The title is the statement of the problem. I want to calculate the volume between the paraboloid and the cone. Over the $xy$ plane with a triple integral.
I started the exercise making a coordenates change:
$x=r \ cos \theta$ and $y=r \ sin \theta$ then
$4z=r^2$ and $(z+1)^2=r^2 → 4z=(z+1)^2 → z=1$
So $4=x^2+y^2$ and $2^2=x^2+y^2$ $→ \ 0≤r≤2$ and $0≤z≤1$ the smallest limit is $0$ because I'm over the $xy$plane and the biggest is $1$ by the previous calculation.
Then the volume can be calculated by
$$V = \int_0^{2\pi} \int_{0}^2 \int_{ 0}^1 r \ dz \ dr \ d \theta$$
I'm not sure if I found rigth the limits $r$ and $ \theta$ and $z$ of this triple integral correctly.
The diagram depicts the cross-section of the volume with $x=0$. The volume (the shaded part) can be integrated with
$$V = \int_0^{2\pi} \int_{0}^2 (z_2-z_1)r drd \theta -\int_0^{2\pi} \int_{0}^1 (0-z_1)r drd \theta$$
where $z_2=\frac14(x^2+y^2)=\frac14r^2$ and $z_1=\sqrt{x^2 + y^2}-1=r-1$. Plug them into the volume integral,
$$V = 2\pi \int_{0}^2 (\frac14r^2-(r-1))r dr -2\pi \int_{0}^1 (1-r)r dr = \frac{2\pi}3-\frac{\pi}3=\frac{\pi}3$$