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: $$ V=2\int _{0}^1\int_{x^2}^{2-x^2}\int_0^{y+3} dzdydx $$