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.
Thanks in advance!
On
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.
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.