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I Surface area of a manifold

  1. Nov 10, 2016 #1
    How do I find the surface area of a f(x) rotated around the y axis?
  2. jcsd
  3. Nov 10, 2016 #2


    Staff: Mentor

    Check the site mathispower4u.com in the calc 3 section there should be a couple short tutorials on surfaces of revolution that should help.

    Heres one video

  4. Nov 16, 2016 #3
    Imagine breaking up the interval on the x-axis (over which the curve lies that is going to be rotated) into small pieces of length Δx each.. Approximating this piece of the curve y = f(x) by a line segment, we see that if it is rotated about the y-axis we get a piece of a cone. The length of that line segment between (x, f(x)) and (x+Δx, f(x+Δx) is approximately

    ΔL = √((Δx)2 + (f'(x)Δx)2))​

    = √(1 + (f'(x)2)) Δx​

    Do you see why? This is a good point to stop and make sure you understand the reason for the last two expressions.

    Therefore the surface area of the piece of surface generated when this line segment is rotated about the y-axis is approximately

    ΔS = 2πx ΔL = 2πx √(1 + (f'(x)2)) Δx.​

    Adding these up for all Δx, we get

    S ≈ k 2πxk √(1 + (f'(x)2)) Δx​

    where xk is (say) the left endpoint of each interval of length Δx on the x-axis.

    So in the limit, this becomes exactly equal to the integral

    S = 2πx √(1 + (f'(x)2)) dx​

    where the integral is taken over the appropriate interval on the x-axis.

    As an example, try to determine the area of the surface generated by rotating the graph of the function

    f(x) = x2

    for 0 ≤ x ≤ 1 about the y-axis.

    (If your exact answer is 5.330 when rounded to three decimal places, then you probably got the same one as I did.)
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