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Volumes of Revolution Question

  1. Dec 11, 2014 #1


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    Gold Member

    1. The problem statement, all variables and given/known data
    Let R be the region bounded by the curves y=0 and y=x^2+x between x=0 and x=1. Compute the volume of the solid of revolution obtained when R is rotated about the axis y=-1.

    2. Relevant equations
    Disc method: integral of pi*r^2 = volume from the bounds

    3. The attempt at a solution

    The bounds are obvious = 0 to 1.

    The hard part is accounting for the empty area inbetween as we rotate around y=-1 rather then y=0.

    What I did to solve the problem was moved the volume up 1 to the origin. I then solved the entire volume of the problem from 0 to 1 using the radius of x^2+x+1 (the +1 to account the entire area).
    My answer here was 37pi/10.

    I then went back and took the volume of the squared middle region which had radius of 1 and then subtracted that from my first answer.

    My final answer was 27pi/10.

    Can anyone confirm my train of thought/answer?

    Thank you.
  2. jcsd
  3. Dec 11, 2014 #2


    Staff: Mentor

    That works for me.

    You could also do the problem without translating the region upward like so:
    ##\Delta V = \pi[(\text{outer radius})^2 - (\text{inner radius})^2]\Delta x##
    where the outer radius is ##x^2 + x - (-1) = x^2 + x + 1## and the inner radius is 1.
    The volume integral is then
    ##\pi \int_0^1 [(x^2 + x + 1)^2 - 1^2] dx##
  4. Dec 11, 2014 #3


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    Gold Member

    Ah, so you managed to get the same answer as well? Confirming my result :D
  5. Dec 12, 2014 #4


    Staff: Mentor

    Yes, same answer.
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