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Surface area of x = ln(y) - y^2/8 from 1 to e

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.


    2. Relevant equations
    SA = [itex]\int 2\pi*(f(x))*\sqrt{1+[f'(x)]^2}dx[/itex]


    3. The attempt at a solution
    SA = [itex]\int 2\pi*(ln(y) - \frac{y^2}{8}))*\sqrt{1+\frac{9}{16*y^2}}dy[/itex] from 1 to e

    I don't know how to solve this integral, which makes me thing I set it up wron.g
     
  2. jcsd
  3. Mar 4, 2012 #2

    Dick

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    I don't think you did the sqrt(1+f'(y)^2) part right. Can you show how you worked that out?
     
  4. Mar 4, 2012 #3
    That was the equation that my prof. gave me, lol
     
  5. Mar 4, 2012 #4

    Dick

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    I'm saying I don't think you simplified 1+f'(y)^2 correctly when f(y)=ln(y)-y^2/8. You should get something inside the square root that becomes a perfect square. I'm asking how you got that part to be 1+9/(16y^2)? I don't think that's right.
     
  6. Mar 4, 2012 #5
    oh ok. I took the derivative of [itex]x = lny - \frac{y^2}{8}[/itex] and got

    [itex]x = \frac{1}{y} - \frac{1}{4y}[/itex] and squared it to get

    [itex]\frac{1}{y^2} - \frac{1}{2y^2} + \frac{1}{16y^2}[/itex] which is

    [itex]\frac{16}{16y^2} - \frac{8}{16y^2} + \frac{1}{16y^2}[/itex] which is

    [itex]\frac{9}{16y^2}[/itex]
     
  7. Mar 4, 2012 #6

    Dick

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    The derivative of that is [itex]\frac{1}{y}-\frac{y}{4}[/itex]. That's where you are starting to go wrong.
     
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