Find the surface area of the portion of the sphere $x^2$ + $y^2$ + $z^2$ = 25 that lies inside the cylinder $x^2$ + $y^2$ = 9.
My attempt at a solution: $z^2$ = 25 - $x^2$ - $y^2$, so z = $\sqrt{25 - x^2 - y^2}$. Then, partials of z: $\frac{-x}{\sqrt{25 - x^2 - y^2}}$ and $\frac{-y}{\sqrt{25 - x^2 - y^2}}$.
So, SA = $\int$$\int$ $\sqrt{1 + \frac{x^2}{25 - x^2 - y^2} + \frac{y^2}{25 - x^2 - y^2}}$dydx. Converting to polar:
$\int$$\int$ $\sqrt{1 + \frac{r^2cos^2\theta}{{25 - r^2}} + \frac{r^2sin^2\theta}{{25 - r^2}}}rdrd\theta$, and simplifying and adding limits of integration gives $$\int_0^{2\pi} \int_0^3 \sqrt{1 + \frac{r^2}{25-r^2}} rdrd\theta$$.
I'm stuck as to where to go from here. I guess I could've made a mistake earlier in the problem but I'm confident with what I have so far.
Let us consider the spherical coordinates: $$x = r \sin(\theta) \cos(\varphi)\\ y = r \sin(\theta) \sin(\varphi)\\ z = r \cos(\theta)\\ $$ $$ r \in [0, + \infty) \\ \theta \in [0, \pi] \\ \varphi \in [0, 2 \pi) $$ Here $(r, \theta, \varphi)$ stand for the radius, inclination, azimuth correspondingly.
Note that the region you are looking for is the union of two hemispheres of the same surface area.
The spherical volume form is $$dV = r^{2} \sin\theta \, dr \, d \theta \, d \varphi$$
Finally, the problem reduces to the following computation: $$V = 2 \int_{R} {d V} = 2 \int_{0}^{\arccos(\frac{3}{5})} \int_{0}^{2 \pi} {25 \sin \theta \, d \theta \, d \varphi} = 50 \cdot 2 \pi \cdot (-(\frac{3}{5}-1)) = 40 \pi$$ (if i'm not mistaken, of course)