How to find this area by integration? Find the area of the portion of the unit sphere $x^2+y^2+z^2=1$ inside the right circular cylinder $x^2+y^2=y$ and the portion of the cylinder inside the sphere.
I know how to find the area of the sphere and the cylinder separately by the formula:
$$A=\int_{\text{Manifold}}\sqrt{\left\|\frac{\partial F}{\partial t_1}^2\right\|\left\|\frac{\partial F}{\partial t_2}^2\right\|- \left(\frac{\partial F}{\partial t_1}\bullet\frac{\partial F}{\partial t_2}\right)^2}dt$$
but I don't know how to restrict each of the areas to just the portion inside the other. How do I do this? Or am I going wrong and there's a simpler way to calculate the areas from scratch?
 A: The surface of the sphere inside the cylinder is formed by two regions (see picture below), one with $z\ge0$, the other with $z\le0$. 
Let's introduce spherical coordinates $\theta$ and $\phi$ so that every point on the sphere has coordinates $(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)$.
In the upper region, the angle $\theta$ varies between 0 and $\pi/2$.
Al the points of a fixed $\theta$ on the sphere lie on a circle $ECF$
whose points satisfy $z=\cos\theta$ and $x^2+y^2=\sin^2\theta$.
One can then find the coordinates of points $E$ and $F$, which are
the intersections between circle and cylinder:
$$
E=(-\sin\theta\cos\theta,\ \sin^2\theta,\ \cos\theta),\quad
F=(\sin\theta\cos\theta,\ \sin^2\theta,\ \cos\theta).
$$
So point $E$ corresponds to $\phi=\pi-\theta$, while point $F$ has $\phi=\theta$.
The area of the upper region can be then computed in spherical coordinates as:
$$
\int_0^{\pi/2}\!\sin\theta\,\int_\theta^{\pi-\theta}d\phi\,d\theta
=\int_0^{\pi/2}\!\sin\theta(\pi-2\theta)\,d\theta=\pi-2.
$$

Notice then that $h=FH=\cos\theta$, for $0\le\theta\le\pi/2$, so the area of the cylinder surface inside the sphere can be evaluated as follows in cylindrical coordinates centered at the cylinder axis, taking into account that 
$\alpha=\angle OGH=2\phi=2\theta$ for $0\le\alpha\le\pi$,
and that the radius of the cylinder is $1/2$:
$$
2\int_0^\pi \int_{-h}^h dz\ {1\over2}d\alpha=
2\int_0^{\pi/2} \int_{-\cos\theta}^{\cos\theta} dz\ d\theta=
2\int_0^{\pi/2} 2{\cos\theta} d\theta=4.
$$
