I am trying to determine the area of integration of the solid represented by:
$$S=\{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2 \le 4; \frac{1}{\sqrt{3}}x \le y \le \sqrt{3}x\}$$
From this I gathered that there are three regions represented:
$$ x^2+y^2+z^2=4 \rightarrow \text{ sphere of radius 2 } \\ y=\frac{1}{\sqrt{3}}x \rightarrow \text{ plane parallel to the z axis } \\ y = \sqrt{3}x \rightarrow \text{ plane parallel to the z axis } $$
With this, I have tried the following points in the conditions of $S$ to determine which area of the solid I will be working with:
$$ (2, 0, 0) \rightarrow \text{ belonging to the sphere } \\ (0, 2, 0) \rightarrow \text{ belonging to the sphere } \\ (0, 0, 2) \rightarrow \text{ belonging to the sphere } \\ $$
From which only the point $(0, 0, 2)$ fulfills all three conditions. How do I proceed from here?
Looking from the +z axis, two planes look like this, and together with sphere they bound the red area.
Insight solution: I can now observe how volume slices (that extend in z direction, blue area on picture) change depending on $r$ (distance from center). Express the volume $dV$ of each slice as a function of $r$ and I would remain with single integral.
$\int_{0}^{R}dV=\int_{0}^{R}(r*\alpha)*h(r)dr$.
Here $\alpha$ is angle of red area, and $h(r)$ is the height of each slice that depends on $r$. $r$ is the variable of integration and $R$ is sphere radius $=2$.
Longer solution: Use double integral with variables of integration $r$ and $\phi$. As height of the volume in question is constant with $\phi$ it resolves as constant factor $\alpha=\phi2-\phi1$ used above.
$\int_{0}^{R}\int_{\phi1}^{\phi2}h(r, \phi)d\phi dr= \int_{0}^{R}h(r)\int_{\phi1}^{\phi2}d\phi dr = \int_{0}^{R}h(r)(\phi2-\phi1)dr $