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Simple Volume

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = sec(x), y = 1, x = 1, x = -1 on the x-axis.

    3. The attempt at a solution

    This should be ridiculously easy but apparently my answer is wrong?!

    To calculate the volume i use the equation [itex]V = \pi \int _{-1}^{1} (sec(x)^{2} - 1)dx[/itex]

    From here it should be trivial since this evaluates to [itex]\pi [ tan^{2} x | _{-1}^{1} ] = 0[/itex]

    But my answer is apparently wrong... Should i be doing something differently? I've followed the formula exactly how it should be followed to my knowledge.
     
  2. jcsd
  3. Mar 11, 2013 #2

    LCKurtz

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    Is the antiderivative of ##\sec^2x## correct? And did you forget the antiderivative of the ##-1##? Or did you use a trig identity and forget to integrate it?
     
  4. Mar 11, 2013 #3
    Whoops. So i apply a trig identity... sec^2(x)-1 = tan^2(x)

    Then i antiderive tan^2(x) which gives me... tan(x) - x

    So, computing (tan(1) - 1) - (tan(-1) + 1) results in... pi(2tan(1)-2) or 3.5022... And that's right! Thanks for the help.
     
  5. Mar 11, 2013 #4

    LCKurtz

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    Notice you could have written the answer tan(x)-x directly by taking the antiderivative of your integrand as it stands since the derivative of tan(x) is sec^2(x).
     
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