1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double integral polar cordiantes

  1. Nov 16, 2012 #1
    Hi, I need help with this problem

    Evaluate the given integral by changing to polar cordinates

    ∫∫xydA where D is the disc with centre the origin and radius.

    My solution so far.

    I believe this would give a circle with radius 3 in xy plane. And then x=r*cos(θ) and y=r*sin(θ)

    So ∫(∫((r*cos(θ)*r*sin(θ))*r,r,0,3),θ,0,2∏)

    But the result is suppose to be zero.
    What am I doing wrong?
     
  2. jcsd
  3. Nov 16, 2012 #2
    Ohh never mind, the result is correct :P
    Just me who needs to learn how to use my calculator.

    I guess it did not make seance that a circle with radius 3 would give 0 as a result, but perhaps it's because z=x*y --> z=0 <> 0=x*y so x=0/y --> x=0 and y=0/x --y=0
    Is that a correct interpretation?
     
  4. Nov 16, 2012 #3

    Mark44

    Staff: Mentor

    And radius what?
    That's what I get.

    Why do you think that z = 0?
    No.
    Edit: Added the inner integral.
    The integral looks like this:
    $$ \int_{\theta = 0}^{2\pi}\int_{r=0}^3 rcos(\theta) \cdot rsin(\theta) r~dr~d\theta$$

    If you carry out the integration, you get a value of 0.
     
    Last edited: Nov 16, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Double integral polar cordiantes
  1. Polar Double Integrals (Replies: 1)

Loading...