# Triple integral spherical

1. Jul 29, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

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

3. 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?

2. Jul 29, 2014

### Matterwave

Looks good to me. :)

3. Jul 29, 2014

### jonroberts74

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

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

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

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

$\frac{\pi}{2}$

4. Jul 29, 2014

### Matterwave

Yep.

5. Jul 29, 2014

### HallsofIvy

Staff Emeritus
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
$$\left(\int_0^{2\pi} d\theta\right)\left(\int_0^\pi sin(\phi) d\phi\right)\left(\int_0^1\rho^7 d\rho\right)$$