Volume of $x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1$ Find the volume of the solid defined by $$x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1.$$

I found the intersection of surfaces:
$$S\equiv\begin{cases}
x^2+y^2&=4\\
2+x^2+y^2&=z\\
z&=-1
\end{cases}
\equiv
\begin{cases}
x^2+y^2&=4\\
2+4&\neq-1\\
z&=-1.
\end{cases}$$
The surfaces are not cut, and therefore do not define a volume in space. Is it right?
Thanks!

FALSE. Not all three have to intersect; notice that the surfaces $x^2+y^2=4$ and $z=x^2+y^2+2$ intersect, and the surfaces $x^2+y^2=4$ and $z= -1$ intersect, forming a cylinder-like volume with a circular base and a concave paraboloid-shaped "lid."
You may calculate the volume by converting to polar coordinates and considering the integral
$$\int_0^{2\pi} \int_{0}^{2} (r^2+3)\cdot r\space dr\space d\theta$$
