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Surface area by revolving a curve problem

  1. Sep 3, 2009 #1
    1. I am suppose to find the surface area of the curve y=sqrt(4-x^2) from -1 to 1 when it is revolving around the x-axis.



    2. Relevant equations: S= 2PIf(x)sqrt(1+(dy/dx)^2)dx



    3. I found the derivative to be -x(4-x^2)^-1/2 and then squared it so the problem is 2Pi -1[tex]\int[/tex]1 sqrt(4-x^2)sqrt(1+[x^2(4-x^2)] but I have no clue where to go from here.
     
  2. jcsd
  3. Sep 4, 2009 #2
    You state that you get

    [tex]2\pi \int_{-1}^{1} \sqrt{4-x^2} \sqrt{1+ \frac{x^2}{4-x^2}} \; dx[/tex]

    Try to simplify the integrand. Remember [tex]\sqrt A \sqrt B = \sqrt{AB}[/tex].

    --Elucidus
     
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