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Homework Help: Integral - I can't get the right answer

  1. Nov 29, 2006 #1
    Find the volume of the solid obtained when the given region is rotated about the x-axis.

    Under [itex]y = (\sin{x})^{\frac{3}{2}}[/itex] between 0 and pi.

    The radius is... [itex]r = (\sin{x})^{\frac{3}{2}}[/itex]

    Then the area for any sample is... [itex]A (x) = \pi((\sin{x})^{\frac{3}{2}})^{2}[/itex]

    Simplifying to... [itex]A (x) = \pi(\sin{x})^{3}[/itex]

    Integrate between 0 and pi to get the volume...

    [tex]V = \pi \int_{0}^{\pi} (\sin{x})^{3}[/tex]

    [tex]V = \pi [ \frac{(\sin{x})^{4}}{4\cos{x}} ]_{0}^{\pi}[/tex]

    But... sin(pi) and sin(0) both equal 0, making the volume 0. But it's actually (4/3)(pi). What am I missing?
  2. jcsd
  3. Nov 29, 2006 #2
    That's not the right way to integrate...

    hint: Split (sin(x))^3 as (sin(x))^2.sin(x)
  4. Nov 29, 2006 #3


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    You are missing the fact that
    is NOT
    [tex] \frac{(\sin{x})^{4}}{4\cos{x}}[/tex]
    Since there was not a cosine in the original integrand, you can't simply put one in. In particular,if you try to differentiate
    [tex]\frac{(\sin{x})^{4}}{4\cos{x}} [/tex]
    you do not get [itex]sin^3(x)[/itex] because you have to use the quotient rule and differentiate that new "cos(x)" also.

    You surely learned how to do integrals of odd powers of sine and cosine long ago: separate out one "sin x" to use with the derivative:
    [tex]\int sin^2(x) sin(x)dx= \int (1- cos^2(x)) sin(x)dx[/itex]
    and let u= cos(x).

    (Blast you, neutrino, you beat me again!)
    Last edited by a moderator: Nov 29, 2006
  5. Nov 29, 2006 #4


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    Homework Helper

    I think you got confused doing the integral.

    It's true that Integral sin^3x cos x dx = 1/4 sin^4 x

    But you can't just "divide by cos x" and say Integral sin^3 x = 1/4 sin^4 x / cos x. Check out the ways you have learned to do integrals like sin^3 x.
  6. Nov 29, 2006 #5
    Ahhh shoot! It didn't even strike me to use substitution. I was being ignorant to the fact that it would require the quotient rule to derivate what I thought was the integral... Thank-you very much! I'm so relieved! :smile:
  7. Nov 29, 2006 #6
    The word is 'differentiate'. :)
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