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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 $\displaystyle 4a^2\arcsin(\frac{b}{a})$.

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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 $e_W$.

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Spherical coordinates are often less hairy than polar when you are working with a 3d surface:

Parametrize surface of sphere: $ \overrightarrow{S}(\phi,\theta)=<a\sin(\phi)\cos(\theta), a\sin(\phi)\sin(\theta), a\cos(\phi)>$ You don't need $\rho$ as a parameter since you are only looking at the surface where $\rho$ always equals a. Then, to find boundaries you examine the cylinder:

$0\le z \le a$ becomes $0 \le a \cos (\phi) \le a $ , so $0 \le \phi \le \pi $

$-b \le y \le b$ becomes $-b \le a\sin(\phi)\sin(\theta) \le b$ , so $-\frac{b}{a} \le \sin(\theta) \le \frac{b}{a}$

We can eliminate the $\sin(\phi)$ because it is between zero and one

$|\overrightarrow{S}_{\phi}\times \overrightarrow{S}_{\theta}|$ $$=|<a\cos(\phi)\cos(\theta),a\cos(\phi)sin(\theta),-a\sin(\phi)>\times <-a\sin(\phi)\sin(\theta),a\sin(\phi)\cos(\theta),0>|$$ $=a^2\sin(\phi)$

So, your surface integral becomes: $$\iint_SdS=\int_{-\arcsin(\frac{b}{a})}^{\arcsin(\frac{b}{a})}\int_0^{\pi}|\overrightarrow{S}_{\phi}\times \overrightarrow{S}_{\theta}|d\phi d\theta$$ $$=\int_{-\arcsin(\frac{b}{a})}^{\arcsin(\frac{b}{a})}\int_0^{\pi}a^2\sin(\phi)d\phi d\theta$$

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