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Double integral: Cartesian to Polar coordinates

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫∫√(x^2+y^2)dxdy with 0<=y<=1 and -SQRT(y-y^2)<=x<=0

    2. Relevant equations

    x=rcos(theta)
    y=rsin(theta)

    3. The attempt at a solution

    0.5<=r=1, we get r=0.5 from -SQRT(y-y^2)<=x by completing the square on the LHS
    then, 0<=theta<=pi

    But, when I calculated the original integral, I got 0.2222...
    When I calculate the integral with the polar coordinates, I get 0.92
     
  2. jcsd
  3. Dec 13, 2013 #2

    ShayanJ

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    Maybe you forgot the determinant of the Jacobian?
    Anyway,without taking a look at your calculations,we can't help!
     
  4. Dec 13, 2013 #3
    Last edited: Dec 13, 2013
  5. Dec 13, 2013 #4
    U need a factor of r. da=rdrdtheta
     
  6. Dec 13, 2013 #5

    CAF123

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    Did you try to plot the region of integration? How did you get those limits?
     
  7. Dec 13, 2013 #6
    Explanation:

    I started out with the graph. In the first quadrant we have a quarter of a circle x^2+y^2=1 (radius = 1. In the second quadrant we have x^2+(y-0.5)^2=0.25 (radius=0.5). Then, since the function exists both in first and second quadrant, 0<=theta<=pi. Then, by my false reasoning, I made a mistake of thinking that 0.5<=r<=1.

    After revising my calculations, I got 0<=theta<=pi/2, 0<=r<=1. I derived the limits for theta in the following way
    0<=rsin(theta)<=1
    -SQRT(rsin(theta)-(rsin(theta))^2)<=rcos(theta)<=0
    rsin(theta)-(rsin(theta))^2=(rcos(theta))^2
    sin(theta)=r
    Then, by substituting into the first line, we have
    0<=(sin(theta))^2<=1
    Then, theta1=0, theta2=pi/2
    Hence, r1=0, r2=1
     
  8. Dec 13, 2013 #7

    CAF123

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    Those bounds describe a portion of the unit circle centered on the origin in the first quadrant.
     
  9. Dec 13, 2013 #8
    So, to get to the right conclusion do I add the area of the half of the circle centered at (0,0.5)?
     
  10. Dec 13, 2013 #9

    CAF123

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    Are you sure you are considering the right half of the curve x = ±√(y-y2)? For sure, this exists in both first and second quadrants, but you are only interested in the half x = -√(y-y2). Since √(y-y2) ≥ 0, what quadrant does x=-√(y-y2) lie?
     
  11. Dec 13, 2013 #10
    In the second quadrant = > (pi/2)<=theta<=pi
    0<=r<=0.5
     
  12. Dec 13, 2013 #11

    CAF123

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    I agree on the bounds for θ, but the values for r suggest a circle of radius 1/2 centered on the origin. Consider the curve x2+y2=y. Convert this to polar coordinates and you will obtain r. This value of r will depend on θ.
     
  13. Dec 13, 2013 #12
    Thank you for correction. If (pi/2)<=theta<=pi, does it meant that sin(pi)<=r<=sin(pi/2); 0<=r<=1.

    that's weird
     
  14. Dec 13, 2013 #13

    CAF123

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    These bounds say that as θ goes from π/2 to π, r goes from 0 to 1. r is measured from the origin and so r is precisely 1 at only one value of θ, namely θ = π/2. (See this from a sketch). So the bounds for r have to depend on θ, otherwise you are describing a portion of the unit circle centered at the origin.
     
  15. Dec 13, 2013 #14
    That's great!

    Does this mean that we can write the integral as a sum of two integrals (first quadrant circle and second quadrant circle)
    0<=theta<=n/2, 0<=r<=1 + n/2<=theta<n, 0<=r<=sin(theta)
    ?

    OR( 0<theta<n (0<=r<=1 + 0<=r<=sin(theta))
     
    Last edited: Dec 13, 2013
  16. Dec 13, 2013 #15

    CAF123

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    Why did you tag on the part '0<=theta<=n/2, 0<=r<=1'? Remember x ≤ 0 so there should be nothing that describes anything in the first quadrant.

    The bounds 'n/2<=theta<n, 0<=r<=sin(theta)' describe the required region. Did you draw a sketch? Convince yourself that this is the correct range for r - try out a few values of θ and see that you get reasonable values for r (r should monotonically decrease to 0 as θ sweeps from π/2 to π)
     
  17. Dec 13, 2013 #16
    RIGHT! Function is undefined for X>0.

    Therefore, n/2<=theta<=n, and 0<=r<=sin(theta)
     
  18. Dec 13, 2013 #17

    CAF123

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    Yes, everything makes sense?
     
  19. Dec 13, 2013 #18
    Out of curiosity, why does not r go between 0 and csc(theta)?
     
  20. Dec 13, 2013 #19

    CAF123

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    The curve in Cartesians was ##x^2 + y^2 = y##. Converting this to polar coordinates gives ##r^2 = r\sin\theta## from which we see that ##r = \sin \theta##. ##\theta = \pi/2## is the maximum value for ##r##, while ##\theta = \pi## is the minimum. In between, ##r## takes values given by the sine of the angle ##\theta##.
     
  21. Dec 13, 2013 #20
    EXCELLENT! THANKS!

    I believe my mistake was looking at 0<=rsin(theta)<=1.

    Thank You for your help and may numbers be in your favor.
     
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