7.4 Arc Length and Surface of Revolution
(Photo not taken by Vickie Kelly)
Greg Kelly, Hanford High School, Richland, Washington
Objectives
• Find the arc length of a smooth curve.
• Find the area of a surface of revolution.
Rectifiable curve: One that has a finite arc length
f is rectifiable on [a,b] if f ' is continuous on [a,b].
If rectifiable, f is continuously differentiable on [a,b] and
its graph is a smooth curve.

Lengths of Curves:
ds
If we want to approximate the
length of a curve, over a short
distance we could measure a
straight line.
dy
dx
By the pythagorean theorem:
ds  dx  dy
2
ds  dx  dy
2
2
2
2
 ds  
S
dx  dy
2
S
2
 dx dy  2
 2  2  dx
 dx dx 
2
2
 dy 2 
1  2  dx
 dx 
Length of Curve (Cartesian)
We need to get dx out from
2
under the radical.
b
dy
S
a
 
1    dx
 dx 

Length of Curve (Cartesian)
(function of x)
S
b
a
1   f   x   dx
2
Length of Curve (Cartesian)
(function of y)
S
d
c
1   g   y   dy
2

Example: Find the arc length of:
3
x
1
y 
6 2x
 1 , 2
 2 
dy 3x 2
1
1 2 1 

 2  x  2 
2
x 
dx
6 2x
S
b
a
2
 dy 
1    dx
 dx 
2
2
S  1
2
 1  2 1 
1    x  2   dx
x 
2 

Example Find the arc length of:
3
x
1
y 
6 2x
 1 , 2
 2 
2
 1  2 1 
1    x  2   dx
x 
2 
2
S  1
2
S  1
2
2
1 4
1
1   x  2  4  dx
4
x 




2
2
1 4
1 2
4
2
S  1
x  x  2 dx  1
x x
dx
4
2
4
2
2
2 1
11 3 1
2
2
S  1
x  x dx   x    33
16
23
x 1
2 2
2


2

Example: Find the arc length of:
Solve for x:
x 
 y  1
x    y  1
x   y  1
dx
S
5
S
5
1
1
dy
3
3
2
y 1
2
(since [0,8])
y  1
2
 x 2 on 0,8
When x=0: 0   y  13
3
3
 y  1
3
1
2
When x=8: 64   y  13
4  y 1
2
1 
3
1    y  1 2  dy
2

y5
5 9
5
9

1    y  1 dy    y   dy
1
4
4

4

Example: Find the arc length of:
S
5
1
1 5
5
9
 4 y  4  dy  2 1


 x 2 on 0,8
u  9y 5
9 y  5dy
du  9dy
 y  1
du  dy
9
40
1
1
  u 2 du
18 4
1 32
 u
27
40
4

3
1

40 2  8
27
3

 9.0734

Surface Area:
ds
Consider a curve rotated about the x-axis:
The surface area of this band is:
r
The radius is the y-value of the
function, so the whole area is given by:
2 r  ds

b
a
2 y ds
This is the same ds that we had in the “length
of curve” formula, so the formula becomes:
To rotate about
the y-axis, just
reverse x and y
in the formula!
Surface Area about x-axis (Cartesian):
2
 dy 
S   2 y 1    dx
a
 dx 
b

If revolving f(x) about x-axis
r(x)=f(x)
or g(y) about the y-axis:
r(y)=f(y)

Example: Find the area of the surface formed by revolving y  x 3
on [0,1] about the x-axis.
2
 dy 
S  2  y 1    dx
0
 dx 
1
r=y
1
S  2  x
3
0
1
S  2  x 1  9 x dx
3
S
18 
10
1
u
1
2
dx
du  36 x3dx
du
du
10

2
u  1  9 x4
4
0

1   3x
2
10
36 x3
 dx
  32 

S   u    u    10 10  1
18  3
 1 27
 1  27
 2
3
2


 3.563

If revolving f(x) about y-axis
r(x)=x
or g(y) about the x-axis:
r(y)=y

Example: Find the area of the surface formed by revolving y  x 2
on 0, 2  about the y-axis.
2


S  2 
2
0
r=x
S  2 
0
 dy 
x 1    dx
 dx 
2
x 1   2 x  dx
2
u  1  4x2
S  2 
0
2
du  8 xdx
x 1  4 x dx
2
du

8x
9
 dx
du
S  2  xu

8x 9 4 1
1
9
  2 32 
3 


S   u    u 2     27  1  13
43
1  6
6
1
3
9
1
2
du

u
1
2

Homework
7.4 (page 485)
#3 – 13 odd,
17 – 25 odd (Don't graph),
37, 39, 43
(Use the calculator to evaluate integrals.)
