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Math Help - Polar Curve Area

  1. #1
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    Polar Curve Area

    Find the area inside both r=5sin(2theta) and r=5cos(2theta).

    I have NO idea how to do it. My book has no examples and I couldn't find anything online.

    I ended up trying to do the integral of (5sin(2theta))^2, evaluated from 0 to Pi/8 + the integral of (5cos(2theta))^2 evaluated from 0 to Pi/8 and then multiplying the whole thing by 8 because of symmetry. I don't believe that to be correct.

    Any explanations would be great!
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  2. #2
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    One point the curves intersect in at Pi/8. If we find the area inside

    5sin(2{\theta}) from 0 to Pi/8 and multiply by 16, we can find the area of all

    the regions inside the two functions.

    8\int_{0}^{\frac{\pi}{8}}\left[5sin(2{\theta})\right]^{2}d{\theta}
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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  3. #3
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    First, I don't understand why you disregard the 5cos(2theta) completely. Why do you only use the sin equation?

    I also don't understand why you multiply by 16. There are only 8 sections that are in BOTH graphs, so wouldn't you multiply by 8?

    Edit: Ah, I see you fixed it. I was wondering why it wasn't squared. Thanks for adding a graph, too.
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