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Double integrals using polar coordinates

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles
    x^2 + y^2 = 4
    and
    x^2 - 2x + y^2 = 0

    2. Relevant equations



    3. The attempt at a solution

    for my integral i got
    0<= theta <=pi/2 for the theta bounds since it lies in the 1st quad
    and i got
    2costheta<= r <= 2 for the bounds of r because 2costheta is the polar version of
    x^2 - 2x + y^2 = 0 and 2 is 'r' in the equation x^2 + y^2 = 4

    and the equation that i integrated is (2- 2costheta)r
    you add the extra r when turning the equation polar and the equation r=2 is over
    r = costheta when i graphed it
    and welll...
    i got pi - 3 but it's wrong...

    am i even setting up the equation right or am i just having problems with the math?
     
  2. jcsd
  3. Nov 12, 2009 #2

    Mark44

    Staff: Mentor

    This is a double integral, so what you want is this:
    [tex]\int_{\theta = 0}^{\pi/2}~\int_{r = ?}^{?} r~dr~d\theta[/tex]

    IOW, the limits of integration describe the region, and all you need for the integrand is 1dA. You seem to have a handle on how to describe the region, so use what you have found for the limits for r.

    I get [itex]\pi/2 - 1/2[/itex] for the area.
     
  4. Nov 15, 2009 #3
    r u serious??
    why is it do i integrate 1dA?
    i saw that in the book for finding the area of one of the clovers of a cosine function
    but it didn't explain why i just integrate 1dA
     
  5. Nov 15, 2009 #4

    Mark44

    Staff: Mentor

    Yes, I'm serious. For a region R, this double integral gives the area:
    [tex]\int \int_R 1 dA[/tex]

    In rectangular coordinates, the iterated integral will look something like this:
    [tex]\int_{y = a}^b \int_{x = f(y)}^{g(y)} 1 dx~ dy[/tex]

    and dA = dx dy or dA = dy dx. If you integrate in the reverse order, the limits of integration will be different.

    In polar coordinates, dA = r dr d[itex]\theta[/itex]. That's where the r came from.
     
  6. Nov 17, 2009 #5
    oh right. I'm looking for the area.
    double integral of 1da gives the area and triple integral of 1 gives the volume of a 3d object.
    thank you
     
  7. Nov 17, 2009 #6

    Mark44

    Staff: Mentor

    double integral of 1dA gives the area and triple integral of 1dV gives the volume of a 3d object.
     
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