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Converting to Polar Coordinates

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Convert ∫ from 0 to 3/√2 ∫ from y to √(9-y^2) of xydxdy to polar form.


    2. Relevant equations

    x2+y2=r2

    3. The attempt at a solution

    I found the equation x2+y^2=9 from the upper range of the second integral. So r=3. Therefore r ranges from 0 to 3. The integrand is now (rcosθ)(rsinθ)rdrdθ.

    I'm having trouble with the range of integration of the first integral (0 to 3/√2). From the other information, I think that the area I'm integrating is from -π/2 to π/2. This is wrong. Any hints? Thanks.
     
  2. jcsd
  3. Nov 14, 2013 #2

    Mark44

    Staff: Mentor

    It's essential in these kinds of problems to have a good understanding of what the region of integration looks like. Can you describe, in words, what this region looks like?
    What does the lower limit of integration, y, represent?
     
  4. Nov 14, 2013 #3
    Oh. Is it the straight line x=y?

    The region I've got so far is a semi-circle from -pi/2 to pi/2. Not really sure how to deal with the x=y.
     
    Last edited: Nov 14, 2013
  5. Nov 14, 2013 #4

    Mark44

    Staff: Mentor

    Here's your integral:
    $$ \int_{y = 0}^{3/\sqrt{2}} \int_{x = y}^{\sqrt{9 - y^2}}~xy~dx~dy$$
    If you click on the integral, you can see the LaTeX I wrote to produce it.
    The inner integration goes across, from left to right, from the line to your half circle. The outer integration goes up from y = 0 to y = 3/√2.

    You should draw a sketch of this region.
     
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