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Problem:

The cylinder \(x^{2}+y^{2}=x\) divides the surface of the upper unit hemisphere \(S\) into two regions \(S_{1}\) and \(S_{2}\) where \(S_{1}\) is inside the cylinder and \(S_{2}\) is outside. Denote the area of \(S_{1}\) by \(A(S_{1})\) and the area of \(S_{2}\) by \(A(S_{2})\). Find \(\frac{A(S_{2})}{A(S_{1})}\).

Solution:

We can write surface of the upper half of the unit sphere as the graph of a function, specifically, \[ \begin{align} f(x,y)=\sqrt{1-x^{2}-y^{2}}. \end{align} \] In this case, our parametrization is \[ \begin{align} \Psi(x,y) = (x,y,f(x,y)) \end{align} \] and thus \[ \begin{align} \Psi_{x}=\frac{\partial \Psi}{\partial x}=(1,0,f_{x}) \quad \textrm{and} \quad \Psi_{y}=\frac{\partial \Psi}{\partial y}=(0,1,f_{y}). \end{align} \] Therefore \[ \begin{align} \Psi_{x}\times \Psi_{y} = (-f_{x}, -f_{y}, 1) \end{align} \] and \[ \begin{align} \|\Psi_{x}\times \Psi_{y}\| = \sqrt{1+(f_{x})^{2}+(f_{y})^{2}}. \end{align} \] Plugging this into the surface area of a parametrized surface formula we get \[ \begin{align} \textrm{Area} = \iint\limits_{R} \sqrt{1+(f_{x})^{2}+(f_{y})^{2}} ~dxdy \end{align} \] where \(R\) is the region we would like to integrate over. In our specific case with the sphere, \[ \begin{align} & f_{x} = -\frac{x}{\sqrt{1-x^{2}-y^{2}}}, \quad f_{x} = -\frac{y}{\sqrt{1-x^{2}-y^{2}}} \\ \implies & \sqrt{1+(f_{x})^{2}+(f_{y})^{2}} = \sqrt{1+\frac{x^{2}+y^{2}}{1-(x^{2}+y^{2})}} \end{align} \] and the region \(R\) is described by the interior of \(x^{2}+y^{2}=x\), which is given by the inequalities \[ \begin{align} \begin{cases} 0\leq x \leq 1 \\ -\sqrt{x-x^{2}} \leq y \leq \sqrt{x-x^{2}} \end{cases} \quad \textrm{(cartesian coordinates)} \end{align} \]

So \(A(S_{1})\) is \[ \begin{align} A(S_{1}) = \int\limits_{0}^{1}\int\limits_{-\sqrt{x-x^{2}}}^{\sqrt{x-x^{2}}} \sqrt{1+\frac{x^{2}+y^{2}}{1-(x^{2}+y^{2})}} ~dxdy. \end{align} \] We can convert to polar coordinates by noting that the bounds will change to \[ \begin{align} \begin{cases} -\pi/2 \leq \theta \leq \pi/2 \\ 0 \leq r \leq \cos\theta \end{cases} \quad \textrm{(polar coordinates)} \end{align} \] and we get \[ \begin{align} A(S_{1}) = \int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{\cos\theta} \sqrt{1+\frac{r^{2}}{1-r^{2}}} ~rdrd\theta = \int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{\cos\theta} r\sqrt{\frac{1}{1-r^{2}}} ~drd\theta \end{align} \] Which is a slightly more manageable integral. After solving this integral (probably with a u-substitution for the \(r\) integral) we find that \[A(S_{1}) = \pi-2\] Since \[A(S_{2})=\textrm{surface area of uppper unit hemisphere}-A(S_{1})=2\pi-(\pi-2) = \pi+2\] We can conclude that \[ \begin{align} \frac{A(S_{2})}{A(S_{1})} = \frac{\pi+2}{\pi-2}. \end{align} \]