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 $R=\{(r, \theta)\mid0\leq r \leq 1, 0\leq \theta \leq 2\pi\}$
$$A(S)= 2\sqrt{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\frac{r}{\sqrt{2-r^2}}\,\mathrm dr\,\mathrm d\theta=4\pi(2-\sqrt{2})$$
 A: 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}
A: 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.)
$$A=\int_0^{\pi\over2}\int_0^1  \sqrt2 \,(2-r^2)^{-{1\over2}}  \,r\,dr\,d\theta$$
$$={\sqrt2\pi\over2}\int_0^1 (2-r^2)^{-{1\over2}} \,r\,dr\,.\ \ \ \ \text{Thus,}$$
$$8A={4\sqrt2\pi}\sqrt{2-r^2}\big|_1^0={4\sqrt2\pi}(\sqrt2-1)={4\pi}(2-\sqrt2)$$
A: The calculus part has already been answered. Your calculations are correct, and you have been offered a different way of arriving to the solution using spherical coordinates. Another way of checking your final answer is using a "visual" aid and the formula for spherical caps.
By visual I mean that I actually constructed the sphere and the cylinder using your description (See my Geogebra construction Intersection of the sphere and cylinder of this problem). I have defined a point $P_C=<1,0,1>$ which satisfies the equations of both surfaces. A point $P_B=<1,0,0>$, which evidently will drop vertically, and from there to that top of the sphere aligned with the length of the cylinder, we just need the point $P_E=<\sqrt{2},0,0>$ (by the definition of the sphere's radius).
The length $h$ from $P_B$ to $P_C$ is
$$h=\overline{P_BP_C}=\sqrt{(1-1)^2+(0-0)^2+(1-0)^2}=1$$
We also need a distance $a$, from $P_B=<1,0,0>$ to $P_E=<\sqrt{2},0,0>$:
$$a=\overline{P_BP_E}=\sqrt{(\sqrt{2}-1)^2+(0-0)^2+(0-0)^2}\approx 0.41$$
Now we are in a position to use the formula for the area of a spherical cap
$$A_{cap}=\pi(a^2 +h^2)=\pi(0.41^2 +1^2)\approx 3.67$$
But since we have two caps being covered by this cylinder
$$3.67\times 2\approx 7.34$$
Comparing with your result, and using the relative error formula
$$4\pi(2-\sqrt{2})\approx 7.36;\; error=\left\lvert \frac{7.34-7.36}{7.36}\right\rvert\times 100 \approx 0.3\%,$$
which entails that your result should be correct.
Granted, this method does not check over the calculus path that you followed, or the method using spherical coordinates; only the final answer, but then again when you ask "did I get it right?" you are at least requiring verification over your final answer, which in my view, if you have an alternate way of obtaining it, this can be very valuable when you want to retrace your steps.
