Find the volume of the region in the first octant bounded by $z=x^2+2$, $x=y=z=0$, and $x=2-y$. Let the order of integration be $dzdxdy$.
Here's what I did: $\int_0^2\int_0^{2-y}\int_2^{x^2+2} dzdxdy$.
Is the triple integral correct so far? I'm unsure of the values for $z$ since it was given that $z=0$ but graph of the parabolic cylinder starts at $z=2$.
The region is bound by $x = 0, y = 0, z = 0$, $z = x^2 + 2$ and $x+y =2 $
So $z$ is bound below by xy-plane $(z = 0)$ and above by the parabolic cylinder. Rest of the bounds are correct.
The correct integral to find volume is,
$ \displaystyle \int_0^2\int_0^{2-y}\int_0^{x^2+2} dz~dx~dy = \frac{16}{3}$