} \rho^2 \sin (\phi) d\rho d\phi d\theta$$

Actually because we have $y=\sqrt{x^2+z^2}$ I would define a modified version of spherical  coordinates. Let $x^2+z^2=r^2$, let $\phi$ be angle with positive $y$ axis of a point to the origin with the restriction that $0 \leq \phi \leq \pi$. Let $\rho$ be the distance from the origin. Let $\theta$ be the counterclockwise angle in the $(x,z)$ plane with the positive $x$ axis. 
Then I would compute the Jacobian and it will turn out to be $\rho^2 \sin (\phi)$. The equation $\frac{r}{y}=\tan (\phi)$ will hold and calculations will be pretty much identical.

Think about why $\frac{r}{z}=\tan (\phi)$. Image from Paul's math.

