<s>[INST] Find volume of region bounded by plane, bounded by $y^2 = 1 - x$ and $x = -1$, and parabolic cylinder, $z = 1 - x^2$ Question:
Let $R$ denote the finite plane region on the $xy$-plane bounded by the line $x = -1$ and the parabola $y^2 = 1 - x$. Let $D$ denote the solid region under the surface of the parabolic cylinder $z = 1 - x^2$ and over the plane region $R$. Find the volume of $D$. 
My steps:
First of all, I went to find the area of region R.
Substitute in $x = -1$, $y^2 = 2 - x$
Define in terms of $y$, $x = 2 - y^2$
When $x = 0$, $y = \pm \sqrt{2}$
Integrate:
$\int^\sqrt{2} _{-\sqrt{2}} \sqrt {2 - y^2} dx$ 
= $[2y - \cfrac{(y)^{3}}{3} + C]^\sqrt{2} _{-\sqrt{2}}$
= $[2\sqrt{2} - \cfrac{2^{1.5}}{3}] - [-2\sqrt{2} - \cfrac{-2^{1.5}}{3}]$
= $4\sqrt{2} - \cfrac{2^{2.5}}{3}$
After this, I'm not sure how I should proceed with finding the region bounded by this area of the plane the parabolic cylinder..
 [/INST] Note: you don't have to find the area of region $R$ first. As $D$ is the volume under the surface of the form $z=f(x,y)$, we first set the limits of $z$ and then find the projection of the given surface on $xy$-plane that is our region $R$ and then we set the limits of $x, y$.
HINT:
The volume of the solid region $D$ can be given by the triple integration as $$V= \int_{y=-\sqrt{2}}^{\sqrt{2}} \int_{x=-1}^{1-y^2} \int_{z=0}^{1-x^2} dxdydz$$
i.e. $$V=\int_{y=-\sqrt{2}}^{\sqrt{2}} \int_{x=-1}^{1-y^2} (1-x^2) dx dy$$
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