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Integral to find the Surface area

  1. Jun 22, 2012 #1
    For a given function f(x)=x, if we rotate this function from [0,1] around the x-axis, we'll have a cone.... why can't I find its surface area by adding the perimeter of all the circunferences from [0,1] with radius = f(x)?Something like 2*pi∫xdx? The formulas says we have to do 2*pi∫x*√(i+y'²)dx .....
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  3. Jun 22, 2012 #2


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    Since the function f(x) is not parallel to the x axis the area differential has to take into account that effect, which for your problem is √2.

    In general, you need the differential of arc length of the revolving figure which is
    Last edited: Jun 22, 2012
  4. Jun 22, 2012 #3
    But to find the volume of the cone, for exemple, the fact that the function isn't parallel to the x-axis doens't influence that "volume is the sum of all areas of all circles".... What changes?
  5. Jun 23, 2012 #4


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    Think of the area of the cone's surface as the sum of a bunch of thin circular strips. The extra factor is the width of the strip.
  6. Jun 24, 2012 #5


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    Consider this fallacious argument: If we divide the interval from x= 0 to x= 1 into n equal subintervals, each of length 1/n, we can "approximate" the line from (0, 0) to (1, 1) by a broken line, the horizontal line from (0, 0) to (1/n, 0), the vertical line from I(1/n, 0) to (1/n, 1/n), the horizontal line from (1/n, 1/n) to (2/n, 1/n), the vertical line from (2/n, 1/n) to (2/n, 2/n), etc.

    The area under the horizontal line from (k/n, k/n) to ((k+1)/n, k/n) is (k/n)(1/n)= k/n^2 so the total area is [itex]\sum_{k=0}^n k/n^2= (1/n^2)\sum_{k=0}^n k= (1/n^2)(n(n+1)/2= (n^2+ n)/2n= n/2+ 1/2. And the limit, as n goes to infinity, is 1/2, the area under the straight line.

    But if we, instead, use the length of the broken line to approximate the length of the line, each segment has length 1/n and there are 2n such segments so that we get a length of 2 which is wrong. The length of the line segment from (k/n, k/n) to ((k+1)/n, (k+1)/n) simply cannot be approximated by the length of those two horizontal and vertical segments. We have to use the Pythagorean theorem to get the length of each segment: [itex]\sqrt{(1/n)^2+ (1/n)^2}= (1/n)\sqrt{2}[/itex]. And that leads to the arclength formula.
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