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

Do the following integral?

  1. Oct 17, 2015 #1
    Do the following integral: I= (double integral) dx dy x^2 with 0<x^2 + y^2<a^2 , y>0?
    Could someone please give me some guidance as to what the inequality means? Should I convert to polar
    coordinates so it gives me 0<r^2<a^2? Thank you.
  2. jcsd
  3. Oct 17, 2015 #2


    User Avatar
    Science Advisor

    [itex]0< x^2+ y^2< a^2[/itex], y> 0 is the half disk with center at (0, 0), radius a, above the x-axis. Yes, I think polar coordinates would be simplest although it would not be that hard to do it in Cartesian coordinates. Since y> 0, [itex]x^2+ y^2= a^2[/itex] is the same as [itex]y= \sqrt{a^2- x^2}[/itex] so the integral would be [itex]\int_{x= -a}^a \int_{y= 0}^{\sqrt{a^2- x^2}} x^2 dy dx[/itex]. What would it be in polar coordinates?
  4. Oct 17, 2015 #3
    How does y >0 lead to x^2 + y^2 = a^2? Using polar coordinates the integral came to be (Sorry for lack of latex) integral of r^3 between 0 and a , times the integral of cos^2(theta) between 0 and pi. That came out to equal ((pi)a^4)/8. Does this sound right? Also could not seem to come to any answer using cartesian coordinates. thanks!
  5. Oct 17, 2015 #4


    Staff: Mentor

    It doesn't lead to ##x^2 + y^2 = a^2##. It just limits the region of integration to the portion of the disk above the x-axis. .

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook