What values of $\rho$ can give the answer? $V=\int_0^{2π}\int_0^{π/2}\int_{?}^{?}\rho^2\sin\phi d\rho d\phi d\theta$

Question

Find the Volume of the region bounded above by sphere $x^2 + y^2 + z^2 = 2a^2$, and below by the paraboloid $az = x^2 + y^2$?

$$V=\int_0^{2π}\int_0^{π/2}\int_{?}^{?}\rho^2\sin\phi d\rho d\phi d\theta$$ I tried This integral setup but I don't know how to find the values of $\rho$?

Answer 1

Easier to use xyz coordinates. Solving the two equations shows that the solids intersect at z = a. Hence the volume is

$$\int_0^a \pi az dz + \int_a^{\sqrt{2}a} \pi (2a^2 -z^2)dz$$

$$= \frac{\pi a^3}{6}(8\sqrt{2} - 7)$$