Question

Find the volume of the solid inside of the sphere x^2+y^2+z^2 = 4z and above the cone z = sqrt(3x^2+3y^2)

Answer 1

My question isn't so much on how to solve it, but why you cannot solve it another way.

Answer 2

\[\Large \int\limits _0^{2\pi }\int\limits _0^{\sqrt{3}}\int\limits_{r\sqrt3}^{\sqrt{4-r^2}}dz ~r \:\:dr\:d\theta \:= \frac{14\pi}{3}\]

Answer 3

The issue I have is that the two curves intersect at a level curve (z= 3)

Answer 4

So why can't I have my lower bound as 3 for the first integral?