Verify process Integral $\iiint_B (x^2+y^2+z^2)^2dV $

Question

I want to know if the integral that I solve it's correct.

$$\iiint_B (x^2+y^2+z^2)^2dV$$ where $B=(x,y,z) \in \mathbb{R}^3 : x^2+y^2+z^2 \leq 1, z \leq 0 $.

So

$x^2+y^2+z^2 =\rho ^2$

$$v=\iiint_B \rho ^2 sin\phi d \rho d \theta d \phi $$

Limits:

$$x^2+y^2+z^2 \leq 1$$

$$\rho ^2 \leq 1 $$

$$\rho \leq \sqrt {1} $$

$$\rho \leq 1 -> 0 \in \rho \leq 1 $$

$$0 \leq \theta \leq 1 $$ Move around axes x,y

$$z \leq 0 -> 0 \leq \phi \leq \frac{\pi}{2} $$

Spherical coordinates

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

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{2}}\int_{0}^{1} \phi ^6 sin\phi d \rho d \theta d \phi $$

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{2}} sin \phi \frac{\rho ^7}{7} |_{0}^{1} d \theta d \phi $$

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{2}} sin \phi \frac{1}{7} d \theta d \phi $$

$$\int_{0}^{\frac{\pi}{2}} \frac{1}{7} \theta |_{0}^{\frac{\pi}{2}} sen \phi d \phi $$

$$\int_{0}^{\frac{\pi}{2}} \frac{2\pi}{7} sen \phi d \phi $$

$$= \frac{2\pi}{7}[-Cos\phi]_{0}^{\frac{\pi}{2}} $$

$$=\frac{\pi}{2} $$

Answer 1

It is much easier to perform integration along shells. We want to integrate $(x^2+y^2+z^2)^2$ over the lower unit hemisphere centered at the origin. The surface area of a sphere with radius $r$ is $4\pi r^2$, hence the given integral equals

$$ \int_{0}^{1}2\pi r^2\cdot r^4 \,dr = \frac{2\pi}{7} $$ nice and easy.

Answer 2

The lower unit ball is characterized by $\phi \in [\frac{\pi}{2},\pi]$,$\rho \in [0,1]$, and $\theta \in [0,2\pi]$ in spherical coordinates. Hence the triple integral $\iiint_{B} (x^2+y^2+z^2)^2 dV$ is equal to,

$$\int_{\frac{\pi}{2}}^{\pi} \int_{0}^{2\pi} \int_{0}^{1} \rho^{6} \sin \phi d\rho d\theta d\phi$$

$$=\int_{0}^{1} \rho^6 d\rho \int_{0}^{2\pi} d\theta \int_{\frac{\pi}{2}}^{\pi} \sin \phi d\phi$$

$$=\frac{1}{7}(2\pi)(1)$$

$$=\frac{2\pi}{7}$$