Finding the Body Volume between two functions: $x^2+y^2+z^2=20$ and $z=x^2+y^2$ ( triangle intergral )

Question

usually I manage to solve those questions, but this one in particular is problematic, I have no other information.
If I try to use $z=x^2+y^2$ in the Ball function, I will get $(x+y)^2+x^4+y^4=20$, which is not a really possible way to solve it...
I tried geogebra, saw the shape it was a flat ball ( I dont know why... ).
Can anyone give me any tip on how to start? I dont know what I need to put at the boundary of the Third integral...
I dont have any radius.
The only think I thought of doing: using polar coordinates: $x^2+y^2=\rho^2$, thus I am getting after a few calculations: $\rho_1=\sqrt(10)$, $\rho_2=-\sqrt(10)$ So I can probably use $\rho=0$ to $\rho=\sqrt10$.
$\theta$ from 0 to $2\pi$ ( just guessing, because it has a circle shape from $x^2+y^2$ ).
but the last angle I dont know what to do...
Can somone tell me if my direction is correct \ What can I do to solve it?
My problem here is the $x^4$ and $y^4$

The answer is:$\frac 83\pi(10\sqrt5-19)$

According to help I got (my try): Okay, I got something close: $\frac {16}3\pi(10\sqrt5-16)$.
This time $\rho$ is between $0$ to $2$.
$\theta$ is between $0$ to $2\pi$
$z$ is between $\sqrt{20-\rho^2})$ with minus and plus

Answer 1

The region is axisymmetric about the $z$ axis. It you make $x=0$ and draw the intersection with the $zy$-plane you'll see very clearly what the integration limits should be. In cylindrical coordinates you would have: \begin{align*} V = &\int_{0}^{2\pi} \int_0^2 \int_{\rho^2}^{\sqrt{20-\rho^2}} \rho \,dz\, d\rho\, d\theta\\ =& 2 \pi \int_0^2 \rho(\sqrt{20-\rho^2}-\rho^2)d\rho = \cdots =\frac{8}{3} \left(10 \sqrt{5}-19\right) \pi \end{align*}

Answer 2

We know that the solid is bounded by the surfaces $z=x^2 + y^2$ and $x^2+y^2+z^2=20$. The first of these conditions also implies that $z\ge0$, and this condition together with the second boundary surface defines a hemisphere.

Substituting the two equations gives $z^2+z-20=0$, leading to $z=4$ where the two surfaces meet.

We now calculate the section of the hemisphere outside the solid.

For each $z\space \epsilon \space [0,4]$ we have an annulus of inner radius $r_1 =\sqrt{z}$ and outer radius $r_2=\sqrt{20 - z^2}$.

The area of the annulus is $A=\pi(r_2^2-r_1^2)=\pi(20-z^2-z)$.

Integrating this area over the range $z\space \epsilon \space [0,4]$ gives $$\int_0^4\pi(20-z^2-z)dz=\frac{\pi\cdot152}{3}$$ The volume of the hemisphere is $$\frac{1}{2}\frac{4\pi}{3}20\sqrt{20}=\frac{80\pi}{3}\sqrt{5}$$ Subtracting these gives $\frac{8\pi}{3}(10\sqrt{5}-19)$ as required.