# Triple integral spherical

## Homework Statement

##\iiint_W (x^2+y^2+z^2)^{5/2}## W is the ball ##x^2+y^2+z^2 \le 1##

## The Attempt at a Solution

changing to spherical

##0 \le \theta \le 2\pi ; 0 \le \phi \le \pi ; 0 \le \rho \le 1##

##(x^2 + y^2 + z^2)^{5/2} \Rightarrow ((\rho \sin \phi \cos \theta)^2 + (\rho sin \phi \sin \theta)^2 + (\rho \cos \phi)^2)^{5/2} = \rho^5##

##\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^5(\rho^2\sin\phi) d\rho d\phi d\theta##

Is that a correct setup?

Matterwave
Gold Member
Looks good to me. :)

1 person
##\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^5(\rho^2\sin\phi) d\rho d\phi d\theta##

##\int_{0}^{2\pi}\int_{0}^{\pi}\rho^7\sin\phi d\rho d\phi d\theta##

##\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{8}\sin\phi d\phi d\theta##

##\int_{0}^{2\pi} \frac{1}{4} d\theta##

##\frac{\pi}{2}##

Matterwave
Gold Member
Yep.

HallsofIvy
Homework Helper
##\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}\rho^5(\rho^2\sin\phi) d\rho d\phi d\theta##

##\int_{0}^{2\pi}\int_{0}^{\pi}\rho^7\sin\phi d\rho d\phi d\theta##

##\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{8}\sin\phi d\phi d\theta##

##\int_{0}^{2\pi} \frac{1}{4} d\theta##

##\frac{\pi}{2}##
Very good. Also note that because the limits of integration on each integral are constants, not depending on the other variables, this is the same as
$$\left(\int_0^{2\pi} d\theta\right)\left(\int_0^\pi sin(\phi) d\phi\right)\left(\int_0^1\rho^7 d\rho\right)$$