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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.

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Curve $z=1-x^2$ is the projection onto $xz$-plane of intersection between plane $y+z=1$ and parabolic cylinder $y=x^2$ – M. Strochyk Dec 22 '12 at 7:47
You can use double dollar signs to get displayed equations that look nicer, especially with integrals, and are centred. – joriki Dec 22 '12 at 8:21
Your edit seems to imply that you answered your own question. If so, please consider writing an answer and accepting it so the question doesn't remain unanswered. If you don't want to invest the time to do that, you could delete the question instead. – joriki Dec 23 '12 at 10:58
up vote 0 down vote accepted

Note that $z=1-x^2$ is the projection of the intersection onto the $xz$-plane

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