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Basic Surface Area of Revolution

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the area of the surface obtained by rotating the curve y=x3, 0≤x≤2 about the x-axis.

    2. Relevant equations
    \begin{equation*}
    SA = \int_{0}^{2} 2 \pi y L
    \end{equation*}
    3. The attempt at a solution

    SORRY, I don't know how to use LaTeX yet.

    ∫2∏y√(1+(dy/dx)2)dx from 0->2
    =∫2∏y√(1+(3x2)2)dx
    =2∏∫x3√(1+9x4)dx
    =2∏∫x3(1+3x2)dx
    =2∏∫x3+3x5dx
    =2∏[x4/4 + x6/2] 0->2
    =plug in 2
    =72∏

    I don't see where I went wrong. The answer is ∏/27*(145√(145)-1)
     
  2. jcsd
  3. Aug 22, 2012 #2

    SammyS

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    Review basic algebra:
    [itex]\displaystyle \sqrt{1+9x^4}\ne1+3x^2[/itex]​

    Notice that [itex]\displaystyle \frac{d}{dx}\ (1+9x^4)=36x^3\,,[/itex] so use a substitution like u = 1 + 9x4 .
     
  4. Aug 23, 2012 #3
    Trying it with that, I now have:

    ∫2∏y√(1+(dy/dx)2)dx from 0->2
    =∫2∏y√(1+(3x2)2)dx
    =2∏∫x3√(1+9x4)dx

    let u = 1+9x4, (1/36)du=x3dx

    =2∏/36∫√udu
    =2∏/36[2u^3/2/3] 0->2
    =2∏/36[2(1+9x4)3/2/3] 0->2
    =plug in 2
    =some ridiculously huge number compared to answer
     
  5. Aug 23, 2012 #4
    No, in fact, it gives the answer. Are you doing some basic arithmetic error?
     
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