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Revolving trig function around y-axis

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/a/emr1n
    01 x= 4tan(pi*y/4) , Find the volume of the solid by revolving shaded region about the y-axis

    2. Relevant equations
    tan^2=sec^2 - 1

    3. The attempt at a solution
    I ended up with 64-16pi using u-substitution, not sure if right, want a confirmation. I only have one last attempt on mathlab so really need this to be right, would like to know how you got to whatever the answer is, or if there's a way to check the answer, that would help a lot. Wolfram hasn't been of much help :/

    Thanks!
     
    Last edited: Sep 5, 2016
  2. jcsd
  3. Sep 5, 2016 #2

    Ray Vickson

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    What formula did you use for the volume? What was your integral before and after u-substitution? We are forbidden from doing the problem for you, so we cannot show you "...how you got to whatever the answer is".
     
  4. Sep 5, 2016 #3
    yeah, sure. It's just that it's kind of long so didn't know if I should post:

    http://imgur.com/a/SaVJF
     
  5. Sep 5, 2016 #4
    basically the area of a circle by its height dx

    substitute the equation f(y) = 4tan (pi*y/4) for r

    so r^2 is tan^2 which is (sec^2 - 1)

    found the integral by u substituting (pi*y/4) with u

    so you can now integrate sec^2 and 1

    which gives tan (u) - u

    plugged those values in multiplied by the scalars I got from before which were 16*4pi

    and after all that, give or take, I got 64-16pi. I'm not sure that's the right answer (it could be) I just only have one more shot on MathLab, and want to get someone who knows what they're talking about to chime in before I submit :/

    Thanks!
     
    Last edited: Sep 5, 2016
  6. Sep 6, 2016 #5

    SammyS

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    It looks OK.

    I get the same result using a somewhat different set of steps.
     
  7. Sep 6, 2016 #6
    bit the bullet and just clicked "final check" and....*drumroll* it's right!

    Ha, thanks guys!
     
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