Volume of a bounded solid in R3 What is the volume of the solid in xyz-space bounded by \begin{align}
    y = 2 - x^2 \\
y = x^2 \\
z = 0 \\
z = y + 3 ?
  \end{align}
I have formatted the problem as follows:
$$\iiint 1 \,dx\,dy\,dx$$
\begin{align}
-1 ≤ x ≤ 1 \\
2 - x^2 ≤ y ≤  x^2 \\
0 ≤ z ≤ y + 3 \\
\end{align}
When I solve the triple integral, though, I get a value of zero. My guess is that my limits of integration are wrong, but I need a nudge in the right direction!

The volume is symmetric with respect to the $yz$-plane, so in the integral:
$$
\int _{-1}^1\int_{x^2}^{2-x^2}\int_0^{y+3} dzdydx  
$$
the two parts $\{-1<x<0\}$ and $\{0<x<1\}$ have opposite sign and its sum is null.  You have to express the volume as:
$$
V=2\int _{0}^1\int_{x^2}^{2-x^2}\int_0^{y+3} dzdydx  
$$
