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Evaluate the double integral

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    As with my other recent posts, I just want to check if I'm right or wrong as I don't have an answer scheme to go by.

    For this question I simply converted to polar to get:

    ∫∫(a+a)r drdθ

    for 0<r<a, 0<θ<2π ,

    which solved to give

    2πa3 .

    Was that correct? I'm convinced I did something wrong because it seems a very easy 7 marks.

  2. jcsd
  3. May 5, 2014 #2
    You almost got it right. You made a silly arithmetical error. The integrand should be (a+r)r not (a+a)r.

    ##\int_0^a\int_0^{2\pi}(a+r)r d\theta dr##
    Last edited: May 5, 2014
  4. May 5, 2014 #3
    Aaah right ok. Hmmm maybe a silly arithmetic error, but I can't see why it should be (a+ar)?

    For converting to polar I did, dxdy = r drdθ,

    which I presumed would then give the expression I used.

    I can't see where I went wrong?
  5. May 5, 2014 #4
    Sorry it should be this ##\int_0^a\int_0^{2\pi}(a+r)r d\theta dr##. What is ##\sqrt{x^2+y^2}## after you transform it into polar coordinates? What is ##x## and ##y## in terms of ##r## and ##\theta##?
  6. May 5, 2014 #5
    Ah yeah I think I follow now. I think the x2 + y2 = a2 thing threw me a bit, making me think I should just swap it for an 'a' when really I can't do that at all.

    It should be

    f(x,y)dxdy → f(rcosθ,rsinθ)r drdθ

    So in the case of the question

    [a + √(x2 + y2)] dxdy → [(a + r)r] drdθ
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