Surface area of sphere within a cylinder I have to 

Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$

My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$. 
$${\delta z\over \delta x}={-x\over \sqrt{a^2-x^2-y^2}}\ ;\ {\delta z\over \delta y}={-y\over \sqrt{a^2-x^2-y^2}}$$
Using the formula $$a(\Bbb{S})=\iint\limits_\Bbb{T}\sqrt{1+\left({\delta z\over \delta x}\right)^2+\left({\delta z\over \delta y}\right)^2}dx \ dy$$ 
I get $$a(\Bbb{S})=\iint\limits_{x^2+y^2=by}\sqrt{1+{by\over a^2-by}}\ dy\  dx\\=\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\sqrt{1+{by\over a^2-by}}\ dy\  dx\\=a\int_0^{b/2}\int\limits_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}{1\over\sqrt{a^2-by}}dy \ dx\\=a\int_0^{b/2}\left\{\left[{-2\sqrt{a^2-by}\over b}\right]_{{b\over 2}-\sqrt{{b^2\over 4}-x^2}}^{{b\over 2}+\sqrt{{b^2\over 4}-x^2}}\right\}\ dx$$
How to proceed now?The integral seems too bad.
OR
Is there a simpler parametrization ?
 A: Another approach in spherical coordinates: parametrize the surface with
\begin{cases}
x=a \sin\phi \cos\theta \\
y=a \sin\phi \sin\theta \\
z=a \cos\phi 
\end{cases}
with $(\theta,\phi)\in [0,\pi]\times[0,\Phi]$, where $\Phi \in ]0,\pi]$ is the solution to
\begin{cases}
x^2+y^2+z^2=a^2 \\
x^2+y^2=by
\end{cases}
Substituting $x,y,z$ by their expressions in spherical coordinates yields
$$
\Phi = \sin^{-1}\left(\frac{b}{a}\sin\theta\right)
$$
It follows that 
$$
S=\int_{0}^{\pi}\int_0^{\Phi} a^2\sin\phi\; d\phi d\theta = a^2 \int_0^{\pi}1-\cos\Phi\; d\theta
$$
With a little trigonometric work ($\cos x =\pm \sqrt{1-\sin^2x}$), we can show that 
$$
\cos\Phi = \sqrt{1-\left(\frac{b}{a}\sin\theta\right)^2}
$$
Therefore
$$
S = \pi a^2 - a \int_0^{\pi} \sqrt{a^2-b^2\sin^2\theta}\; d\theta
$$
A: Your derivatives are wrong by a factor of $2$. This causes your integrand to be 
$$
\sqrt{1-{x^2+y^2\over a^2-x^2-y^2}}=\frac a{\sqrt{a^2-x^2-y^2}}
$$ 
If we represent the cylinder in polar coordinates, we have $$r^2=b r \sin\theta.$$After cancelling $r$, we get $r=b\sin\theta$. So the interior of the cylinder is the region
$$
0\leq\theta\leq\pi,\ \ \ 0\leq r\leq b\sin\theta
$$
(the restriction on $\theta$ comes from $\sin\theta\geq r/b\geq0$). So the area is 
\begin{align}
a(\Bbb{S})&=\iint\limits_\Bbb{T}\sqrt{1+\left({\partial z\over \partial x}\right)^2+\left({\partial z\over \partial y}\right)^2}dx \ dy\\ \ \\
&=\int_0^\pi\int_0^{b\sin\theta}\frac {ar}{\sqrt{a^2-r^2}}\,dr\,d\theta\\ \ \\
&=-a\,\int_0^\pi \left.\vphantom{\int}{\sqrt{a^2-r^2}}\,\right|_0^{b\sin\theta}\,d\theta\\ \ \\
&=a\,\int_0^\pi \left(a-{\sqrt{a^2-b^2\sin^2\theta}}\right)\,d\theta\\ \ \\
&=a^2\pi-a\int_0^\pi {\sqrt{a^2-b^2\sin^2\theta}}\,d\theta.\\ \ \\
\end{align}
This last integral is an elliptic integral (unless $b=a$). It looks to me like all approaches to this problem will lead to this integral (I will be happy to be proven wrong, though). 
