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Homework Help: Arc length/ surface area with integrals

  1. Jun 8, 2008 #1
    I have a question on the formulas for arc length and surface area.

    Do you use the formula: [tex]s= \int_{c}^{d}\sqrt{1+[g'(y)]^2}dy[/tex] only when you are provided with a function x=g(y)?? Can you convert that to y=g(x) and solve it by replacing g'(y) with y(x), changing the bounds and the dy to dx?

    Like for example if you wanted to solve:
    Find the area of the surface formed by revolving the graph of [tex]f(x)=x^2[/tex] on the interval [0,2] about the y-axis.

    How would you know whether to find the surface area with respect to x or y? Are these independent on the axis of revolution? Why does the book give the Surface area formula with respect to x and y?
    Last edited: Jun 8, 2008
  2. jcsd
  3. Jun 8, 2008 #2
    Before finding the surface area, you will need to find the arc length that is traced out on that interval, and then rotate it (ie, pretend it is a cylinder with radius = f(x)). Then you can probably use the equation A = 2*pi*r*l with some help from the calculus. Perhaps the easiest way to find the correct arc length formula is to make it yourself imagining you have a series of small triangles with lengths dx and df (this can be dy too but this might be easier) (and you can add dz if you're doing it in 3d), with the pythagorian theorem giving you
    ds = \sqrt{dx^{2}+df^{2}}
    now, you want to turn this into something that you can integrate. So you know what f(x) is, so you will want to integrate with respect to x. So you pull the dx out:
    ds = \sqrt{1 + (\frac{df}{dx})^{2}}dx
    Once you have found this on your interval, you can rotate it. Its been a while since ive done these, but I believe this is a workable method.
  4. Jun 8, 2008 #3


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    Yes, you can. In fact, if you have both x and y as functions of some parameter t, say x= f(t), y= g(t) then [itex]s= \int \sqrt{f'(t)^2+ g'(t)^2}dt[/itex]. That reduces your first integral if x= t, y= g(t) and to your second if x= f(t), y= t.

    I'm not sure I understand your last question- you don't say what formula you are talking about.

    If, for example, you had a straight line x= a, rotated around the y-axis, you would get a circular cylindar of radius a. Its surface area would be the circumference of the circle, [itex]2\pi x= 2 \pi a[/itex] time the height (measured in the y direction) h. For a curved figure rotated around the y-axis, you would, of course, imagine very thin "layers", each a very short cylinder, stacked on top of one another. In each cylinder, since x does not change MUCH, you can approximate the surface area by [itex]2\pi x dy[/itex] where "dy" is the height. The total surface area can be approximated by the Riemann sum [itex]\Sum (2\pi x dy)[/itex] and can be made exact by the limit process that gives the integral [itex]2\pi\int x dy[/itex]. Of course, in the case that x= g(y) that becomes [itex]2\pi\int g(y)dy[/itex]. If the graph y= f(x) is rotated about the x-axis, that would, of course, just swap x and y ("of course" because you can just swap x and y in the "Riemann sum" argument): [itex]2\pi\int ydx= 2\pi\int f(x)dx[/itex].

    In both of those, the curve is rotated around the x or y axis. But you can use the same argument for any axis. Suppose you have the curve x= g(y) rotated around an axis x= a rather than the x-axis, x= 0. First you would draw a picture and decide whether the graph is on the left or right of x= a. That's important because the radius of the circle formed has to be positive: r= |x- a| which is either x- a or a- x depending upon whether x is larger than or less than a. (If the graph x= g(y) crosses the line x= a, you will need to do separate integrals for g(y)< a and g(y)> a.) If x> a, then the circle has radius x- a rather than just x and so the circumference is [itex]2\pi (x- a)[/itex] and the surface area integral becomes
    [itex]\int 2\pi (x-a) dy= 2\pi \int (g(y)-a)dy[/itex].

    Again, if you are given y= f(x) rotated around the line y= b, you just swap x and y: [itex]2\pi\int |f(x)- b|dx[/itex].
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