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.

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 $R^3/3$
$$= \pi(2 - \frac{ 1}{\sqrt2}) \, \frac{7^3}{3}. $$
