Arc Length and Surfaces of
Revolution
1
Arc Length for y = f(x)
Suppose a curve is given by y = f(x), where f is a function
with a continuous first derivative on the interval [a, b]. The
goal is to determine how far one would travel if one
walked along the curve from (a, f(a)) to (b, f(b)). This
distance is the arc length, which we denote as L.
As shown in the figure on the next slide, we divide [a, b]
into n subintervals of length ∆x = (b - a)/n, where xk is
the right endpoint of the kth subinterval, for k = 1,…,n.
Joining the corresponding points on the curve by line
segments, we obtain a polygonal line with n line
segments.
2
If n is large and ∆x is small, the length of the polygonal line is a good
approximation to the length of the actual curve. The strategy is to find the
length of the polygonal line and then let n increase, while ∆x goes to zero, to
get the exact length of the curve.
3
Consider the kth subinterval [xk-1, xk] and the line segment
between the points (xk-1, f(xk-1)) and (xk, f(xk)). We let the
change in the y-coordinate between these points be
y k  f x k   f x k1.
The kth line segment is the hypotenuse of a right triangle
with sides of length ∆x and |∆yk| = |f(xk) - f(xk-1)|. The
lengthof each line segment is
x 
2
2
 y k ,k  1,2,...,n.
Summing these lengths, we obtain the length of the
polygonal line, which approximates the length L of the
curve 
n
L 
k1
x 
2
2
 y k .
4
In previous applications of the integral, we would, at this
point, take the limit as n  ∞ and ∆x  0 to obtain a
definite integral. However, due to the presence of the ∆yk
term, we must complete one additional step before taking
a limit.
Notice the slope of the line segment on the kth
subinterval is ∆yk/ ∆x (rise over run). By the Mean Value
Theorem, this slope equals f ´(xk*) for some point xk* on
the kth subinterval. Therefore
n
L 
x   y k
2
2
k1
n


k1
 y 2
2.
2
k
Factor
out
(∆x)
x  1   

  y  

5
n

k1
2
y k 
1   x
 y 
n
  1 f x

 x.
* 2
k
Bring ∆x out of the
square root.
Mean Value Theorem
k1
Now we have a Riemann sum. As n increases and as ∆x
approaches zero, the sum approaches a definite integral,
which is 
also the exact length of the curve. We have
n
L  lim  1 f x
n 
k1
 x 
* 2
k
b

1 f (x) 2 dx.
a
6

Note: 1 + f´(x)2 is positive, so the square root in the
integrand is defined whenever f´ exists. To ensure √(1 +
f´(x)2) is integrable on [a, b], we require f´ to be continuous.
7
Example 1: Find the length of the curve f(x) = x3/2 between
x = 0 and x = 4.
Notice f´(x) = (3/2)x1/2,
which is continuous on the
interval [0, 4]. Using the
arc length formula, we
have….
8
L
b

1 f (x) 2 dx 
a
3 12 2
1  x  dx
2

4

0

4
9
1 xdx
4

0



4

9
Substitute for f´(x).
Simplify.
10

udu
u = 1 + (9/4)x, du = (9/4) dx
1
10
4 2 3 2 
  u 
9 3 1
8  3 2 
 10 1

27 
 9.1 units.
Fundamental Theorem
Simplify.
9
Example 2: Find the length of the curve f(x) = x3 + 1/(12x)
on the interval [1/2, 2].
We first calculate f´(x) = 3x2 - 1/(12x2) and f´(x)2 = 9x4 (1/2) + 1/(144x4). The length of the curve on [1/2, 2] is
L
2

1
1 f (x) dx 
2
2

1
2

2

1
2
2
 4 1
1 
1 9x  
dx
4 

2 144 x 
2
 2
1 
dx
3x 
2 

12x 
 2
1 
  3x 
dx
2 
12x 
1 
2

2
Substitute.
Factor.
Simplify.
2
 3
1 
 x 
  8.

12x 1
2
Evaluate the integral.
10
Example 3: Consider the segment of the parabola f(x) = x2
on the interval [0, 2].
a. Write the integral for the length of the curve.
Noting that f´(x) = 2x, the arc length integral is
2

2
1 f (x) 2 dx 
0

1 4 x 2 dx.
0
b. Use a calculator to evaluate the integral.

2

1 4 x 2 dx  4.647.
0
11
Note: When using technology, it is a good idea to check
whether an answer is plausible. We found the arc
length of y = x2 on [0, 2] is approximately 4.647. The
straight-line distance between (0, 0) and (2, 4) is √20 ≈
4.472, so our answer is reasonable.
Arc Length for x = g(y)
Sometimes it is advantageous to describe a curve as a
function of y, that is, x = g(y). The arc length formula in
this case is derived exactly as in the case of y = f(x),
switching the roles of x and y. The result is the following
arc length formula.
12
Example 4: Find the length of the curve y = f(x) = x2/3
between x = 0 and x = 8.
13
The derivative of f(x) = x2/3 is f´(x) = (2/3)x-1/3, which is
undefined at x = 0. Therefore, the arc length formula with
respect to x cannot be used, yet the curve certainly
appears to have a well-defined length.
The key is to describe the curve with y as the independent
variable. Solving y = x2/3 for x, we have x = g(y) = y3/2.
Notice when x = 8, y = 82/3 = 4, which says we should use
the positive branch of y3/2. Therefore, finding the length of
the curve y = f(x) = x2/3 from x = 0 to x = 8 is equivalent to
finding the length of the curve x = g(y) = y3/2 from y = 0 to y
= 4.
This is precisely the problem solved in Example 1.
14
Example 5: A climber anchors a rope at two points of equal
height, separated by a distance of 100 feet, in order to
perform a Tyrolean traverse. The rope follows the catenary
f(x) = 200 cosh(x/200) over the interval [-50, 50]. Find the
length of the rope between the two anchor points.
15
Note that f´(x) = 200 sinh(x/200)(1/200) = sinh(x/200).
Therefore, the length of the rope is
L
50

50
50
 2

 x 
1 sinh 
dx
200 
2
0
 x 
1 sinh 
dx
200 
2
Arc length formula.
Use symmetry.
Let u = x/200, the 400 du = dx. u(0) = 0 and u(50) = 1/4.

1
4
 400  1 sinh 2 udu
0
We know that 1 + sinh2 u = cosh2 u. So…..
16
1
4
 400  cosh udu
0
1
4
 400sinh u 0




1 
 400sinh   sinh 0
4 


≈ 101 feet.

17
Area of a Surface of Revolution
The area of a surface of revolution is derived from the
formula for the lateral surface area of the frustum of a
right circular cone.
18
Area of a Surface of Revolution
Consider the line segment in Figure 7.43, where L is the
length of the line segment, r1 is the radius at the left end of
the line segment, and r2 is the radius at the right end of the
line segment.
Figure 7.43
19
Area of a Surface of Revolution
When the line segment is revolved about its axis of
revolution, it forms a frustum of a right circular
cone, with
Lateral surface area of frustum
S = 2 rL
where
Average radius of frustum
20
Area of a Surface of Revolution
Suppose the graph of a function f, having a
continuous derivative on the interval [a, b], is
revolved about the x-axis to form a surface
of revolution, as shown in Figure 7.44.
Figure 7.44
21
Area of a Surface of Revolution
Let  be a partition of [a, b], with subintervals of width xi.
Then the line segment of length
generates a frustum of a cone.
Let ri be the average radius of this frustum. By the
Intermediate Value Theorem, a point di exists (in the i th
subinterval) such that ri = f (di).
The lateral surface area Si of the frustum is
22
Area of a Surface of Revolution
By the Mean Value Theorem, a point ci exists in (xi
– 1, xi ) such that
So,
and the total
surface area can be approximated by
23
Area of a Surface of Revolution
It can be shown that the limit of the right side as
is
In a similar manner, if the graph of f is revolved about the
y-axis, then S is
24
Area of a Surface of Revolution
In these two formulas for S, you can regard the products
2 f (x) and 2 x as the circumferences of the circles traced
by a point (x, y) on the graph of f as it is revolved about the
x-axis and the y-axis (Figure 7.45). In one case the radius
is r = f (x), and in the other case the radius is r = x.
Figure 7.45
25
Area of a Surface of Revolution
26
Example 6: The graph of f(x) = 2√x on the interval [1, 3] is
revolved about the x-axis. What is the area of the surface
generated?
27
Noting that f´(x) = 1/√x, the surface area formula gives
S
b
2

2

f
(x)
1
f
(x)
dx

a
3
1
 2  2 x 1 dx
x
1

 4

x  1dx
1





3
3
8
x  1
3
2
16
4 2
3

3

1
≈ 43.325
28

Example 7: The curve y = √(4 - x2), on [-1, 1], is an arc of
the circle x2 + y2 = 4. Find the area of the surface obtained
by rotating this arc about the x-axis. (The surface is a
portion of a sphere of radius 2).
29
Note that
1
dy 1
x
2  2
 4  x  2x  
dx 2
4  x2
S
b
2

2

f
(x)
1
f
(x)
dx

a

 2

1

4x
2
4x
2
1
 2

1

1
1
x2
1
dx
2
4x
2
4x
2
dx
 4   1dx
1

 4 (2)  8
30
Example 8: The arc of the parabola y = x2 from (1, 1) to
(2, 4) is rotated about the y-axis. Find the area of the
resulting surface.
31
dx
1

dy 2 y
x  y and
Using

S  2
d
 r(y)
1 r(y) dy  2
2
c
we have
4

1

4

1
y 1
dy
4y
4 y  1dy
1






4

17

u du
where u = 1 + 4y
5
17

6
17  5 5

≈ 30.846

32
Example 9: The curved surface of a funnel is generated by
revolving the graph of y = f(x) = x3 + 1/(12x) on the interval
[1, 2] about the x-axis. Approximately what volume of paint
is needed to cover the outside of the funnel with a layer of
paint 0.05 cm thick? Assume x and y are measured in
centimeters.
33
Note f´(x) = 3x2 - 1/(12x2). Therefore, the surface area of
the funnel in cm2 is
S  2
b

f (x) 1 f (x) 2 dx
a
2
 3
 2
1 
1 
 2  x 
dx
 1 3x 
2 


12x 
12x 
1
2


1 
 2  x 3 



12x
1
2

2
 2
1 
dx
3x 
2 

12x 

1  2
1 
 2  x 3 
3x

dx

2 

12x 
12x 
1
Substitute for f and f´
Expand and factor
under square root.
2

Simplify.

x
1 
 2  3x 5  
dx
3 

3 144 x 
1
2

34
2
1 6 x
1 
 2  x 

2 
6 288x 1
2
2

12,289

  201.078
192
Since the paint layer is 0.05 cm thick, a good approximation
to the volume of paint is

12,289 2 
3
cm
0.05cm
10.1cm



 192


35
Example 10: Consider the curve defined by the
equation 4y3 - 10x√y + 5 = 0, for 1 ≤ y ≤ 2. Find the
area of the surface generated when this curve is
revolved about the y-axis.
36
We solve the equation 4y3 - 10x√y + 5 = 0 for x so that the
curve is expressed as a function of y:
4y 3 10x y  5  0
10x y  4y 3  5
2 5 2 1  12
x  g(y)  y  y
5
2

Note g´(y)
= y3/2 - (1/4)y-3/2 and the interval of integration
on the y-axis is [1, 2]. The area of the surface is

S  2
d

g(y) 1 g(y) 2 dy
c
2 5 2 1  12 
 3 2 1  3 2 2
 2   y  y  1 y  y  dy




2
4
1 5
2

37
2 5 2 1  12 
 2   y  y 


2
1 5
2
 3 2 1  3 2 2
y  y  dy


4
2 4 3
1 2 
 2   y  y  y dy

5
8 
1 5
2

2


 2 5 3 2 1 
 2  y  y  
10
8y 1
25
1,377

  21.63
200

38