Find the volume of the region: $$\iiint(x^2+y^2+z^2)dV $$ where $R$ is the region above the cone $z = a\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2=b^2$.
I am trying to use spherical coordinates to solve this with the bounds of rho as 0 to b and the bounds of theta from 0 to 2 Pi but I am not sure what the bounds of phi would be when the cone and the sphere intersect.
I also know the integral expression should be $\rho^4\sin\phi d\rho d\phi d\theta$
Hint. Where the two surfaces intersect, we have that $(1+a^2)(x^2+y^2)=b^2,$ or $$\rho=\frac{b}{\sqrt{1+a^2}}.$$ This gives the limits for $\rho.$ So the radius doesn't go all the way to $b.$
Thus in cylindrical coordinates the integral simplifies to $$2π\int_0^{b/\sqrt{1+a^2}} \rho^3+\frac{a^3}{3}\rho^4-\frac13\rho(a^2-\rho^2)\sqrt{a^2-\rho^2}\,\mathrm d\rho,$$ which may be evaluated easily. The first part is a polynomial in $\rho.$ For the second part take $\rho=a\sin\psi.$