<s>[INST] Finding the surface area of a sphere within a cylinder and above the $xy$-plane using double integral I'm trying to find the surface area of the part of the sphere $x^2+y^2+z^2=a^2$ that lies within the cylinder $x^2+y^2=ax$ and above the xy-plane.
I want to do this using $$A(S)=\iint_D \sqrt{[f_{x}(x,y)]^2+[f_{y}(x,y)]^2+1} \; dA.$$
My understanding is that $f=z=\sqrt{a^2-x^2-y^2}$, $z_x=-x(a^2-x^2-y^2)^{-1/2}$,  $z_y=-y(a^2-x^2-y^2)^{-1/2}$.
What I don't understand is what the bounds for the integrals should be. I think it has to be done using polar coordinates, but I'm lost as to what the radius and angle bounds would be -- I'm not visualizing the cylinder and sphere very well.
Any help would be appreciated.
 [/INST] Yes it is easier to evaluate using polar coordinates. Based on your working,
$\displaystyle \sqrt{1+z_x^2 + z_y^2} = \cfrac{a}{\sqrt{a^2-x^2-y^2}}$
$\displaystyle dS = \cfrac{a}{\sqrt{a^2-x^2-y^2}} \ dx \ dy$
So integral to find surface area of sphere inside the cylinder above xy-plane is,
$\displaystyle \int_{x^2+y^2 \leq ax} \cfrac{a}{\sqrt{a^2-x^2-y^2}} \ dx \ dy$
Converting to polar coordinates,
$x = r \cos\theta, y = r\sin\theta, x^2 + y^2 = r^2$
$x^2 + y^2 \leq ax \implies r \leq a \cos\theta, - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
[please note the bounds of $\theta$. As we are using polar coordinates centered at the origin, the circle $r = a \cos\theta$ forms for $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right) \ $]
So the integral is,
$\displaystyle \int_{-\pi/2}^{\pi/2} \int_0^{a\cos\theta} \cfrac{a}{\sqrt{a^2-r^2}} \ r \ dr \ d\theta$
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