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Integral calculus: volume of a solid of revolution

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the first quadrant region bounded by x=y-y3, x=1 and y=1 that is revolved about the line x=1.


    3. The attempt at a solution

    dV=∏R2t

    where :

    t=dy
    R=1-(y-y3)
    =1-y+y3

    so..
    dV=∏(1-y+y3)2dy
    dV=∏(1-2y+y2+2y3-2y4+y6)dy
    V=∏∫ from 0 to 1 of (1-2y+y2+2y3-2y4+y6)dy
    V=∏(y-y2+1/3(y3)+1/2(y4)-2/5(y5)+1/7(y7) from 0 to 1

    V= 121∏/210 cubic units
    V= 1.81 cubic units


    was my solution and final answer correct?
     
  2. jcsd
  3. Mar 13, 2012 #2
    The setup and answer are both fine. I'm a little confused by the t, but I assume t takes the place of either dx or dy. Only thing I might suggest is to leave your answer as [itex]\frac{121}{210} \pi[/itex] instead of rounding, it's usually standard procedure in math courses!
     
  4. Mar 13, 2012 #3
    ok i'll do that. you are right, 't' takes the place of either dx or dy, it is the thickness. thank you very much for your reply and time. appreciate it. thanks.. :))
     
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