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Math Help - Polar Graph Area

  1. #1
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    Polar Graph Area

    Find the area for one loop of r^{2} = 64sin2\theta

    I plotted three graphs; r^{2} vs. [tex]\theta]/math], where it looks like a normal sin curve repeating twice between 0 and 2\pi ranging from 64 to -64. Another of r vs. \theta where it looks kind of like two elliptical shapes ranging from -8 to 8, restricted to 0 to \frac{pi}{2} and \pi to \frac{3\pi}{2}. Finally, the polar graph on x and y axis, where it looks like a lemniscate reaching a limit of 8 at \frac{\pi}{4} and \frac{5\pi}{4}.

    I did this:

    A=\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(64sin2\theta) d\theta

    Solving that I got 8\pi. I think the answer is 8\pi - 16, though. Someone suggested I set the interval to 2 \int_{0}^{\frac{\pi}{4}}. I understand what they're doing... integrating to the maximum r and doubling that, but I don't understand why that gives the correct answer and doing it the other way does not.

    I think I may have a couple of different computational errors going on, despite trying and retrying the problem, as well as a conceptual block that I'm just not getting.

    Any help, much appreciated.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by DarrenM View Post
    Find the area for one loop of r^{2} = 64sin2\theta

    I plotted three graphs; r^{2} vs. [tex]\theta]/math], where it looks like a normal sin curve repeating twice between 0 and 2\pi ranging from 64 to -64. Another of r vs. \theta where it looks kind of like two elliptical shapes ranging from -8 to 8, restricted to 0 to \frac{pi}{2} and \pi to \frac{3\pi}{2}. Finally, the polar graph on x and y axis, where it looks like a lemniscate reaching a limit of 8 at \frac{\pi}{4} and \frac{5\pi}{4}.

    I did this:

    A=\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(64sin2\theta) d\theta
    Looks ok to me.

    Solving that I got 8\pi.
    No, I think that's wrong: I get \int_0^{\pi/4}32\sin(2\theta)\, d\theta = \Big[-16\cos(2\theta)\Big]_{\theta=0}^{\pi/4}=16+16=32

    I think the answer is 8\pi - 16, though.
    The difference to 32 is so large that it should be possible to see it quite easily in your plot of the graph.
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  3. #3
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    Wow. Idiotic mistake. sin2\theta is not even remotely the same as sin^{2}\theta. As such, using the double-angle identity would probably be an incredibly stupid thing to do.
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