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Iterated integrals converted to polar

  1. Dec 8, 2013 #1
    The problem statement, all variables and given/known data
    ∫[itex]^{4}_{0}[/itex] ∫[itex]^{√(4y-y^{2})}_{0}[/itex] (x2) dx dy


    The attempt at a solution

    I'm confused on how to convert the bounds into polar coordinates.

    I believe x2 just becomes r2cos2θ

    0≤x≤√(4y-y2)
    0≤y≤4

    but i don't know how to convert the bounds
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    Mark44

    Staff: Mentor

    Have you drawn a sketch of the region over which integration is taking place? That's very important when you're changing from one coordinate system to another.
     
  4. Dec 8, 2013 #3
    ya I came up with a half circle in the 1st quadrant thats center is on 2 and it's radius is 2 but i can't figure out how to use that to find my bounds. If the center was on 0 I would be a lot more comfortable
     
  5. Dec 8, 2013 #4
    I plugged in the values of x and y for polar coordinates and came up with r=4sinθ so would my r bounds simply go from 0→4sinθ?
     
  6. Dec 8, 2013 #5

    Mark44

    Staff: Mentor

    Your description isn't very clear, but I think you understand what the region looks like.

    The equation of your circle is x2 + y2 - 4y = 0, where x ≥ 0 (for the right half).

    Converting to polar form, we get r2 - 4rsin(θ) = 0, or r(r - 4sin(θ)) = 0. Since there is a value of θ that is paired with r = 0, we don't lose any solutions by dividing by r.

    The polar equation of your circle is r = 4sin(θ).
     
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