41. A family of exponential functions
a. Show that the arc length integral for the function f 1×2 = Aeax + 1
4Aa2 e-ax, where a 7 0 and A 7 0, may be
integrated using methods you already know. b. Verify that the arc length of the curve y = f 1×2 on the interval 30, ln 24 is
A12a – 12 – 1 4a2 A
12-a – 12. 42. Bernoulli’s “parabolas” Johann Bernoulli (1667–1748) evalu-
ated the arc length of curves of the form y = x12n + 12>2n, where n is a positive integer, on the interval 30, a4. a. Write the arc length integral.
b. Make the change of variables u2 = 1 + a2n + 1 2n
b2x1>n to obtain a new integral with respect to u.
c. Use the Binomial Theorem to expand this integrand and evalu- ate the integral.
d. The case n = 1 1y = x3>22 was done in Example 1. With a = 1, compute the arc length in the cases n = 2 and n = 3. Does the arc length increase or decrease with n?
e. Graph the arc length of the curves for a = 1 as a function of n.
QUICK CHECK ANSWERS 1. 12a (The length of the line segment joining the points) 2. 121d – c2 (The length of the line segment joining the points) 3. L = 1p0 21 + cos2 y dy
6.6 Surface Area In Sections 6.3 and 6.4, we introduced solids of revolution and presented methods for computing the volume of such solids. We now consider a related problem: computing the area of the surface of a solid of revolution. Surface area calculations are important in aero- dynamics (computing the lift on an airplane wing) and biology (computing transport rates across cell membranes), to name just two applications. Here is an interesting observation: A surface area problem is “between” a volume problem (which is three-dimensional) and an arc length problem (which is one-dimensional). For this reason, you will see ideas that appear in both volume and arc length calculations as we develop the surface area integral.
Some Preliminary Calculations Consider a curve y = f 1×2 on an interval 3a, b4, where f is a nonnegative function with a continuous first derivative on 3a, b4. Now imagine revolving the curve about the x-axis to generate a surface of revolution (Figure 6.60). Our objective is to find the area of this surface.
y ! f (x) y
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452 Chapter 6 Applications of Integration
Before tackling this problem, we consider a preliminary problem upon which we build a general surface area formula. First consider the graph of f 1×2 = rx>h on the interval 30, h4, where h 7 0 and r 7 0. When this line segment is revolved about the x-axis, it generates the surface of a cone of radius r and height h (Figure 6.61). A formula from geometry states that the surface area of a right circular cone of radius r and height h (excluding the base) is pr2r2 + h2 = pr/, where / is the slant height of the cone (the length of the slanted “edge” of the cone).
➤ One way to derive the formula for the surface area of a cone (not including the base) is to cut the cone on a line from its base to its vertex. When the cone is unfolded, it forms a sector of a circular disk of radius / with a curved edge of length 2pr. This sector is a fraction 2pr 2p/
= r /
of a full circular disk of radius /.
So the area of the sector, which is also the surface area of the cone, is p/2 # r
/ = pr/.
f (x) ! cx
(b, f (b))
Surface area of large cone ”
(a, f (a))
Surface area of small cone Surface area of frustrum
Surface area Sb Surface area S ! Sa ” SbSurface area Sa
y ! x
r r h
Surface area ! !rl ! !r!r2 ” h2
l !!r2 ” h2 r
QUICK CHECK 1 Which is greater, the surface area of a cone of height 10 and radius 20 or the surface area of a cone of height 20 and radius 10 (excluding the bases)?
With this result, we can solve a preliminary problem that will be useful. Consider the linear function f 1×2 = cx on the interval 3a, b4, where 0 6 a 6 b and c 7 0. When this line segment is revolved about the x-axis, it generates a frustum of a cone (a cone whose top has been sliced off ). The goal is to find S, the surface area of the frustum. Figure 6.62 shows that S is the difference between the surface area Sb of the cone that extends over the interval 30, b4 and the surface area Sa of the cone that extends over the interval 30, a4.
Notice that the radius of the cone on 30, b4 is r = f 1b2 = cb, and its height is h = b. Therefore, this cone has surface area
Sb = pr2r2 + h2 = p1bc221bc22 + b2 = pb2c2c2 + 1. Similarly, the cone on 30, a4 has radius r = f 1a2 = ca and height h = a, so its surface area is
Sa = p1ac221ac22 + a2 = pa2c2c2 + 1. The difference of the surface areas Sb – Sa is the surface area S of the frustum on 3a, b4:
S = Sb – Sa = pb2c2c2 + 1 – pa2c2c2 + 1 = pc1b2 – a222c2 + 1.
Curved edge length ! 2!r
Curved edge length ! 2!r
S ! ! ( f (b) ” f (a)) Surface area of frustrum:
! ! (r2 ” r1)
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6.6 Surface Area 453
y ! f (x) y
x1x0 x2 xk”1 xn”1 xnxk
! (!x )2 ” |!yk |2
x1 x2 xk#1 xn
|!yk | $ | f (xk ) # f (xk#1)| !x
A slightly different form of this surface area formula will be useful. Observe that the line segment from 1a, f 1a22 to 1b, f 1b22 (which is the slant height of the frustum in Fig- ure 6.62) has length
/ = 21b – a22 + 1bc – ac22 = 1b – a22c2 + 1. Therefore, the surface area of the frustum can also be written
S = pc1b2 – a222c2 + 1 = pc1b + a21b – a22c2 + 1 Factor b2 – a2. = p¢cb + ca≤1b – a22c2 + 1 Expand c1b + a2.
f 1b2 f 1a2 / = p1 f 1b2 + f 1a22/.
This result can be generalized to any linear function g1x2 = cx + d that is positive on the interval 3a, b4. That is, the surface area of the frustum generated by revolving the line segment between 1a, g1a22 and 1b, g1b22 about the x-axis is given by p1g1b2 + g1a22/ (Exercise 36).
QUICK CHECK 2 What is the surface area of the frustum of a cone generated when the graph of f 1×2 = 3x on the interval 32, 54 is revolved about the x-axis? Surface Area Formula With the surface area formula for a frustum of a cone, we now derive a general area for- mula for a surface of revolution. We assume the surface is generated by revolving the graph of a positive, differentiable function f on the interval 3a, b4 about the x-axis. We begin by subdividing the interval 3a, b4 into n subintervals of equal length ∆x = b – a
The grid points in this partition are
x0 = a, x1, x2, c, xn – 1, xn = b. Now consider the kth subinterval 3xk – 1, xk4 and the line segment between the points 1xk – 1, f 1xk – 122 and 1xk, f 1xk22 (Figure 6.63). We let the change in the y-coordinates between these points be ∆yk = f 1xk2 – f 1xk – 12.
When this line segment is revolved about the x-axis, it generates a frustum of a cone (Fig- ure 6.64). The slant height of this frustum is the length of the hypotenuse of a right triangle whose sides have lengths ∆x and ” ∆yk ” . Therefore, the slant height of the kth frustum is21∆x22 + ” ∆yk “2 = 21∆x22 + 1∆yk22
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454 Chapter 6 Applications of Integration
and its surface area is
Sk = p1 f 1xk2 + f 1xk – 12221∆x22 + 1∆yk22. It follows that the area S of the entire surface of revolution is approximately the sum of the surface areas of the individual frustums Sk, for k = 1, c, n; that is,
S ≈ a n
k = 1 Sk = a
k = 1 p1 f 1xk2 + f 1xk – 12221∆x22 + 1∆yk22.
We would like to identify this sum as a Riemann sum. However, one more step is required to put it in the correct form. We apply the Mean Value Theorem on the kth subinterval 3xk – 1, xk4 and observe that
f 1xk2 – f 1xk – 12 ∆x
= f ′1xk*2, for some number xk
* in the interval 1xk – 1, xk2, for k = 1, c, n. It follows that ∆yk = f 1xk2 – f 1xk – 12 = f ′1xk*2∆x.
We now replace ∆yk with f ′1xk*2∆x in the expression for the approximate surface area. The result is
S ≈ a n
k = 1 Sk = a