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Arc Length and Surface Area Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction Arc Length and Surface Area

Just as the integral may be used to calculate planar area and spatial volume, so this tool may also be used to calculate the arc length of a curve and surface area. The basic idea is to approximate the length of a curve by the length of its piecewise linear approximation. A similar comment applies to the surface area. We begin by describing the basic rubric.

Arc Length

Suppose that f (x) is a function on the interval [ a, b ]. Let us see how to calculate the length of the curve consisting of the graph of f over this interval (Fig. 8.28). We partition the interval:

a = x 0x 1x 2 ≤ ... ≤ x k −1x k = b.

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.28

Look at Fig. 8.29. Corresponding to each pair of points x j −1, x j in the partition is a segment connecting two points on the curve; the segment has endpoints (x j −1 , f ( x j −1 )) and (x j, f (x j)). The length l j of this segment is given by the usual planar distance formula:

l j = ([x jx j −1 ] 2 + [ f (x j ) − f (x j −1)] 2 )1/2.

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.29

We denote the quantity x j − x j −1 by Δx and apply the definition of the derivative to obtain

Applications of the Integral 8.5 Arc Length and Surface Area

Now we may rewrite the formula for l j as

Applications of the Integral 8.5 Arc Length and Surface Area

Summing up the lengths l j (Fig. 8.30) gives an approximate length for the curve:

Applications of the Integral 8.5 Arc Length and Surface Area

Applications of the Integral 8.5 Arc Length and Surface Area

Fig. 8.30

But this last is a Riemann sum for the integral

Applications of the Integral 8.5 Arc Length and Surface Area

As the mesh of the partition becomes finer, the approximating sum is ever closer to what we think of as the length of the curve, and it also converges to this integral. Thus the integral represents the length of the curve.

Examples

Example 1

Let us calculate the arc length of the graph of f ( x ) = 4 x 3/2 over the interval [0,3].

Solution 1

The length is

Applications of the Integral 8.5 Arc Length and Surface Area

Example 2

Let us calculate the length of the graph of the function f ( x ) = (1/2) × (e x + e −x) over the interval [1, ln 8].

Solution 2

We calculate that

f′ (x) = (1/2)(e x − e −x).

Therefore the length of the curve is

Applications of the Integral 8.5 Arc Length and Surface Area

You Try It: Set up, but do not evaluate, the integral for the arc length of the graph of Applications of the Integral 8.5 Arc Length and Surface Area on the interval π/4 ≤ x ≤ 3π/4.

Sometimes an arc length problem is more conveniently solved if we think of the curve as being the graph of x = g (y). Here is an example.

Example

Calculate the length of that portion of the graph of the curve 16 x 2 = 9 y 3 between the points (0,0) and (6,4).

Solution

We express the curve as

Applications of the Integral 8.5 Arc Length and Surface Area

Then Applications of the Integral 8.5 Arc Length and Surface Area. Now, reversing the roles of x and y in (*), we find that the requested length is

Applications of the Integral 8.5 Arc Length and Surface Area

This integral is easily evaluated and we see that it has value [2 · (97) 3/2 − 128]/243.

Notice that the last example would have been considerably more difficult (the integral would have been harder to evaluate) had we expressed the curve in the form y = f (x).

You Try It: Write the integral that represents the length of a semi-circle and evaluate it.

Find practice problems and solutions for these concepts at: Applications of the Integral Practice Test.

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