Triple integral: Find the volume bounded above paraboloid $z= x^2 + y^2$ and below sphere $x^2 + y^2 + z^2 = 2$?

Question

can anyone help me solve this problem for my math homework? I'm desperate and need help.

Answer 1

I'll work with cylindrical coordinates because the symmetry of the problem.

So you have the paraboloid $z = r^2$ and above it the sphere $ r^2 +z^2 = 2 \Rightarrow z^2 = 2-r^2 $

When do they intersect?

$$ z = r^2 \Rightarrow z^2=r^4 , z^2 = 2-r^2 \Rightarrow r^4 = 2-r^2 \Rightarrow r^4+r^2 - 2 =0 $$ $$ (r^2+2)(r^2-1)=0 \Rightarrow r=1$$

So when $ 0\leq r \leq 1 $ we need to integrate in the region where $ r^2 \leq z \leq \sqrt{2-r^2} $:

$$\int_0^1 \int_{r^2}^{\sqrt{2-r^2}} \int_0^{2\pi} rd\theta dz dr =\int_0^1 \int_{r^2}^{\sqrt{2-r^2}} 2\pi r dz dr = \int_0^1 2\pi (r\sqrt{2-r^2}-r^3)dr $$

And that is a simple 1D integral. Hope it helped you understand!