Compute $\iiint_A x dV$ where A is bounded in the first octant by $y = x^2$ and the plane $y + z = 1$
So basically I drew the region and the integration set up (what I care about, I don't care about the results)
$\int_{0}^1 \int_{0}^{x^2}\int_{0}^{1-y}x \;dz dydx$
In my progress I wrote some scratch work that $y + z = x^2 + z = 1 \implies z = 1 - x^2$
So somehow the intersection of the plane and the parabolic cylinder is a curve that lies on the xz-plane, but in my sketch the intersection is "inside" the parabolic cylinder and lines on the plane $y+z=1$. So what is the significance of $z = 1-x^2$?
EDIT: figured it out, $z = 1-x^2$ is the projection of the intersection onto the xz-plane.