How Do You Set Up a Triple Integral in Spherical Coordinates for a Unit Ball?

[jonroberts74]

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]

Looks good to me. :)

 

[jonroberts74]

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


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

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

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

$\frac{\pi}{2}$

 

[Matterwave]

Yep.

 

[HallsofIvy]

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


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

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

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

$\frac{\pi}{2}$

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)