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Math Help - area of polar curve

  1. #1
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    area of polar curve

    find the area bounded by the spiral r=ln(\theta) on the interval \pi \theta 2 \pi

    (A) 2.405
    (B) 2.931
    (C) 3.743
    (D) 4.810
    (E) 7.487
    Last edited by yoman360; April 28th 2010 at 03:18 PM.
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  2. #2
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    The area of a polar curve is given by A=\frac{1}{2}\int\mid{r}\mid^2\ d\theta. You'll have to integrate by parts twice to evaluate the integral.

    As a check, the area can be approximated by the area of a semicircle of radius \frac{3\pi}{2}, which is \frac{1}{2}\pi\left(\frac{3\pi}{2}\right)^2=\frac{  9}{8}\pi^3.

    Post again in this thread if you're still having trouble.

    - Hollywood
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  3. #3
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    Quote Originally Posted by hollywood View Post
    The area of a polar curve is given by A=\frac{1}{2}\int\mid{r}\mid^2\ d\theta. You'll have to integrate by parts twice to evaluate the integral.

    As a check, the area can be approximated by the area of a semicircle of radius \frac{3\pi}{2}, which is \frac{1}{2}\pi\left(\frac{3\pi}{2}\right)^2=\frac{  9}{8}\pi^3.

    Post again in this thread if you're still having trouble.

    - Hollywood
    \frac{9}{8}\pi^3 ≈ 34.882

    problem is its not in one of the multiple choice options
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  4. #4
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    My mistake - I left out the "ln". The correct value should be \frac{1}{2}\pi\left(\ln\frac{3}{2}\pi\right)^2\app  rox3.77. But I think you're supposed to evaluate the integral:

    A=\frac{1}{2}\int_\pi^{2\pi}(\ln{\theta})^2\ d\theta

    which you can do by integrating by parts twice.

    - Hollywood
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