Finding the boundary of a domain

Recently I've been practicing on the divergence's theorem and although I get the usual method I often find myself stuck in exercices when it comes to find the bounds of my integration. I'm going to give an exemple but I'm more interested in the general sense of it :

The domain is $$\Omega = \{(x,y,z)\in \mathbb{R}^3 : x^2+y^2+z^2 < 4,x^2+y^2 <3z \}$$

Then in order to apply the divergence's theorem, I need to find $$\partial\Omega$$ which is where I usually struggle.

I do understand that if you go into the cylindrical coordinates you get $$\Omega = \{(rcos(\theta),rsin(\theta),z) : \theta \in(0,2\pi), r \in(0,\sqrt3) , \frac{r^2}{3} < z < \sqrt{4-r^2} \}$$

But can't understand how you find the boundary of this. On the correction of the exercice, they draw a plane with r and z so they can place the two curve : $$z =r^2/3$$ and $$r^2+z^2= 4$$ but what is the point of drawing it ?

Also they do $$\partial\Omega = E_1 \cup E_2$$ and I get that both are a part of the boundary, but I definitely can't visualize how to get them. I've noticed that often you fix a variable so you travel on the other and vice-versa but on more complex cases like this I would like to know the usual method to arrive to finding $$E_1$$ and $$E_2$$

I hope I've made my problem clear enough and would gladly accept any help comming my way

• The boundary of the region $R_1 = \{(r,\theta,z)_{\text{cyl}} : r^2/3 < z\}$ is $B_1 = \{(r,\theta,z)_{\text{cyl}} : r^2/3 = z\}$. The boundary of the region $R_2 = \{(r,\theta,z)_{\text{cyl}} : z < \sqrt{4 - r^2}\}$ is $R_2 = \{(r,\theta,z)_{\text{cyl}} : z = \sqrt{4 - r^2}\}$. The point of drawing the regions is to see which pieces of these boundaries form the boundary of the intersection $R_1 \cap R_2$ Dec 24, 2022 at 16:39
• In general: the boundary of a region that is defined by a single inequality can be found by changing this inequality into an equation. The boundary of an intersection $R_1 \cap R_2$ of regions is some subset of the union of the boundaries $(\partial R_1) \cup (\partial R_2)$. Dec 24, 2022 at 16:41