<s>[INST] Find the area of the part of the sphere $x^2+y^2+z^2=2$ that lies inside the cylinder $y^2+z^2=1$ This is what I have tried so far. I am just not sure whether I got it right or not.
We have $y^2+z^2=1$, then $r=1,\quad x=\sqrt{2-y^2-z^2}\quad\text{and}\quad x=\sqrt{2-r^2}$
$$\frac{\partial{x}}{\partial{y}}=-\frac{y}{\sqrt{2-y^2-z^2}}\quad\text{and}\quad
\frac{\partial{x}}{\partial{z}}=-\frac{z}{\sqrt{2-y^2-z^2}}
$$
$$\begin{align}
A(S)
&=2\iint\limits_{R}\sqrt{1+\left(-\frac{y}{\sqrt{2-y^2-z^2}}\right)^2+\left(-\frac{z}{\sqrt{2-y^2-z^2}}\right)^2}\,\mathrm dA\\
&=2\iint\limits_{R}\sqrt{1+\frac{y^2+z^2}{2-y^2-z^2}}\,\mathrm dA\\&=2\sqrt{2}\iint\limits_{R}\frac{1}{\sqrt{2-y^2-z^2}}\,\mathrm dA
\end{align}
$$
Converting to polar coordinate $R=\{(r, \theta)\mid0\leq r \leq 1, 0\leq \theta \leq 2\pi\}$
$$A(S)= 2\sqrt{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\frac{r}{\sqrt{2-r^2}}\,\mathrm dr\,\mathrm d\theta=4\pi(2-\sqrt{2})$$
 [/INST] My calculations are easy, but I use the
arc length of parametric curves. The question is to
find the area of the sphere $x^2+y^2+z^2=2$ inside the cylinder $x^2+y^2=1.$ (The changed variables give the same area.)
The total area of two caps equals $8A,$ where $A$ is one-quarter of the area
of the top cap.  I use cylindrical coordinates and
consider an arbitrary point $P=(r\cos\theta, r\sin\theta, \sqrt{2-r^2})$ on the top cap. Observe
that the length of
$${\partial\over\partial r}P=(\cos\theta, \sin\theta, {{-r}\over{\sqrt{2-r^2}}})$$
equals $\sqrt{2\over2-r^2}$ and is independent of $\theta.$
(Geometrically, this length times $dr$ is the length  of a piece of arc on the great circle
for fixed $\theta.$ The small piece sweeps out an area when rotated about the $Z$-axis.)
$$A=\int_0^{\pi\over2}\int_0^1  \sqrt2 \,(2-r^2)^{-{1\over2}}  \,r\,dr\,d\theta$$
$$={\sqrt2\pi\over2}\int_0^1 (2-r^2)^{-{1\over2}} \,r\,dr\,.\ \ \ \ \text{Thus,}$$
$$8A={4\sqrt2\pi}\sqrt{2-r^2}\big|_1^0={4\sqrt2\pi}(\sqrt2-1)={4\pi}(2-\sqrt2)$$
</s>