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Double Integral in polar coordinates

  1. Apr 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫∫D (3x + 4y 2 ) dA, where D = {(x, y) : y ≥ 0, 1 ≤ x 2 + y 2 ≤ 4} with the use of polar coordinates.

    2. Relevant equations

    3. The attempt at a solution
    I made a sketch of the circle. It's radius is = 1 and it's lowest point is at (0,0), highest at (0,2), leftmost point at (-1,1) and rightmost point at (1,1).

    Converting this integral into polar coordinates:
    ∫∫D (3rcosθ+4(rsinθ)2) r dr dθ

    then distributing the r

    ∫∫D (3r2cosθ+4r(rsinθ)2) dr dθ

    The outside integral goes from 0≤θ≤2π
    I'm not totally sure with the inside integral though.
    0≤r≤1? Because the maximum that the radius is is 1?

    I haven't evaluated it yet. I want to make sure I've set it up correctly first. :)
     
  2. jcsd
  3. Apr 23, 2016 #2

    LCKurtz

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    Have you drawn a picture of the full region yet? Isn't there more than one circle? What does the region ##D## look like?
     
  4. Apr 23, 2016 #3
    Sorry, I just realised there are a few formatting errors in my original post.

    Evaluate ∫∫D (3x + 4y2 ) dA, where D = {(x, y) : y ≥ 0, 1 ≤ x2 + y2 ≤ 4} with the use of polar coordinates.

    I only got one circle. But I've seen that I've made a mistake. I'm having a lot of trouble interpreting what D means and plotting the graph. :/

    I take D to mean that y is always greater than or equal to 0, so the region isn't below 0, and that x2 + y2 is between 1 and 4. But with this interpretation I get a 4 triangles, which doesn't seem right...
     
  5. Apr 23, 2016 #4

    ehild

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    Can you sketch x2+y2=1? what curve is it? And what is x2+y2=4?
     
  6. Apr 23, 2016 #5
    I can sketch both, I don't understand what you mean by 'what curve is it?' It's a circle, but D states that y has to be equal to or greater than 0, and that x2+y2 is between 1 and 4.
     
  7. Apr 23, 2016 #6

    ehild

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    Both are circles. What are the centres and the radii?
     
  8. Apr 23, 2016 #7
    The radius of x2+y2=1 is 1, and for the other the radius = 2.

    So the upper limit is 2 and the lower limit is 1. I should get two circles with this D?
     
  9. Apr 23, 2016 #8
    Ok, I understand. d is saying:

    x2 + y2 =1
    x2 + y2 =4

    Then the upper limit of the radius = 2 and lower limit = 1
     
  10. Apr 23, 2016 #9

    ehild

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    The limits for what?
    D is two-dimensional, you have an other variable, the angle θ. What are the limits for it?
    Have yo sketched those circles?
     
  11. Apr 23, 2016 #10
    The limit of integration.
    For the angle θ, because y ≥ 0, the limit of integration is between 0≤θ≤π
     
  12. Apr 23, 2016 #11

    ehild

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    View attachment 99614
    Correct. So the blue region is D.
    Convert x, y into polar coordinates.
    upload_2016-4-23_22-30-3.png
     
  13. Apr 23, 2016 #12
    3x + 4y2
    ∫∫(3rcosθ+4(rsinθ)2) r dr dθ
    distribute the r
    ∫∫(3r2cosθ+4r(rsinθ)2) dr dθ
     
  14. Apr 23, 2016 #13

    ehild

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    Correct, go ahead. You want to integrate with respect to r first, what do you get?
     
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