Sphere Coordinates

Question

How do you convert the set $$ K = \{((x, y, z) \in \mathbb{R^3} | x^2 + y^2 + z^2 ≤ 25, -4 ≤ x ≤ 4, -4 ≤ y ≤ 4, -4 ≤ z ≤ 4)\} $$ into spherical coordinates? It's evident that the radius ranges from 0 to 5, but what are the restrictions over phi and theta?

in my spherical conversion I got $$ \; 0 ≤ r ≤ 5 \\ -4 \le r\sin\phi \sin\theta \le 4 \\ -4 \le r\sin\phi \cos\theta \le 4 \\ -4 \le r\cos\phi\le4$$

How do I proceed from here?

Thanks

Answer 1

A few more hints:

The integral of $f(x,y,z)=3(x^2+y^2+z^2)$ over the whole ball with radius $5$ is \begin{align} I_1:=\int\limits_{x^2+y^2+z^2\le 25}3(x^2+y^2+z^2)\,dz\,dy\,dx\,. \end{align} This is easy to calculate using 3D polar coordinates. Looking at the picture it is clear from symmetries that from $I_1$ you have to subtract six times the integral

\begin{align} I_2=\int\limits_{x^2+y^2\le 9}\quad\int\limits_4^{\sqrt{25-x^2-y^2}} 3(x^2+y^2+z^2)\,dz\,dy\,dx \end{align} which covers the portions of the ball sticking out of the cube. The outer $dy\,dx$-integral in $I_2$ should be easy to calculate using 2D polar coordinates.

\begin{align}I_1&=7500\pi\,,&I_2&=2\pi\int_0^3\int_4^{\sqrt{25-r^2}}3(r^2+z^2)\,r\,dz\,dr=\frac{1536\pi}{5}\,.\end{align}

Answer 2

From the inequalities you wrote, using $\rho$ for $r$, parametrization of $K$ is $$\rho\leq\min\left\{5,\frac{4}{|\cos\theta\sin\phi|}, \frac{4}{|\sin\theta\sin\phi|}, \frac{4}{|\cos\phi|} \right\}$$ $$0\leq\theta\leq2\pi$$ $$0\leq\phi\leq\pi$$