Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$ The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.
The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.
We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:
$$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$
where $F$ is the equation of the sphere.
Plugging in the expressions and simplifying ($z \ge 0)$, we get:
$$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$
Converting to polar coordinates, we have:
$$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$
Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I  going wrong?
 A: Given the equations
$$
x^2+y^2+z^2=a^2,
$$
and
$$
x^2+y^2 = ay,
$$
we obtain
$$
ay + z^2 = a^2.
$$

Using
$$
\begin{eqnarray}
x &=& a \sin(\theta) \cos(\phi),\\
y &=& a \sin(\theta) \sin(\phi),\\
z &=& a \cos(\theta),\\
\end{eqnarray}
$$
we obtain
$$
a^2 \sin(\theta) \sin(\phi) + a^2 \cos^2(\theta) = a^2
\Rightarrow \sin(\theta) = \sin(\phi) \Rightarrow \theta=\phi \vee \theta=\pi-\phi.
$$

For the surface we have
$$
\begin{eqnarray}
\int d\phi \int d\theta \sin(\theta) &=&
\int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta) +
\int_{\phi/2}^{\pi}d\phi \int_0^{\pi-\phi} d\theta \sin(\theta)\\
&=& 2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta).
\end{eqnarray}
$$

We can calculate the surface as
$$
\begin{eqnarray}
4 a^2 \int_0^{\pi/2}d\phi \int_0^\phi d\theta \sin(\theta)
&=& 4 a^2 \int_0^{\pi/2}d\phi \Big( 1 - \cos(\phi) \Big)\\
&=& 4 a^2 \Big( \pi/2 - 1 \Big)\\
&=& a^2 \Big( 2\pi - 4 \Big).
\end{eqnarray}
$$
A: I know what you r doing rong, I solved this a week ago, the same way you did.
You forgot that $\sqrt{\sin^2\theta} = |\sin\theta|$, not $\sin\theta$ 
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$ $
This is how you might have done 
$$ 2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}\dfrac{a}{\sqrt{a^2-r^2}}\cdot r \cdot  drd\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}\left[ -a\sqrt{a^2-r^2}\right]_0^{a\cos\theta}d\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}- a^2\sqrt{\sin^2\theta}-\left(-a^2 \right)d\theta$$
$ $
$$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2\sin\theta d\theta$$
MISTAKE !!!!
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INSTEAD  $$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2|\sin\theta| d\theta$$
$ $
$$= 4\int_{0}^{\pi/2}a^2- a^2\sin\theta d\theta$$
$ $
$$= 4a^2 \left[1-  \sin\theta \right] _{0}^{\pi/2}$$
$ $
$$= a^2\left(2\pi-4 \right)$$
