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 frustum:

 A = 2 \pi r l \,

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

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 frustum 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.

Definition (Surface of Revolution)[edit]

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

The Surface Area Formula[edit]

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 frustum  r_i approaches a single value

2. the slant height of each conical frustum  l_i 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