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Double integral for area evaluation

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Use an appropriate double integral and the substitution

    [tex] y = br\sin \theta \text{\ \ \ } x = ar\cos \theta [/tex]

    to calculate the bounded area inside the curve:

    [tex] {\left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right)}^2 = \frac{x^2}{a^2} - \frac{y^2}{b^2} [/tex]

    (you can assume that [tex] a,b > 0 [/tex])

    2. Relevant equations



    3. The attempt at a solution

    I began with the substitution and I got the following representation of the curve:
    [tex] r = \sqrt{ \cos^2 \theta - \sin^2 \theta } [/tex].

    That means that [tex]0 \leq r \leq \sqrt{ \cos^2 \theta - \sin^2 \theta } [/tex].

    Now I have to find from where to where [tex] \theta [/tex] goes. But according to the polar representation I get a very segmented range which is:

    [tex] 0 \leq \theta \leq \pi / 4 \text{\ \ \ } 7\pi / 4 \leq \theta \leq 2\pi \text{\ \ \ } 3\pi / 4 \leq \theta \leq 5\pi / 4 [/tex]

    And this is a problem because I need [tex] \theta [/tex] to range in a continuous range. So it can be used as a boundaries of an integral.

    So what should I do?

    Edit:
    Fixed substitution. It was x = [tex] br\cos \theta [/tex] instead of what it is now.
     
    Last edited: Jun 2, 2009
  2. jcsd
  3. Jun 2, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can integrate over the ranges separately and then add them. No law against that. It might be easier to do two ranges [-pi/4,pi/4] and [3pi/4,5pi/4] rather than three. You might also notice your r^2=cos(2*theta), so it should be pretty clear you are going to get the same result on each of those two intervals.
     
  4. Jun 2, 2009 #3

    Mark44

    Staff: Mentor

    Are you sure that the substitution is as you wrote it? That seems to be what you actually used.

    Assuming that's the case, you'll get r2 = cos2(theta) - sin2(theta) = cos(2theta).
     
  5. Jun 2, 2009 #4
    Oh no!
    You are right.
    The substitution I wrote is a mistake.
    I will edit my post to fix that.
     
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