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

  1. #1
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    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.
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  2. #2
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    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 $\displaystyle V = \iiint_{f(x,y)}^{g(x,y}dxdydz$
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  3. #3
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    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 $\displaystyle \pi$.
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