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..
Best Answer
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. $$V=\int_{y=-\sqrt{2}}^{\sqrt{2}} \int_{x=-1}^{1-y^2} (1-x^2) dx dy$$