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Finding the surface area of a curved object using calculus

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data
    I need some help with a surface area of a solid. The solid is made from rotating the line y=x^2 around the x axis. So it's sort of like a cone or a horn. Here are my steps:

    2. Relevant equations
    Surface of revolution formula
    Integrate 2∏r times the square root of 1 plus the derivative squared (dx).

    3. The attempt at a solution
    2[itex]\pi[/itex] [itex]\int[/itex] x[itex]^{2}[/itex] [itex]\sqrt{1+2x^2}[/itex]
    This is the surface of revolution concept of course. How do I integrate this? Should I make the square root a power of .5??
     
  2. jcsd
  3. Nov 11, 2012 #2

    SteamKing

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    If your derivative = 2x, then the derivative squared = (2x)^2, NOT 2x^2

    As to the integral itself, try u-substitution with integration by parts.
     
  4. Nov 11, 2012 #3
    Thanks!
     
  5. Nov 11, 2012 #4
    As for the curve y = x[itex]^{2}[/itex] , it is an upward parabola with the centre on the origin and x-axis.So when rotated about the x-axis the solid should look something like the attachement image I did.

    Now you can integrate the figure using area under curves method.
     

    Attached Files:

  6. Nov 11, 2012 #5
    No that's not correct sorry
     
  7. Nov 11, 2012 #6
    But I think the parabola statement was right,because y = x^2 is an upward parabola right?
     
  8. Nov 11, 2012 #7
    The parabola's base or curve is at the origin, the lines point up left and right. I'm focusing on the parabola's part that is to the right of the y axis. So half of a curve rotated around the origin looks like a curvy cone.

    Thanks for you guys' help, I now have the answers. Thanks!
     
  9. Nov 11, 2012 #8
    Great!! :wink:
     
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