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Math Help - [SOLVED] Trig integral w/ volume of rotation

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Trig integral w/ volume of rotation

    I know I have my integral correct:

    V = 41\pi  \int\limits^{\pi}_{\frac{\pi}{2}} sin^2x dx

    So then I did this:
    <br />
41\pi  \int\limits^{\pi}_{\frac{\pi}{2}}  \frac{1}{2} (1-cos (2x))

    <br />
= \frac{41\pi}{2}  \int\limits^{\pi}_{\frac{\pi}{2}}  1-cos (2x)

    <br />
= \frac{41\pi}{2}  \int\limits^{\pi}_{\frac{\pi}{2}}  1 - cos (u) du

    = \frac{41\pi}{4} \cdot (x - \frac{sin2x}{2}) \|^{\pi}_{\frac{\pi}{2}}

    After I evaluate I get \frac{41\pi^2}{4} which is the wrong answer.
    I know this is a lot of work, but can someone work this and just point out where my error is? Don't need all the steps, unless I screwed something up.
    Thanks!!
    Molly
    Last edited by mollymcf2009; February 8th 2009 at 10:01 PM.
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  2. #2
    Member Abu-Khalil's Avatar
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    You didn't change the integration limits when changed the variable. By the way, that's the correct value for the integral.

    Could you post the "correct answer"?
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  3. #3
    Senior Member mollymcf2009's Avatar
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    But I put 2x back in for "u" and then evaluated at the old limits. Should I not have done that?

    This is from that SOLVED problem you helped me with before. Guess I didn't really solve it

    Wish I could post the correct answer, but it is in webassign, so I cant get it. I found one in the book that was almost identical:

    Find Vol of solid for area btw. curves: y = sinx y=0 with same limits

    the book answer for that one is \frac{\pi^2}{4} which is almost exactly what my answer is.
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  4. #4
    Member Abu-Khalil's Avatar
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    The integral: \int_\frac{\pi}{2}^\pi\sin^2 xdx =\frac{1}{2}\int_\frac{\pi}{2}^\pi 1-\cos 2xdx=\frac{1}{4}\int_{\pi}^{2\pi} 1-\cos tdt=<br />
\left.\frac{1}{4}\left(t-\sin t\right)\right|^{2\pi}_\pi=\frac{\pi}{4}. The problem was that you keep the \frac{1}{4} when it should be \frac{1}{2} if you were reversing the change.

    It was (41\sin x)^2, so the answer should be 41^2\frac{\pi^2}{4}.
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  5. #5
    Senior Member mollymcf2009's Avatar
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    OMG! Thank you SOOOOO much! Now I can go to bed! Thanks for all your help tonight!!
    Have a great week!
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    There's even a quick way of calculating that integral.

    Having \int_{\frac{\pi }{2}}^{\pi }{{{\sin }^{2}}x\,dx}, then the substitution u=\pi-x shows that the integral equals \int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx}=\int_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}x\,dx} since \int_{0}^{\frac{\pi }{2}}{f(\sin x)\,dx}=\int_{0}^{\frac{\pi }{2}}{f(\cos x)\,dx}.

    Finally, \int_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}x\,dx}+\int_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}x\,dx}=\frac{\pi }{2} which implies that both integrals achieve to the same value, \frac\pi4, and the rest follows.
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