Calculate the area of a sphere drilled by two cylinders. 
Let $S$ be the sphere given by the equation $x^2+y^2 +z^2 =4$ cut with
  $z  \geq 0$. Now, we drill the semisphere that is left with two
  vertical cylinders of radius $1$, whose axes are respectively on the
  points $(0,1,0)$ and $(0,-1,0)$. Calculate the area of the surface
  that is left.

I know that the area of the semisphere of radius two is $8\pi$, but I don't know how to compute the area of the intersection between the cylinders and the semisphere. Any help with that would be highly appreciate. 
Thanks in advance!
 A: 
Parametrization of the surface cut out by one cylinder


The image above shows the intersection of the hemisphere with the cylinder whose axis passes through $(0,1,0)$ and whose equation is 
$$
x^2+(y-1)^2=1
$$
The surface cut out by the cylinder can be parametrized with Cartesian coordinates as
$$
\vec\Sigma=\left(x,y,\sqrt{4-x^2-y^2}\right),\quad 0\le x^2+(y-1)^2\le1\\
\vec\Sigma_x\times\vec\Sigma_y=\begin{vmatrix}
\hat i&\hat j&\hat k\\
1& 0&-{x\over\sqrt{4-x^2-y^2}}\\
0& 1&-{y\over\sqrt{4-x^2-y^2}}\\
\end{vmatrix}\\
|\vec\Sigma_x\times\vec\Sigma_y|={2\over\sqrt{4-x^2-y^2}}$$

Calculation of surface area cut out by one cylinder

Area of the surface cut out by one cylinder is
$$\mathrm{
\int_0^{2}\int^{\sqrt{1-(y-1)^2}}_{-\sqrt{1-(y-1)^2}}{2\over\sqrt{4-x^2-y^2}}\,dx\,dy\\
=2\int_0^{2}\int^{\sqrt{2y-y^2}}_{-\sqrt{2y-y^2}}{1\over\sqrt{4-x^2-y^2}}\,dx\,dy\\
=4\int_0^{2}\int^{\sqrt{2y-y^2}}_{0}{1\over\sqrt{4-y^2-x^2}}\,dx\,dy\\
=4\int_0^2\arcsin{\sqrt{2y-y^2\over4-y^2}}\,dy\\
=4\int_0^2\arctan{\sqrt{2y-y^2\over4-2y}}\,dy\\
=4\int_0^2\arctan\sqrt{y\over2}\,dy\\
=16\int_0^1v\arctan v\,dv\quad\left(v=\sqrt{y\over2}\right)\\
=16\left[{v^2\over2}\arctan v\Big{|}_0^1-\int_0^1{v^2\over2(1+v^2)}dv\right]\\
=16\left[{\pi\over8}-{1\over2}\int_0^1\left(1-{1\over1+v^2}\right)dv\right]\\
=8\left[{\pi\over4}-\left(v-\arctan v\right)\Big{|}_0^1\right]\\=\color{blue}{4(\pi-2)}
}$$

Final answer

As $2$ identical holes have been drilled and the area of the hemisphere is ${1\over2}4\pi\times2^2$ sq units, the area of the surface that is left is 
$$
{1\over2}4\pi\times2^2-2\times\color{blue}{4(\pi-2)}=16
$$
