The Surface Area of a Rotational CurveDate: 04/17/98 at 09:32:58 From: Katie Shannon Subject: The formula for the surface area of a curve rotated about the y-axis My math professor defined the surface area for the curve given by: x = f(t), y = g(t), a<t<b, rotated about the y-axis and I have been having some difficulty understanding how he proved the formula. I would like to have full knowledge of this formula. If you could please explain it in greater detail I would appreciate it. Date: 04/17/98 at 12:02:03 From: Doctor Rob Subject: Re: The formula for the surface area of a curve rotated about the y-axis The idea is to compute the surface area as the limit of the sum of the surface areas of an approximating set of frustums of cones. Divide the t-interval [a,b] into n parts, n some positive integer, using t_k = a + k*(b-a)/n, for k = 0, 1, ..., n. Observe t_0 = a and t_n = b. Let delta-t = (b-a)/n be the length of each subinterval [t_0,t_1], [t_1,t_2], ... . Draw a picture of the curve in the first quadrant. Plot the points with coordinates: x_k = f(t_k) y_k = g(t_k) This divides the curve into n pieces. Connect these points in order with line segments. These approximate the curve in that small t-interval. Draw horizontal lines y = y_k from these points to the y-axis. For any two adjacent points (x_k,y_y) and (x_[k+1],y_[k+1]), the line segment connecting them, the two horizontal lines through them y = y_k and y = y_[k+1], and the y-axis x = 0 form a trapezoid. If you rotate it about the y-axis you will form the kth frustum to which I refer. Then for any value of t, put: delta-x = f(t + delta-t) - f(t) delta-y = g(t + delta-t) - g(t) These frustums have height delta-y, which is very small if n is very large, and radii of the two bases x_k and x_(k+1) = x_k + delta-x. The area of the curved surface of this frustum is about 2*Pi*x_k*s_k, where s_k is the slant-height of the kth frustum. By the Pythagorean Theorem, s_k^2 = delta-x^2 + delta-y^2, so s_k = sqrt(delta-x^2 + delta-y^2). Then the total area is about: A(n) = Sum 2*Pi*f(t_k)*sqrt(delta-x^2 + delta-y^2) where the sum ranges over 0 <= k <= n-1. We can convert everything to be in terms of t quite readily, using: delta-x = delta-t*[f(t_k + delta-t) - f(t_k)]/[(t_k + delta-t) - t_k] delta-y = delta-t*[g(t_k + delta-t) - g(t_k)]/[(t_k + delta-t) - t_k] Then substitute these expressions into the previous formula for A(n). Now we let n -> infinity, so delta-t -> 0, the line segments more and moreclosely approximate the curve, and the height of the frustums -> 0. Note that: lim [f(t_k + delta-t) - f(t_k)]/[(t_k + delta-t) - t_k] = f'(t_k) lim [g(t_k + delta-t) - g(t_k)]/[(t_k + delta-t) - t_k] = g'(t_k) Then we get in the limit: A = Integral from a to b [2*Pi*f(t)*sqrt[f'(t)^2+g'(t)^2]*dt] Wherever I have written approximations above, there are small correction terms that can be proved to go to zero as n -> infinity, so that the above answer is correct and exact. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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