Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Polar double integral

  1. Nov 1, 2006 #1
    I'm having trouble finding the function and/or the limits to this problem:

    Using polar coordinates, evaluate the integral http://ada.math.uga.edu/webwork2_files/tmp/equations/01/19aeef09224e0fca11ef9d6e45fb311.png [Broken] where R is the region http://ada.math.uga.edu/webwork2_files/tmp/equations/21/9cc2c610adbc73bdcbe1922b3dea321.png [Broken]

    I've tried the function as sin(r^2) with the limits as 3 to 6 and 0 to 2pi, but that does not give the right answer.

    That would give -1/2*cos(r^2) which gives .0893 dtheta, which is then
    .0893*2pi.

    I'm pretty sure I'm doing the integrals correctly, I think I just have the function wrong. If anyone can help me out I'd appreciate it.

    Thanks
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 1, 2006 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You realize you need to change it from dA to [tex]drd\theta[/tex] right?
     
    Last edited: Nov 1, 2006
  4. Nov 1, 2006 #3
    [tex]r dr d\theta[/tex]. But he did that, and -1/2*cos(r^2) looks right to me. Your limits look good too. I don't see where you are getting .0893 though.
     
    Last edited: Nov 1, 2006
  5. Nov 1, 2006 #4
    I evaluated 1/2*cos(r^2) from 6 to 3

    -1/2(cos(36)-cos(9))
     
    Last edited: Nov 1, 2006
  6. Nov 1, 2006 #5
    He is using degrees rather than radians?
     
  7. Nov 1, 2006 #6
    It's in radians, all the other homework has been and I just tried the answer with 360 instead of 2pi.
     
  8. Nov 1, 2006 #7
    Max meant that your calculation used degrees, when you should have used radians... and I think he's right.
     
  9. Nov 1, 2006 #8
    Ah, you are right. That worked, thanks a lot.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook