Results 1 to 4 of 4

Math Help - Setting up triple integral

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    12

    Setting up triple integral

    Edit: Title should read double integral... oops

    Having a little trouble figuring this one out.

    Region bounded by 4x^2 + y^2 = 4z and z = 2

    Thanks.
    Last edited by PMatt; December 9th 2009 at 04:19 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    The two surfaces intersect when

    4x^2+y^2=4\cdot 2=8.

    What shape does this define? Would it be easier to use Cartesian or polar coordinates?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    12
    I was assuming that I should use polar coordinates, but I'm still not quite sure how to set it up.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    The equation that bounds the region of integration can be rewritten

    \left(\frac{x}{\sqrt{2}}\right)^2+\left(\frac{y}{2  \sqrt{2}}\right)^2=1.

    This defines an ellipse whose equation in polar coordinates is

    r=\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 18th 2010, 06:04 AM
  2. Need some help setting up a triple integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 13th 2010, 06:05 PM
  3. Replies: 3
    Last Post: July 6th 2010, 01:00 PM
  4. setting up a triple integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 5th 2009, 05:56 PM
  5. Replies: 1
    Last Post: April 19th 2009, 01:13 AM

Search Tags


/mathhelpforum @mathhelpforum