Surface integral, area of a part of a sphere inside a cylinder

Question

I'm stuck with the following question, "find the surface area of the part of the sphere $x^2+y^2+z^2=a^2$ that lies inside the cylinder $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $b \leq a$, $z\geq 0$". First, I took the parametrization $\varphi(x,y)=(x,y,\sqrt{a^2-x^2-y^2})$, then I switch to polar coordinates to calculate the surface integral, but at the end I have the integral $$ab \int_0^{2 \pi} \frac{1}{1+\sqrt{1-(\cos^2(\theta)+\frac{b^2}{a^2}\sin^2(\theta))}} \,d\theta,$$ and I don't have idea how to calculate it. Maybe I have to take another parametrization of the sphere. The answer should be $\displaystyle 4a^2\arcsin(\frac{b}{a})$.

Answer 1

Geometry here provides a much simpler solution than Calculus.

Since the given region $S$ lies on a sphere, its surface area just depends on the volume of the 3d-region $R$ given by joining $\partial S$ with the center of the sphere, $3\cdot V(R) = a\cdot A(S)$. You can check$^{(*)}$ that the intersection of the given elliptic cylinder with the given sphere is bounded by two equators of the sphere, hence the exercise boils down to computing the volume of a spherical wedge, pretty simple:

$$ V = \frac{4}{3}\pi a^3\cdot\frac{2\arcsin\frac{b}{a}}{2\pi} \quad \Longrightarrow\quad A=\frac{3V}{a}=\color{red}{4a^2\arcsin\frac{b}{a}}.$$

$(*)$ Let $W$ be a spherical wedge in a sphere centered at $O$. Let $M$ be the center of the spherical lune and $e_W$ be the edge of $W$. Let $\pi$ be the plane containing $e$ that is orthogonal to $OM$. By symmetry, the projection of $\partial W$ on $\pi$ is an ellipse with major axis given by $e_W$.