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!


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 
$$
{1\over2}4\pi\times2^2-2\times\color{blue}{4(\pi-2)}=16
$$
