Results 1 to 3 of 3

Math Help - Find the volume of the region R between two surfaces

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    232

    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 R is bounded above, and which surface the region R is bounded below. Anyhow, here is the question.

    Find the volume of the region R between the surfaces z=3x^2+y^2 and 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:
    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 z=0 (I think), I got the following:
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    177
    If you have f(x,y) \leq z \leq g(x,y), then the volume can be calculated as V = \iint_R (g(x,y)-f(x,y))dxdy

    where R is x^2+y^2 = 1

    It comes from the fact that V = \iiint_{f(x,y)}^{g(x,y}dxdydz
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    232
    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:
    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 R.
    The surfaces intersect on the cylinder x^2+y^2=1.
    As such, the top surface is z=2+x^2-y^2, and the bottom surface is z=3x^2+y^2.
    Thus, the integrand is as follows:
    (2+x^2-y^2)-(3x^2+y^2)=2(1-x^2-y^2)
    Converting to polar coordinates, the integrand becomes the following:
    2\pi\cdot 2(1-r^2)r=4\pi (r-r^3)
    Evaluating at 0\leq r\leq 1, we integrate and obtain the following:
    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 R is \pi.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 23rd 2010, 03:50 AM
  2. find the volume of the region...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 4th 2010, 04:51 PM
  3. Replies: 7
    Last Post: March 27th 2010, 01:18 PM
  4. Find the volume of this region!
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: June 23rd 2009, 01:24 PM
  5. volume of a region bounded by two surfaces
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2008, 02:06 PM

Search Tags


/mathhelpforum @mathhelpforum