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The Surface Area of a Rotational Curve

Date: 04/17/98 at 09:32:58
From: Katie Shannon
Subject: The formula for the surface area of a curve rotated about the 

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
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Associated Topics:
College Calculus
High School Calculus

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