# Volume of Cone and Sphere Question

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 $$\rho=a\sin\psi.$$
• So would the answer be $(-2πb^5)* a/\sqrt{a^2+1}-1/5$ – FreshDough Apr 8 at 18:22