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Math Help - Double integration to find the area

  1. #1
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    Double integration to find the area

    Use a double integral to find the area of the region.
    The region inside the cardioid r = 1 + cos(θ) and outside the circle r = 3cos(θ).

    I found the boundaries to be from pi/3 to 5pi/3 and 3costheta to 1+costheta. But I'm not getting the right answer.. after integrating everything, I get the equation- 1/2 (3theta+2sin2theta-2sintheta). Plugging in the boundaries (5pi/3 and pi/3) I get 10pi/6, but that's not correct. Did I integrate wrong?
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  2. #2
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    It looks like you've integrated correctly, but inserting those limits does not give \frac{10\pi}{6}
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  3. #3
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    I keep getting -2pi, but that isn't right either. Am I doing something wrong here?

    The boundaries for r: 3cos theta to 1+cos theta
    The boundaries for theta: pi/3 to 5pi/3, found by setting the two r's equal to eachother.

    Then, I set up the integral, and got:
    the integral from pi/3 to 5pi/3 (1/2)(1+2cos theta-4theta-4cos 2theta.
    Plugging in pi/3 and 5pi/3 I keep getting -2pi...

    [\frac{5\pi}{3} + 2 \sin (\frac{5\pi}{3}) - 4(\frac{5\pi}{3}) - 2 \sin(\frac{10\pi}{3})] - [\frac{\pi}{3} + 2 \sin (\frac{\pi}{3}) - 4(\frac{\pi}{3}) - 2 \sin(\frac{2\pi}{3})]
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  4. #4
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    If you get a negative area that means that you've got the two r functions in the wrong order. Recall that


    \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx

    What answer does your textbook give?
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  5. #5
    Super Member Random Variable's Avatar
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    Is the answer  \pi \over 4 ?
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