Calculus/Surface area

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Surface area


Suppose we are given a function f and we want to calculate the surface area of the function f rotated around a given line. The calculation of surface area of revolution is related to the arc length calculation.

If the function f is a straight line, other methods such as surface area formulas for cylinders and conical frustra can be used. However, if f is not linear, an integration technique must be used.

Recall the formula for the lateral surface area of a conical frustrum:

 A = 2 \pi r l \,

where r is the average radius and l is the slant height of the frustrum.

For y=f(x) and a\le x\le b, we divide [a,b] into subintervals with equal width Δx and endpoints x_0,x_1,\ldots,x_n. We map each point y_i= f(x_i) \, to a conical frustrum of width Δx and lateral surface area A_i \,.

We can estimate the surface area of revolution with the sum

 A = \sum_{i=0}^{n} A_i

As we divide [a,b] into smaller and smaller pieces, the estimate gives a better value for the surface area.

[edit] Definition (Surface of Revolution)

The surface area of revolution of the curve y=f(x) about a line for a\le x\le b is defined to be

 A = \lim_{n\to \infty} \sum_{i=0}^{n} A_i

[edit] The Surface Area Formula

Suppose f is a continuous function on the interval [a,b] and r(x) represents the distance from f(x) to the axis of rotation. Then the lateral surface area of revolution about a line is given by

 A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx

And in Leibniz notation

 A = 2 \pi \int_a^b r(x) \sqrt{ 1+ \left(\frac{dy}{dx}\right)^2 } dx

Proof:

 A \ =  \lim_{n \to \infty} \sum_{i=1}^n A_i
=  \lim_{n \to \infty} \sum_{i=1}^n 2 \pi r_i l_i
=  2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i

As  n \rightarrow \infty and  \Delta x \rightarrow 0 , we know two things:

1. the average radius of each conical frustrum ri approaches a single value

2. the slant height of each conical frustrum li equals an infitesmal segment of arc length

From the arc length formula discussed in the previous section, we know that

 l_i = \sqrt{ 1+ (f'(x_i))^2 }

Therefore

 A \ =  2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i
=  2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i \sqrt{ 1+ (f'(x_i))^2} \Delta x

Because of the definition of an integral  \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x_i , we can simplify the sigma operation to an integral.

 A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx

Or if f is in terms of y on the interval [c,d]

 A = 2 \pi \int_c^d r(y) \sqrt{ 1+ (f'(y))^2} dy