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Solids of revolution, y axis

  1. Mar 9, 2010 #1
    The area
    Is rotated about the y axis, bounded by y=1, x=0 and x=ln2 find the volume of the solid.

    And i am clearly making something wrong, so if anyone could verify my work.

    [tex]~ 2\pi\int_0^{ln2}x\cdot(1-(e^x-1)[/tex]

    u=x, du=1
    dv=e^x, v=e^x

    [tex]\int xe^x -2x= xe^x -e^x -x^2[/tex]
    [tex]e^x(x-1)-x^2\bigg|_0^{ln2} = 2\pi \cdot ln2-1 -(ln2)^2[/tex]

    The answer is supposed to be [tex]2\pi(ln2-1)^2[/tex]
  2. jcsd
  3. Mar 9, 2010 #2
    where did that 1 come from in your shell method formula?
  4. Mar 9, 2010 #3
    When you integrate you should obtain : 2*pi*(ln22-2*ln2+1)

    That then can be the factored to the book's answer.

    You lost a 2 with the ln2
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