# Solve the triple integral $\iiint_D (x^2 + y^2 + z^2)\, dxdydz$

How does one go about solving the integral: $$\iiint_D (x^2 + y^2 + z^2)\, dxdydz,$$ where $$D=\{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \le 9\}.$$ I believe I am supposed to convert to spherical coordinates but I would need some help with how this is done and what the answer to this integral would be.

• Well, what is this in spherical coordinates? What does that domain describe? What is e.g. $x^2 +y^2 \leq 9$? – M.B. Oct 15 '13 at 20:10
• There is a hint saying that: x^2+y^2+z^2 = r^2 and dxdydz = r^2 sin t dr dtheta dphi, but I am having difficulties visionalizing it. – Soakr Oct 15 '13 at 20:39

A quick way to evaluate it is to note that the volume of the spherical shell from radius $r$ to radius $r + \Delta r$ is approximately $4\pi r^2 \Delta r$, so your result should be $$\int_0^3 r^2 (4\pi r^2) \,dr$$ $$= {4 \over 5} \pi r^5\bigg|_{r=0}^3$$ $$= {4 \over 5} 3^5 \pi$$ $$={972 \pi \over 5}$$ To do it properly you should do spherical coordinates like mathematics2x2life is trying to do.

The easiest way to do this is to make a switch to spherical coordinates. There $\rho^2=x^2+y^2+z^2$ and $dxdydz=\rho^2 \sin \phi \,\,d\rho d\theta d\varphi$. So we have $$\iiint_Dx^2+y^2+z^2 dxdydz=\iiint_D \rho^2 \cdot \rho^2 \sin \varphi\,d\rho\theta d\varphi$$ Now we are integrating over a region $D$. What is $D$? It is a sphere of radius $3$ centered at the origin. So $0\leq \rho \leq 3$, $0\leq \theta\leq 2\pi$, and $0\leq \varphi \leq \pi$.

$$\iiint_D \rho^4\sin \varphi \, d\rho d \theta d\varphi=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^3 \rho^4 \sin \varphi d\rho d\theta d\varphi$$

Which is more easily integrated as $$\int_{0}^{2\pi}d\theta \int_{0}^\pi \sin \varphi \,d \varphi\int_0^3 \rho^4 d\rho=\frac{972\pi}{5}$$

Edit: I must have had a stroke. The answer has been corrected.

• Sanity check: you integrate something positive (except for one point). Can the integral be $0$? – Daniel Fischer Oct 15 '13 at 20:37

$$\int_{0}^{3}{\rm d}r\,r^{2}\times r^{2}\quad \overbrace{\int{\rm d}\Omega_{\,\vec{r}}}^{4\pi}\ =\ 4\pi\,{3^{5} \over 5} = {972 \over 5}\,\pi$$

• can you explain your $\Omega$ notation , namely this integral $\overbrace{\int{\rm d}\Omega_{\,\vec{r}}}^{4\pi}\$ – john Nov 1 '17 at 11:22
• It's the solid angle: $\int\mathrm{d}\Omega_{\vec{r}}$ means "integration over the solid angle". For instance, in spherical coordinates it becomes $\mathrm{d}\Omega_{\vec{r}} = \sin\left(\theta\right)\,\mathrm{d}\theta\,\mathrm{d}\phi$. Thanks. – Felix Marin Nov 3 '17 at 17:48