1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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].

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook