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Help with volume of revolution question

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the volume obtained by rotating the region between the graph of y=0.5(sin(x^2)^2) and the x-axis (from 0 to squareroot pi) about the y-axis.

    The answer pi^2/4, but I don't understand how to get the answer, I can set up the integral but can't simplify it to that answer.


    2. Relevant equations

    y=0.5(sin(x^2)^2)


    3. The attempt at a solution

    First to get the radius in terms of y -> y=0.5sin^2(x^2), thus √arcsin(√2y)=x

    The volume is thus calculated via the disk method and is equal to the integral of pi√arcsin(√2y), from 0 to √pi. I don't know where to go form there,maybe I didn't set up the integral correctly?
     
  2. jcsd
  3. Jan 26, 2014 #2
    Hi thisisssu!

    I suggest using method of cylinders instead of disk as we are rotating about the y-axis.
     
  4. Jan 26, 2014 #3

    vela

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    The radius of what?

    I think you're saying that
    $$V = \int_0^\sqrt\pi \pi r\,dy,$$ where r is the radius you describe above. This isn't a correct application of the disk method. Also, I'm assuming you're trying to integrate with respect to y because you wrote the integrand in terms of y. If that's the case, your limits aren't correct because those are x values.

    In any case, the disk method is not a good choice for this particular problem. Take Pranav-Arora's advice and use cylinders.
     
  5. Jan 26, 2014 #4
    I've attached the problem in this post to avoid any confusion, also by "cylinder method", I'm assuming you're actually referring to the shells method?
     

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  6. Jan 26, 2014 #5
    Yes!
     
  7. Jan 26, 2014 #6
    So I tried the shell method and ended up getting pi/2 which is not the correct answer. I was able to set up the integral which was the integral of 2pi0.5sin^2(x^2)xdx from 0 to √pi etc., with the height and radius being 0.5sin^2(x^2) and x respectively. Is my setup wrong?
     
  8. Jan 26, 2014 #7

    vela

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    No, that looks right. It seems you didn't evaluate the integral correctly.
     
  9. Jan 26, 2014 #8
    I verified via wolframaalpha, and it doesn't come out to pi^2/4, but 1.56 which is essentially pi/2, so I'm assuming I didn't set up integral correctly or maybe I did screw up somewhere while trying to solve the integral? Could someone help me evaluate this integral.
     
  10. Jan 26, 2014 #9

    vela

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    You're typing it in wrong to Wolfram Alpha then. Show us your work so we can point out where you're going astray.
     
  11. Jan 26, 2014 #10
    V=2π∫(1/2)(sin^2(x^2))(x)dx, from 0 to √π
    V=π/2∫sin^2(u)du, from 0 to √π, with u=x^2 and du=2xdx (I started having trouble simplifying after getting up to here).
     
  12. Jan 26, 2014 #11

    vela

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    That's almost correct. You should have
    $$ V = \frac{\pi}{2} \int_0^\pi \sin^2 u\,du.$$ The upper limit is ##\pi##, not ##\sqrt\pi##. You want to use a trig identity for ##\sin^2 u##.
     
  13. Jan 27, 2014 #12
    Could you please explain why the upper limit is π? Also how did you calculate the upper limit as π?
     
  14. Jan 27, 2014 #13

    haruspex

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    The limits are x=0 to x = √π. (Without the "x=" they're incomplete, technically.) With u = x2, that's u=x2=0 to u=x2=π.
     
  15. Jan 28, 2014 #14
    Thanks, I finally ended up getting the correct answer after using the correct limits.
     
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