Question

Find the volume of the region that lies above the cone z= sqrt(x^2+y^2) and within the unit sphere x^2+y^2+z^2= 1 using spherical coordinates

Answer 1

The region of integration is a cone in spherical coordinates where radius changes from 0 to 1,

azimuthal angle $\phi$ changes from 0 to $2\pi$ and polar angle $\theta$ changes from 0 to$\frac{\pi}{4}$ .

$V=\iiint_{V}{dV}=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{1}r^2\sin{\theta} dr d\theta d\phi=$

$=\int_{0}^{1}r^2dr \int_{0}^{\frac{\pi}{4}} \sin\theta d\theta \int_{0}^{2\pi}d\phi$ =

$=\frac{r^3}{3}|_{0}^{1}* (-\cos{\theta})|_{0}^{\frac{\pi}{4}}*\phi|_{0}^{2\pi}=$

$=\frac{1}{3}(-\frac{\sqrt{2}}{2}-(-1))*2\pi=$

$=\frac{\pi}{3}(2-\sqrt{2})$