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Finding volume of axis of revolution

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data

    2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi
    That's the integral i need solved

    2. The attempt at a solution[/b]
    so first i subbed u=y-1
    took the 2 pi out of the integral
    that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
    the first integral = (2*pi*24^(3/2))/3

    the second i used a trig sub and my final answer is
    (25pi^2)/2 +

    The final answer is suppose to be 25pi^2 + 500pi/3

    but i got25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
    which is approx
    25pi^2 + 499.89pi/3

    how do i get the exact answer
  2. jcsd
  3. Aug 25, 2010 #2


    Staff: Mentor

    I infer from the problem title that you are rotating some curve around some axis. Are you sure that your integral is the right one for your problem? I don't want to spend any time with the integral you have if I'm not sure it represents the problem you're working.
  4. Aug 25, 2010 #3
    Yes I am 100% sure I am using the right integrel. The integral itself is odd and a longer and harder way to find the volume of this object. I would normally solve this in respect to x but the instrucotr wants it in respect to y.
    And yeah I am 100% sure it is the right integrel
  5. Aug 26, 2010 #4


    Staff: Mentor

    That looks fine.
    I'm not sure about this part, especially sqrt(24). My answer for the 2nd integral, before being multiplied by 2pi and without adding 5pi was
    25pi/4 - 25/2 * sin(2arcsin(-1/5)) - 25/2 * arcsin(-1/5).

    I don't have time to check that before going to work, so I might have made a mistake.

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