Find the surface ara of the sphere $\displaystyle x^2 + y^2 + z^2=4z$ that lies inside the paraboloid $\displaystyle z=x^2+y^2$, using cylindrical co-ords.

What I've done:

equations of the sphere and paraboloid in cylindrical co-ordinates are $\displaystyle r^2+z^2=4z$ and $\displaystyle z=r^2$.

substituting and rearranging these gives

$\displaystyle z+z^2=4z\; \; \Rightarrow z^2-3z=0\; \; \Rightarrow z(z-3)=0\; \; \Rightarrow z=(0,3)$

$\displaystyle r^2+r^4=4r^2\; \; \Rightarrow r^4-3r^2=0\; \; \Rightarrow r^2(r^2-3)=0\; \; \Rightarrow r=(0,\sqrt{3})$

this gives the points of intersections $\displaystyle (r,z)=(0,0) \; and\; (\sqrt{3},3)$

And that's as far as I can get. I'm having a problem with the partial derivatives.

To get them I've done this:

$\displaystyle x^2 + y^2 + z^2=4z\; \; \Rightarrow z^2=4z-x^2 - y^2\; \; \Rightarrow z=\sqrt{4z-x^2-y^2}$

When I differentiate, am I right in assuming the 'z' on the RHS disappears as well? Does it go to 1 or 0?

ie.

$\displaystyle \frac{\partial z}{\partial x}=\frac{-x}{\sqrt{4-x^2-y^2}}$

or:

$\displaystyle \frac{\partial z}{\partial x}=\frac{-x}{\sqrt{-x^2-y^2}}$