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Double Integrals in Polar Coordinates

  1. Mar 16, 2015 #1
    I'm in the middle of the Great Courses Multivariable Calculus course. A double integral example involves a quarter circle, in the first quadrant, of radius 2. In Cartesian coordinates, the integrand is y dx dy and the outer integral goes from 0 to 2 and the inner from 0 to sqrt(4-y^2). In polar, the integrand is (r sin theta) r dr dtheta, with the outer going from 0 to pi/2 and inner from 0 to 2. The final result, in both forms, is 8/3.

    My thinking was that this would describe the area of the quarter circle, but elementary geometry tells me the area should be pi. I'm trying to visualize what this value of 8/3 represents. Maybe this sounds stupid, but I could use some help.
  2. jcsd
  3. Mar 16, 2015 #2
    When you calculate a integral like that you ain't calculating the area. you are calculating the volume beneath ##f(x,y) = y## in the quarter circle.
    ##\iint 1dxdy## would be the area.
  4. Mar 16, 2015 #3


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    If you were calculating just the area of a quarter circle, the double integral would just be:

    A = ∫∫ dx dy, since dA = dx dy and ∫∫ dA ≡ A

    But, the integral which is actually being evaluated here is

    I = ∫∫ y dx dy = ∫∫ y dA

    In effect, I is the first moment of the area of the quarter circle about the x-axis.
  5. Mar 16, 2015 #4
    Thanks for the quick reply. I'm still unable to visualize what the "area under y" looks like. What would it's graph be?
  6. Mar 16, 2015 #5
    sorry, edited my post quickly after. meant volume!
  7. Mar 16, 2015 #6

    Can this be graphed in 2 dimensions?
  8. Mar 16, 2015 #7
    So this involves the z axis?
  9. Mar 16, 2015 #8


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    Not really. Since f(x,y) = y for this integral, you would need to add a third dimension. The base of the figure is the quarter circle of radius 2 centered at the origin, with the heights of the surface above the base being z = y.

    I think it's better to interpret this double integral as representing the first moment of area of the quarter circle about the x-axis. You don't give yourself a headache trying to visualize a 3-dimensional surface. There are more complicated integrals than this for which there are no easy visualizations.
  10. Mar 16, 2015 #9
    Not really no, or at least i don't think so. But the surface f(x,y) = y would just be a plane. and you simply sum up the volume beneath it in the quarter circle.
    Think of it as a circle in the plane with ##x^2+y^2=2## and a plane above it ##z-y=0 , x \in R##.

    When doing multivariable calculus I find you really need to learn to draw 3d graphs, it really helps (I just finished the course myself only a week ago). If you got a lecturer, watch how he/she draws it on the board and try to copy it.
  11. Mar 16, 2015 #10
    Thanks, but I'm still trying to visualize. It sounds like, if this value is a volume, its average height on the z axis would be about .85. Is this in the right ballpark?
  12. Mar 16, 2015 #11
    Thanks much. I have some 3D graphing software, so I'll spend some more time with it. It's just good to know that the integral was not as simple as a 2D area. I was losing some marbles trying to reconcile 8/3 with pi :-)
  13. Mar 16, 2015 #12
    Yes that would be about right. 8/(3pi) should be the average height. you could think of it that y goes up to 2. and the average would be about 1 but since its a circle and not a square theres less area where y is big then where it is small so we get a bit less.
  14. Mar 16, 2015 #13


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    For a quarter circle, the centroid is located at x-bar = y-bar = 4 R / (3 π), so the first moment of area M = A * y-bar = [π R2/ 4] * [4 R / (3 π)]

    M = R3 / 3

    When R = 2, then M = 23 / 3 = 8/3

    The π's cancel out, which is why π doesn't show up in the evaluation of this double integral.
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