1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    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


    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


    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


    Staff: Mentor

    double integral of 1dA gives the area and triple integral of 1dV gives the volume of a 3d object.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook