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Math Help - Finding Limits of r^2=4sin2 (Polar Coordinates)

  1. #1
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    Finding Limits of r^2=4sin2 (Polar Coordinates)

    I have to find the area of this region, but I'm not quite sure what my limits are for it.

    Thanks for your help!

    r^2=4sin2
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  2. #2
    MHF Contributor
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    Plot points until you see it. Watch out for multiple laps and duplicity and crossing over itself.

    \theta = 0

    Please pardon the 2nd grade algebra lesson. It is more confusing than may be realized. It helps to back up to fundamentals and go one tiny step at a time.

    r^{2} = 4\sin(2*0) \implies r^{2} = 4\sin(0) \implies r^{2} = 4*0 \implies r^{2} = 0 \implies r = 0

    I was worried we were going to get two values. This is an excellent warning that we WILL for future values.

    So far, we have (0,0)

    \theta = \frac{\pi}{6}

    r^{2} = 4\sin(2*\frac{\pi}{6}) \implies r^{2} = 4\sin(\frac{\pi}{3}) \implies r^{2} = 4*\frac{\sqrt{3}}{2}
    \implies r^{2} = 2\sqrt{3} \implies r = \pm \sqrt{2}\sqrt[4]{3}

    ...or about 1.861

    So we jumped rather far from the Origin and spawned two parts of the graph. This is VERY important. If nothing else, it might mean that limits of [0,\pi] will cover the whole figure, rather than requiring [0,2\pi]

    This same effort with maybe \frac{\pi}{4}, \frac{\pi}{3}, and \frac{\pi}{2} will show that this is all it takes to trace the ENTIRE curve. A little thought may suggest that continuing will retrace the curve another time, perhaps a little differently, but trace nonetheless. In this case, it seems that the result goes off the Real plane and you do not get a retrace on [\frac{\pi}{2},\pi], but it comes back for another tracing after that.

    With a little practice, you WILL get better at it.
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