Question

A solid is bounded below by the cone, z=x2+y2, and bounded above by the sphere of radius 2 centered at the origin. Find integrals that compute its volume using Cartesian and cylindrical coordinates. For your answers use θ= theta.

Answer 1

The cone $z=\sqrt{x^2+y^2}$ and the sphere $z=\sqrt{4-x^2-y^2}$ intersect in a circle of radius $\sqrt 2$ in the plane $z=\sqrt2$:

$\sqrt{x^2+y^2}=\sqrt{4-x^2-y^2}\implies 2x^2+2y^2=4\implies x^2+y^2=2$

$\implies z=\sqrt{x^2+y^2}=\sqrt2$

In Cartesian coordinates, the volume is then given by the integral

$\displaystyle\int_{-\sqrt2}^{\sqrt2}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

In cylindrical coordinates, the integral is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt2}\int_r^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$