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 $R=\{(r, \theta)\mid0\leq r \leq 1, 0\leq \theta \leq 2\pi\}$
$$A(S)= 2\sqrt{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\frac{r}{\sqrt{2-r^2}}\,\mathrm dr\,\mathrm d\theta=4\pi(2-\sqrt{2})$$

Yes it is correct.
Another method, that gives the answer faster id to use the spherical coordinates $(r,\theta,\phi)$ oriented not around $z$ axis as usual, but around $x$ axis (so that $y^2+z^2 = r^2 \sin^2\theta$).
The sphere is given by the condition $r=\sqrt{2}$. The element of the area of a sphere is $dA = r^2\sin\theta\, d\theta\, d\phi = 2 \sin\theta\, d\theta\, d\phi$.
The inside of the cylinder is given by the condition $r^2\sin^2\theta \le 1$, which with the condition $r=\sqrt{2}$ gives the condition $$\sin^2 \theta < \frac{1}{2} $$
that is (remembering that for spherical coordinates $\theta\in[0,\pi]$)
$$\theta\in[0,{\pi/ 4}] \cup [{3\pi/ 4},\pi] =: I$$
We have then
\begin{align} A(S) &= \int_{\theta\in I} \int_{\phi\in[0,2\pi]} dA = 2\left(\int_0^{\pi/4}\sin\theta d\theta + \int_{3\pi/4}^{\pi}\sin\theta d\theta\right) \int_0^{2\pi} d\phi = \\
&= 2\cdot\big((1-\frac{\sqrt{2}}{2})+(1-\frac{\sqrt{2}}{2})\big)\cdot 2\pi = 4\pi(2-\sqrt{2})\end{align}
