If I have a sphere $T: x^{2}+y^{2}+z^{2}\leqslant 10z$ by transformation to the spherical coordinate system by the:
$
x=r\cos\theta\sin\varphi\\
y=r\sin\theta\sin\varphi\\
z=r\cos\varphi
$
What is values for $\varphi$ I will get?
$0\leqslant\varphi\leqslant\pi$ or $0\leqslant\varphi\leqslant\pi/2$ and why??
Thanks!
Best Answer
$x^2+y^2+z^2 \leq 10z$ is the same as $x^2+y^2+(z-5)^2\leq 25$ so we are dealing with a solid ball of radius 5 centered at $(0,0,5)$.
Naively we can translate this inequality to $\rho^2 \leq 10\rho \cos(\phi)$ so that $\rho \leq 10\cos(\phi)$.
I have drawn a ray emanating from the origin out to the sphere [whose equation is $\rho=10\cos(\phi)$]. The angle $\phi$ should sweep from the $z$-axis down to $\phi=\pi/2$. Notice that at this point $\rho=10\cos(\pi/2)=0$ (so we should stop).
Notice that if you had allowed $\phi$ to continue to go past $\pi/2$, cosine and thus $\rho$ would become negative (which indicates something isn't quite right).
Therefore, the bounds in spherical coordinates are:
$0 \leq \rho \leq 10\cos(\phi)$, $0 \leq \phi \leq \pi/2$, and $0 \leq \theta \leq 2\pi$.