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Thread: Find area bounded by loop

  1. #1
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    Find area bounded by loop

    I can do the integral for these problems all day long but cannot determine the boundaries to save my life. Polar coordinates are my arch nemesis at the moment.

    The problem:
    find the area bounded by =$\displaystyle r=3sin(3\theta)$

    1. $\displaystyle 1/2\int_a^b(3sin(3\theta))^2d\theta$
    2. find boundaries
    3. do integral
    4. win

    I must have been catatonic when boundaries were explained because I simply do not get how to find them.

    thanks for any help!
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  2. #2
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    Re: Find area bounded by loop

    First, note that $\displaystyle sin(\theta)$ goes from 0, up to +1, then down to 0 again as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$. For $\displaystyle \theta$ from $\displaystyle \pi$ to $\displaystyle 2\pi$, sine is negative so you just go over the same path again. That is, $\displaystyle sin(\theta)$ covers one loop as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$. Here, $\displaystyle \theta$ is replaced by $\displaystyle 3\theta$ so this function covers one loop as $\displaystyle 3\theta$ goes from 0 to $\displaystyle \pi$. That is, as $\displaystyle \theta$ goes from 0/3 to $\displaystyle \pi/3$.
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  3. #3
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    Re: Find area bounded by loop

    I follow that. Does 3 at the beginning of the function not make a difference then?
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  4. #4
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    Re: Find area bounded by loop

    No, that does not affect the $\displaystyle \theta$ values. In fact, it could be taken outside the integral.
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  5. #5
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    Re: Find area bounded by loop

    Oh, okay. So if I have a function like$\displaystyle r=ksin(3\theta)cos(3\theta)$ I only need to look at $\displaystyle sin(3\theta)cos(3\theta)$? But this is only for multiplications of functions correct? In other words I could not pull out "n" or "k" in this example:$\displaystyle r=n-ksin(3\theta)cos(3\theta)$. Do I have this right?

    And, in general form I could solve $\displaystyle r=n-ksin(3\theta)cos(3\theta)$ by $\displaystyle n=ksin(3\theta)cos(3\theta)$ and the solution to $\displaystyle \theta$ would just be between 0 and $\displaystyle 2\pi$.

    Am I thinking of this correctly?
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