Volume using triple Integrals, cylindrical coordinates I want to calculate the volume of a solid with $z+1\ge x^2+y^2$ and $3\left(z-1\right)\le -\left(x^2+y^2\right)$. After cylindrical coordinates 
x=rcosϕ, y=rsinϕ
I got $r^2-1\le z$, and $z\le \frac{\left(3-r^2\right)}{3}$ , so $r^2-1\le z\le \frac{\left(3-r^2\right)}{3}$.
Then $r^2-1\le \frac{\left(3-r^2\right)}{3}$  ==> $0\le r\le 1$, $0\le \phi \le 2\pi $. After calculating the integral I got $\frac{4\pi }{3}$, which doesn't check the answer considered correct by the one who made the exercise, which is $\frac{3\pi }{2}$. I don't understand what I'm doing wrong. Or could that answer be wrong? Thanks.

Since $r^2-1=1-\frac{r^2}{3}\implies r^2=\frac{3}{2}\implies r=\sqrt{\frac{3}{2}}$,
$V=\displaystyle\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\int_{r^2-1}^{1-\frac{r^2}{3}} r\;dz dr d\theta=\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{2}}}\left(2r-\frac{4}{3}r^3\right)\;dr d\theta$
$\displaystyle\;\;=\int_0^{2\pi}\left[r^2-\frac{1}{3}r^4\right]_0^{\sqrt{\frac{3}{2}}}=2\pi\left(\frac{3}{2}-\frac{3}{4}\right)=\frac{3\pi}{2}$.
