Results 1 to 2 of 2

Math Help - Polar Coordinate Double Integral Area Problem

  1. #1
    Junior Member
    Joined
    Jun 2009
    Posts
    68

    Polar Coordinate Double Integral Area Problem

    The answer in the book is 1/4. I googled, and secant cubed is integration by parts. Before I went through all that, suspecting I set something up incorrectly, I used my ti-89 to calculate a numeric estimate. The number is negative and is not of the magnitude 1/4. What did I do wrong? Edit: My picture says tan when it should say arctan.



    Last edited by 1005; July 18th 2009 at 08:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Try it this way:

    \frac{1}{((rcos{\theta})^{2}+(rsin{\theta})^{2})^{  2}}=\frac{1}{r^{4}}

    \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\int_{sec(\theta)}^{\sqrt{  2}}\frac{1}{r^{3}}drd{\theta}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Polar Double Integral Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 9th 2011, 11:31 AM
  2. Double Integrals (polar coordinate)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2011, 10:19 AM
  3. Replies: 5
    Last Post: August 10th 2010, 11:15 PM
  4. Polar Area using a Double Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 25th 2010, 11:26 PM
  5. Replies: 1
    Last Post: August 12th 2009, 08:26 AM

Search Tags


/mathhelpforum @mathhelpforum