I have two inequalities:
$x^2 + y^2 + z^2 \le 8$ and $z \ge \sqrt{x^2+y^2}$
These make up a body $K$ in 3D space. I would like to represent this body $K$ using spherical coordinates, but I'm unsure on how to do it.
Basically I want it on the form:
$x = r\sin(\theta)\cos(\phi)$
$y = r\sin(\theta)\sin(\phi)$
$z = r\cos(\theta)$
And determine how $r$, $\theta$ and $\phi$ change.
I can set the inequalities to equal eachother, and get the projection that they make on the $xy$ plane, which is $x^2+y^2 \le4$ but I'm not sure if that is useful for this problem.
Since $x^2+y^2+z^2=r^2$, the first inequality translates into $0\le r\le\sqrt8$.
For the second one observe that $x^2+y^2=r^2\sin^2\theta$, so that the inequality becomes $r\cos\theta\ge r\sin\theta$, or equivalently $\tan\theta\le1$, that is $0\le\theta\le\pi/4$.
The solid is a cone with vertex at the origin and aperture $\pi/4$.
Note: the usual notation has $\theta$ and $\phi$ interchanged.