How to do this rather basic Surface area question I am having a bit of difficulty evaluating the surface area of the region that consists of the part of the sphere $$x^2+y^2+z^2=3c^2,$$ within the paraboloid $$2cz=x^2+y^2,$$ where $c \gt 0$.
I know that I need to evaluate
$$Sa= \iint_{D} \sqrt{(1+f_{x}^2+f_{y}^2)}dA,$$ where $z=f(x,y)$ is the unique projection.
Setting $z=\sqrt{3c^2-x^2-y^2}$ and computing gives me $$\sqrt{(1+f_{x}^2+f_{y}^2)}=\sqrt{1+\frac{x^2+y^2}{3c^2-x^2-y^2}}.$$
Or in polar coordinates, $$\iint_{D'}\sqrt{1+\frac{r^2}{3c^2-r^2}}rdrd\theta$$
But I am having my issues when it comes to determine the limits of $r$ and $\theta$.
Since we are also bound by $2cz=x^2+y^2$, I think $r$ would still be going from $0$ to $\sqrt3c$, but I don't know about $\theta$.
The only thing I thought of was setting $3c^2-z^2=2cz$ and solving for the quadratic for z to get $z=c$ and $z=-3c$ but I don't really think this will help.
I also wanted to try doing it in spherical coordinates, but I cannot figure out the variation of $\psi$, I get $$\iint_{D} 3c^2\sin(\psi)d\psi d\theta$$ with $\theta$ going from $0$ to $2\pi$, but I can't figure out the final value for $\psi$.
Any help? Thanks.
 A: The sphere and the paraboloid are both formed by rotating curves about the $z$-axis. When you take slices through either parallel to the $xy$-plane, you get circles. Imagine the top of the sphere as a bunch of horizontal circles that shrink to a point as you move your slice closer to the top. Imaging the paraboloid as a bunch of horizontal circles that shrink as you move your slice downwards. The surfaces intersect in a circle, and the projection of that circle onto the $xy$-plane is ... a circle.
You can find the maximum value of $r$ by solving for $z$ in both surfaces then substituting back to get $r$.  Equating $x^2+y^2$ gives $2cz+z^2=3c^2$, or $z^2+2cz-3c^2=0$, $z=\dfrac{-2c\pm\sqrt{4c^2+12c^2}}{2}=\dfrac{-2c\pm4c}{2}=c$ (ignoring nonsensical $-3c$). Then $r^2=x^2+y^2=2cz=2c^2$, so $r=\sqrt{2}c$.
So the region on the $xy$-plane of the circle of radius $\sqrt2c$. $r$ varies from $0$ to $\sqrt2c$. $\theta$ ...I'm sorry, I can't work out how to disguise this as a hint! ... $\theta$ varies from $0$ to $2\pi$.
