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Double integral in polar coordinate

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data
    With a > 0, b > 0, and D the area defined by

    [tex]D: \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1[/tex]

    Change the integral expression below:

    [tex]\iint\limits_D (x^2+y^2) dx\,dy[/tex]

    by using x = a r cos θ, y = b r sin θ. After that evaluate the integral.

    3. The attempt at a solution

    First of all, I can already visualize that the area is an ellipse, and because of that if we want to express the outside boundary in polar coordinates, it is

    [tex]r_1(\theta) = \frac{ab}{\sqrt{(b \, cos \theta)^2+(a \, \sin \theta)^2}}[/tex]

    Therefore, since x^2 + y^2 = r^2, the integral can be expressed in terms of θ and r as:

    [tex]\int_0^{2\pi} \int_0^{r_1(\theta)} \! (r^2)r \, dr\,d\theta[/tex]

    I've also understood how this construct comes from the Riemann sums.

    However, I don't understand the substitution hinted from the problem. For every random point (r, θ), clearly x = r cos θ and not x = a r cos θ! Can someone give a geometric interpretation/explanation, and from that show how one changes the integral expression?

    Thanks a lot.
  2. jcsd
  3. Jul 31, 2010 #2
    x = r cos θ and y = r sin θ maps out a circle, not an ellipse.

    The equation of an ellipse is:

    x = a cos θ
    y = b sin θ

    Substitute these into your integral.
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