<s>[INST] Surface of sphere above/below ellipse I am struggling with the following problem: Find the surface area of $x^2+y^2+z^2=a^2$ enclosed by the cylinder $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$ $(a>b>0)$. The solution of the problem is supposed to be $4 \pi a^2 - 8a^2 \arcsin\left(\frac{\sqrt{a^2-b^2}}{a}\right)$.
I tried polar coordinates, I tried $x= a \, r \, cos(\theta)$, $y= b \, r \, sin(\theta)$ (with the Jacobian of this transformation equal to $abr$, and the limits of integration $0≤\theta ≤ \frac{\pi}{2}$ and $0≤ r ≤1$).
Time and time again, I get integranda which are impossible to integrate with respect to $\theta$ (and which also definitely won't give me an $\arcsin$ as a result).
Any help would be greatly appreciated.
 [/INST] It turns out we don't really need any calculus to solve this problem.
For completeness, let us first solve this problem the calculus way.
Method 1 - using spherical polar coordinates.
By symmetry, we only need to compute the area on the upper hemisphere and then multiply the result by $2$.
Let $c^2 = a^2-b^2$ and parametrize the sphere by spherical polar coordinate 
$$(x,y,z) = (a\sin\theta\cos\phi,a\sin\theta\sin\phi,a\cos\theta)$$
In the upper hemisphere, the portion of unit sphere within the ellipsoidal cylinder $$\left\{ (x,y,z) : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \right\}$$
is given by
$$\frac{(a\sin\theta\cos\phi)^2}{a^2} + \frac{(a\sin\theta\sin\phi)^2}{b^2} \le 1
\iff \sin\theta \le \frac{1}{\sqrt{\cos^2\phi + \frac{a^2}{b^2}\sin^2\phi}}
= \frac{1}{\sqrt{1 + \frac{c^2}{b^2}\sin^2\phi}}
$$
This means as $\phi$ varies from $0$ to $2\pi$, $\theta$ can take values from $0$ to $\theta_0$ (dependent of $\phi$) where
$\theta_0 \in [ 0, \frac{\pi}{2} ]$ is determined by
$$\sin\theta_0 = \frac{1}{\sqrt{1+\frac{c^2}{b^2}\sin^2\phi}}
\quad\iff\quad
\cos\theta_0 =
\frac{\frac{c}{b}|\sin\phi|}{\sqrt{1+\frac{c^2}{b^2}\sin^2\phi}}
= \frac{\frac{c}{a}|\sin\phi|}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}}
$$ 
The area on the upper hemisphere we seek is then given by the integral
$$\begin{align}
\verb/Area/_{up} 
&= a^2\int_0^{2\pi}\int_0^{\theta_0} \sin\theta d\theta d\phi
= a^2\int_0^{2\pi}(1-\cos\theta_0) d\phi\\
&= a^2\left[ 2\pi - \int_0^{2\pi}\frac{\frac{c}{a}|\sin\phi|}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}} d\phi \right]
= a^2\left[ 2\pi - 4\int_0^{\pi/2}
\frac{\frac{c}{a}\sin\phi}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}} d\phi \right]
\end{align}
$$
Change variable to $t = \frac{c}{a}\cos\phi$, we get
$$\verb/Area/_{up} 
= a^2\left[ 2\pi - 4\int_0^{c/a} \frac{dt}{\sqrt{1-t^2}} \right]
= a^2\left[ 2\pi - 4\sin^{-1}\left(\frac{c}{a}\right)\right]
$$
As a result, the area we seek is
$$\verb/Area/ = 2\verb/Area/_{up} = 4\pi a^2 - 8a^2\sin^{-1}\left(\frac{\sqrt{a^2-b^2}}{a}\right)$$
Method 2 - the geometric way.
As mentioned in beginning of this answer, we don't need any calculus to derive 
the result.
For any point $(x,y,z)$ on the intersection of the sphere and the ellipsoidal cylinder, we have
$$\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{a^2} = 1 = \frac{x^2}{a^2} + \frac{y^2}{b^2}
\iff \frac{z^2}{a^2} = \frac{y^2}{b^2} - \frac{y^2}{a^2} = \frac{y^2c^2}{a^2b^2}
\iff z = \pm \frac{c}{b} y
$$
The RHS is the equation for a pair of planes.
What this means is the portion of upper hemisphere within the ellipsoidal cylinder is the portion of sphere above the two planes $z \ge \pm \frac{c}{b} y$.
Since this two planes are making an angle $\tan^{-1}\left(\frac{c}{b}\right)$ with $xy$-plane, we find:
$$\begin{align}
\verb/Area/ = 2\verb/Area/_{up} 
&= 4a^2\left[\pi - 2\tan^{-1}\left(\frac{c}{b}\right)\right]
= a^2\left[4\pi - 8\sin^{-1}\left(\frac{\frac{c}{b}}{\sqrt{1+\frac{c^2}{b^2}}}\right)\right]\\
&= a^2\left[4\pi - 8\sin^{-1}\left(\frac{c}{a}\right)\right]
\end{align}
$$
The same result we obtained by Method 1 above.
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