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Math Help - Volume...finding k

  1. #1
    Senior Member polymerase's Avatar
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    Volume...finding k

    Let k be a real number between 0 and \pi. Let R be the region in the first quadrant bounded by the x-axis, the line x=k and the curve 0\le" alt="y=sin\:x\0\le" /> \:x\: \le k). If the volume of the solid generated by revolving the region R about the x-axis is \frac{\pi}{24}(5\pi-3), find the value of k.

    Thanks!
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  2. #2
    Eater of Worlds
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    Use the washers method and find k.

    {\pi}\int_{0}^{k}sin^{2}(x)dx=\frac{(k-sin(k)cos(k)){\pi}}{2}

    \frac{(k-sin(k)cos(k)){\pi}}{2}=\frac{\pi}{24}(5\pi-3)

    k-sin(k)cos(k)=\frac{1}{12}(5\pi-3)

    2k-sin(2k)=\frac{1}{6}(5\pi-3)

    Can you solve for k?. Try Newton'e method if all else fails.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by galactus View Post
    Use the washers method and find k.

    {\pi}\int_{0}^{k}sin^{2}(x)dx=\frac{(k-sin(k)cos(k)){\pi}}{2}

    \frac{(k-sin(k)cos(k)){\pi}}{2}=\frac{\pi}{24}(5\pi-3)

    k-sin(k)cos(k)=\frac{1}{12}(5\pi-3)

    2k-sin(2k)=\frac{1}{6}(5\pi-3)

    Can you solve for k?. Try Newton'e method if all else fails.
    how would you isolated for k, or dont you, is it just inspection?
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  4. #4
    Eater of Worlds
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    This one converges rather quickly, so Newton's method is a decent choice.

    Let your initial guess be 1.5.

    f'(k)=2-2cos(2k)

    x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Let k=1.5:

    k-\frac{2k-sin(2k)-\frac{1}{6}(5\pi-3)}{2-2cos(2k)}=1.31384700816

    Now, do it again with the answer we got above and we get 1.30900315524

    Again: 1.30899693901

    That's good enough.

    Looks like x=1.309. There abouts.

    There are numerous ways to go about it, but Newton's method can be handy to know.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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