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Area of a rotated surface

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the exact value when [itex] y = \frac{x^3}{2} + \frac{1}{6x} [/itex] in domain [itex] \frac{1}{2} \leq x \leq 1 [/itex] is rotated around the X axis


    2. Relevant equations
    Area is given by
    [tex] A = \int \pi \left(f(x)\right)^2 [/tex]


    3. The attempt at a solution
    If the above formula is correct, then I should be integrating the following

    [tex] A = \int_{\frac{1}{2}}^{1} ( \frac{x^3}{2} + \frac{1}{6x})^2 dx [/tex]

    Is this set up alone correct?
     
  2. jcsd
  3. Feb 27, 2013 #2

    CompuChip

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    I think the formula you are using is for the volume of the solid of rotation: you can cut it up into thin slices of surface area [itex]\pi r^2 = \pi f(x)^2[/itex] and "thickness" dx.

    Do you have a formula with a [itex]2\pi[/itex] and an f'? :)
     
  4. Feb 27, 2013 #3
    Hmm i think I am mistaken...
    I was trying to look this up and foudn the following formula:

    [tex] A = \int_{a}^{b} 2 \pi f(x) \sqrt{1 + (f'(x))^2} dx [/tex]

    is this the fomula I need to be using?
     
    Last edited: Feb 27, 2013
  5. Feb 27, 2013 #4

    Dick

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    If it's surface area you want, then that looks right.
     
  6. Feb 27, 2013 #5
    Ok thats perfect...

    Now just to clarify, the integral becomes:

    [tex] 2\pi \int_{\frac{1}{2}}^{1} \left( \frac{3x^4+1}{6x^2} \right) \sqrt{1+\left(\frac{9x^4-1}{6x^2}\right)^2} dx [/tex]


    So I was a bit lazy so I plugged into wolfram alpha and got the answer [itex] 445\pi/768 [/itex]

    is this correct? is there anything that I missed to consider?
     
  7. Feb 27, 2013 #6

    Dick

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    Conceptually you aren't missing anything. In detail, it might be wrong. [Edit: my mistake, it is correct].
     
    Last edited: Feb 27, 2013
  8. Feb 27, 2013 #7
    Thank you :)
    Been stressing about these questions for quite some time now
     
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