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Calculate the volume of the solid of revolution of f(x)=e^x

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the volume of the solid of revolution formed by rotating the region around the y-axis. Apply the shell method.

    f(x)=e^x, x=0, y=8



    2. Relevant equations

    V=∫2∏x((f(x))-g(x))dx

    3. The attempt at a solution

    This is what I did: (I integrated from 0 to 8)

    V=∫ 2∏x(8-e^x)dx
    =2∏∫ (8x-xe^x)

    I used integration by parts with u=x, du=1dx, v=e^x, and dv e^x(dx)

    giving:

    2∏[8∫ xdx-(xe^x-∫ e^x(dx)]

    my final answer was -129507.1677

    When I apply the disk method using x=ln(y) I get 48.13407626.

    These two answers should be the same and I think there's an error in my shell method that I can't figure out???
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 31, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Did you subtract your integral from a cylinder with length (e^8-1) and radius 8?
    Anyway, this number is too small.

    I don't think the volume should be negative, by the way.
     
  4. Mar 31, 2013 #3
    This what I did for the disk method (about the y-axis)

    V=∫(lny)^2-(0)^2 dy........from 1 to 8

    used two integration by parts

    1st: u=(lny)^2, du=2lny/y dy, dv=dy, v=y

    giving: v=∏[(lny)^2-∫(y)(2lny/y)]

    2nd: u=lny, du= 1/y dy, dv=dy, v=y

    giving: ∏[y(lny)^2-2ylny+∫y(1/y)dy

    answer: ∏[y(lny)^2 -2ylny+2y], evaluated from 1 to 8=∏[8(ln8)^2-16ln8+14]=48.13407626
     
  5. Mar 31, 2013 #4

    mfb

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    2016 Award

    Staff: Mentor

    You are calculating two different things like this - one volume is between the function and the y-axis, the other volume is between the function and the x-axis.
     
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