Suppose a sphere is described by the equation $x^2+y^2+z^2=4z$, How do we find the area of the part of sphere that lies in a parabloid $z=x^2+y^2$ if we have to use spherical parametrization?
First I have issue setting up spherical coordinate
I think we rewrite the equation as $x^2+y^2+(z-2)^2=4$, but then how do we adjust the $x=\rho\sin\phi\cos\theta, y=\rho\sin\phi\sin\theta, z=\rho\cos\phi$?
On
Alternatively, note that the area does not change if you shift the origin $2$ units along the $z$ axis so that you have the sphere $$ x^2+y^2+z^2 = 4 $$ and the paraboloid $$ z+2=x^2+y^2 $$ With this new coordinate system, you can use the classical spherical coordinates, but the bounds in $\phi$ will differ.
In spherical coordinates, the sphere has equation $\rho = 4 \cos \phi$.
So you need to parametrize the surface as follows: $$ \begin{cases} x=\overbrace{(4\cos \phi)}^{=\rho} \sin \phi \cos \theta \\ y=(4cos \phi) \sin \phi \sin \theta \quad \text{with}\quad (\phi,\theta)\in D=[0,\frac{\pi}{2}] \times [0,2\pi]\\ z=(4cos \phi) \cos \phi \\ \end{cases} $$ The area then equals $$ A = \iint_D \mid\mid \vec{r}_{\phi}\times \vec{r}_{\theta} \mid \mid \; d\phi d\theta $$