# Thread: Find the volume of the region R between two surfaces

1. ## Find the volume of the region R between two surfaces

My Prof. probably didn't think this one through too much. I found a similar question to this one here, yet from it I discerned an issue with the question (which is listed below, word-for-word as usual): it doesn't state which surface the region $\displaystyle R$ is bounded above, and which surface the region $\displaystyle R$ is bounded below. Anyhow, here is the question.

Find the volume of the region $\displaystyle R$ between the surfaces $\displaystyle z=3x^2+y^2$ and $\displaystyle z=2+x^2-y^2$.
(I know, the wording is horrible; so open to loopholes)

All I really know is that I have to get to here somehow:
$\displaystyle V=\iint_R f(x,y) dx dy$?

So far, here is what I have, borrowing bits from the link above.

For the point of intersection between the two surfaces, setting $\displaystyle z=0$ (I think), I got the following:
$\displaystyle 3x^2+y^2=2+x^2-y^2\Rightarrow 0=2x^2+2y^2-2\Rightarrow 0=x^2+y^2-1$
From here, however, I'm not sure what to do. It's very little, I know, but I'm just not getting this material drilled in right because of a lack of legible examples.

2. If you have $\displaystyle f(x,y) \leq z \leq g(x,y)$, then the volume can be calculated as $\displaystyle V = \iint_R (g(x,y)-f(x,y))dxdy$

where $\displaystyle R$ is $\displaystyle x^2+y^2 = 1$

It comes from the fact that $\displaystyle V = \iiint_{f(x,y)}^{g(x,y}dxdydz$

3. Okay, I've figured this one out thanks to another source. Here's my answer (I might've skipped a step or two, or gotten a step mixed up):

First, we determine the point of intersection between the two surfaces:
$\displaystyle 3x^2+y^2=2+x^2-y^2\Rightarrow 2=2x^2+2y^2\Rightarrow 1=x^2+y^2$, which is the region $\displaystyle R$.
The surfaces intersect on the cylinder $\displaystyle x^2+y^2=1$.
As such, the top surface is $\displaystyle z=2+x^2-y^2$, and the bottom surface is $\displaystyle z=3x^2+y^2$.
Thus, the integrand is as follows:
$\displaystyle (2+x^2-y^2)-(3x^2+y^2)=2(1-x^2-y^2)$
Converting to polar coordinates, the integrand becomes the following:
$\displaystyle 2\pi\cdot 2(1-r^2)r=4\pi (r-r^3)$
Evaluating at $\displaystyle 0\leq r\leq 1$, we integrate and obtain the following:
$\displaystyle V=4\pi \int_0^1 (r-r^3)dr = 4\pi [\frac{1}{2}r^2-\frac{1}{4}r^4]_0^1=2\pi (1-\frac{1}{2})=\pi$

Therefore, the volume of the region $\displaystyle R$ is $\displaystyle \pi$.