Volume of $x^2+y^2\leq4,\quad z=2+x^2+y^2,\quad z\geq-1$

Question

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

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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!

Answer 1

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 $$\int_0^{2\pi} \int_{0}^{2} (r^2+3)\cdot r\space dr\space d\theta$$