Find the area of the lateral surface defined by two sets Let $S$ be the lateral surface of the set
$$
\left\{x^2 + y^2 +z^2 \leq 1, y \leq \frac{|x|}{\sqrt{3}}\right\} \bigcup
\left\{x^2 + y^2 +z^2 \leq 1, y \geq \frac{1}{2}\right\} = S_1 \cup S_2.\ \mbox{Evaluate} \left\vert\,{S}\,\right\vert.
$$

*

*I think I need to define a curve and then evaluate the surface integral over this curve, but I don't know how to parametrize it.


*In spherical coordinates, $x=r\sin\left(\phi\right)\cos\left(\theta\right), y = r\sin\left(\phi\right)\sin\theta, z = r\cos\left(\phi\right)$.
I came to the conclusion that

*

*in $S_1$ whe have $0 \leq \phi \leq \pi,\ -7\pi/6 \leq \theta \leq \pi/6$

*and in $S_2$ we have $0 \leq \phi \leq \pi/3,\ 0 \leq \theta \leq 2\pi$.



 A: First check if there is any common surface between $S_1$ and $S_2$ as $S = S_1 \cup S_2$.
Please note that if $y \geq \frac{1}{2}$, $|x| \geq y \sqrt3 \geq \frac{\sqrt3}{2}$ but $x^2+y^2+z^2 = 1$ and that shows $z = 0$ for $x = \frac{\sqrt3}{2}, y = \frac{1}{2}$. So we cannot simultaneously have $y \gt \frac{1}{2}, x \gt \frac{\sqrt2}{2}$.
Hence there is no common surface between $S_1$ and $S_2$. Once we find their surface area, we can simply add them.
You already found the right bounds. Saw your edit later.
To find $S_1$, please note that $y \leq \frac{|x|}{\sqrt3} \implies \sin\theta \leq \frac{|\cos\theta|}{\sqrt3}$. That leads to,
$0 \leq \theta \leq \frac{\pi}{6}, \frac{5\pi}{6} \leq \theta \leq 2\pi$. Bound of $\phi$ does not change: $0 \leq \phi \leq \pi$.
For $S_2$, my suggestion is to use $y-$axis for polar angle instead of $z$ given the plane is $y = \frac{1}{2}$. At $\rho = 1, \cos\phi = \frac{1}{2}, \phi = \frac{\pi}{3}$
So bounds are $0 \leq \phi \leq \frac{\pi}{3}, 0 \leq \theta \leq 2\pi$.
Edit:
As far as your question on integral, for a sphere it is simply $\iint \sin\phi \ d\phi \ d\theta$ as $\rho = 1$. You can find it as below,
for $S_1$,
$r(\theta, \phi) = (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$
$r'_{\theta} = (-\sin\theta \sin\phi, \cos\theta \sin\phi,0)$
$r'_{\phi} = (\cos\theta \cos\phi, \sin\theta \cos\phi,-\sin\phi)$
$|r'_{\theta} \times r'_{\phi}| = \sin\phi$
As you can see, it will be same for $S_2$.
