
Surface area of a sphere
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 coords.
What I've done:
equations of the sphere and paraboloid in cylindrical coordinates are $\displaystyle r^2+z^2=4z$ and $\displaystyle z=r^2$.
substituting and rearranging these gives
$\displaystyle z+z^2=4z\; \; \Rightarrow z^23z=0\; \; \Rightarrow z(z3)=0\; \; \Rightarrow z=(0,3)$
$\displaystyle r^2+r^4=4r^2\; \; \Rightarrow r^43r^2=0\; \; \Rightarrow r^2(r^23)=0\; \; \Rightarrow r=(0,\sqrt{3})$
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:
$\displaystyle x^2 + y^2 + z^2=4z\; \; \Rightarrow z^2=4zx^2  y^2\; \; \Rightarrow z=\sqrt{4zx^2y^2}$
When I differentiate, am I right in assuming the 'z' on the RHS disappears as well? Does it go to 1 or 0?
ie.
$\displaystyle \frac{\partial z}{\partial x}=\frac{x}{\sqrt{4x^2y^2}}$
or:
$\displaystyle \frac{\partial z}{\partial x}=\frac{x}{\sqrt{x^2y^2}}$

Hint: Use implicit differentiation with respect to x.
What you did is not correct.