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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

    HallsofIvy

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    Staff Emeritus
    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

    Mark44

    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. .


     
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