8
FURTHER APPLICATIONS
OF INTEGRATION
FURTHER APPLICATIONS OF INTEGRATION
8.2
Area of a Surface
of Revolution
In this section, we will learn about:
The area of a surface curved out by a revolving arc.
SURFACE OF REVOLUTION
A surface of revolution is formed when
a curve is rotated about a line.
 Such a surface is the lateral boundary of a solid
of revolution of the type discussed in Sections 6.2
and 6.3
AREA OF A SURFACE OF REVOLUTION
We want to define the area of a surface
of revolution in such a way that it corresponds
to our intuition.
 If the surface area is A, we can imagine that
painting the surface would require the same
amount of paint as does a flat region with area A .
AREA OF A SURFACE OF REVOLUTION
Let’s start with some
simple surfaces.
CIRCULAR CYLINDERS
The lateral surface area of a circular cylinder
with radius r and height h is taken to be:
A = 2πrh
 We can imagine
cutting the cylinder
and unrolling it to
obtain a rectangle
with dimensions of
2πrh and h.
CIRCULAR CONES
We can take a circular cone with base radius r
and slant height l, cut it along the dashed line
as shown, and flatten it to form a sector of a
circle with radius and central angle θ = 2πr/l.
CIRCULAR CONES
We know that, in general, the area
of a sector of a circle with radius l and
angle θ is ½ l2 θ.
CIRCULAR CONES
So, the area is:
 2 r 
A l   l 
   rl
 l 
1 2
2
1 2
2
 Thus, we define the lateral surface area
of a cone to be A = πrl.
AREA OF A SURFACE OF REVOLUTION
What about more
complicated surfaces of
revolution?
AREA OF A SURFACE OF REVOLUTION
If we follow the strategy we used with arc
length, we can approximate the original curve
by a polygon.
 When this is rotated about an axis, it creates
a simpler surface whose surface area approximates
the actual surface area.
 By taking a limit, we can determine the exact
surface area.
BANDS
Then, the approximating surface consists
of a number of bands—each formed by
rotating a line segment about an axis.
BANDS
To find the surface area, each of these
bands can be considered a portion of
a circular cone.
BANDS
Equation 1
The area of the band (or frustum of a cone)
with slant height l and upper and lower radii r1
and r2 is found by
subtracting the areas of
two cones:
A   r (l1  l )   r1l1
   ( r2  r1 )l1  r2l 
BANDS
From similar triangles, we have:
This gives:
r2l1  rl
1 1  rl
1
or
(r2  r1 )l1  rl
1
l1 l1  l

r1
r2
Formula 2
BANDS
Putting this in Equation 1, we get
A   (rl
1  r2l )
or
A  2 rl
where r = ½(r1 + r2) is the average radius of
the band.
AREA OF A SURFACE OF REVOLUTION
Now, we apply this formula
to our strategy.
SURFACE AREA
Consider the surface shown here.
 It is obtained by rotating the curve y = f(x), a ≤ x ≤ b,
about the x-axis, where f is positive and has
a continuous derivative.
SURFACE AREA
To define its surface area, we divide
the interval [a, b] into n subintervals with
endpoints x0, x1, . . . , xn and equal width Δx,
as we did in determining arc length.
SURFACE AREA
If yi = f(xi), then the point Pi(xi, yi) lies
on the curve.
 The part of the surface between xi–1 and xi
is approximated by taking the line segment Pi–1 Pi
and rotating it about the x-axis.
SURFACE AREA
The result is a band with
slant height l = | Pi–1Pi |
and average radius r = ½(yi–1 + yi).
 So, by Formula 2, its surface area is:
yi 1  yi
2
| Pi 1 Pi |
2
SURFACE AREA
As in the proof of Theorem 2 in Section 8.1,
we have
Pi 1Pi  1   f '( x*) x
2
where xi* is some number in [xi–1, xi].
SURFACE AREA
When Δx is small, we have yi = f(xi) ≈ f(xi*)
and yi–1 = f(xi–1) ≈ f(xi*), since f is continuous.
Therefore,
yi 1  yi
*
* 2
2
Pi 1 Pi  2 f ( xi ) 1   f '( xi )  x
2
Formula 3
SURFACE AREA
Thus, an approximation to what we think
of as the area of the complete surface of
revolution is:
n
 2 f ( x
i 1
i
*
2
) 1   f '( xi )  x
*
SURFACE AREA
The approximation appears to
become better as n → ∞.
SURFACE AREA
Then, recognizing Formula 3 as a Riemann
sum for the function g ( x)  2 f ( x) 1   f '( x)
we have:
2
n
lim  2 f ( xi ) 1   f '( xi )  x
n 
*
*
2
i 1
  2 f ( x) 1   f '( x) dx
b
a
2
SURFACE AREA—DEFINITION
Formula 4
Thus, in the case where f is positive and has
a continuous derivative, we define the surface
area of the surface obtained by rotating the
curve y = f(x), a ≤ x≤ b, about the x-axis as:
S   2 f ( x) 1   f '( x) dx
b
a
2
Formula 5
SURFACE AREA
With the Leibniz notation for derivatives,
this formula becomes:
2
 dy 
S   2 y 1    dx
a
 dx 
b
Formula 6
SURFACE AREA
If the curve is described as x = g(y),
c ≤ y ≤ d, then the formula for surface area
becomes:
2
S
d
c
 dx 
2 y 1    dx
 dy 
Formula 7
SURFACE AREA
Then, both Formulas 5 and 6 can be
summarized symbolically—using the notation
for arc length given in Section 8.1—as:
S   2 y ds
Formula 8
SURFACE AREA
For rotation about the y-axis, the formula
becomes:
S   2 x ds
 Here, as before, we can use either
2
2
 dy 
ds  1    dx
 dx 
or
 dx 
ds  1    dy
 dy 
SURFACE AREA—FORMULAS
You can remember
these formulas in the following
ways.
SURFACE AREA—FORMULAS
Think of 2πy as the circumference of a circle
traced out by the point (x, y) on the curve as
it is rotated about the x-axis.
SURFACE AREA—FORMULAS
Think of 2πx s the circumference of a circle
traced out by the point (x, y) on the curve as
it is rotated about the y-axis.
SURFACE AREA
Example 1
The curve y  4  x 2, –1 ≤ x ≤ 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.
SURFACE AREA
Example 1
We have:
dy 1
2 1 2
 2 (4  x ) ( 2 x)
dx
x

2
4 x
Example 1
SURFACE AREA
So, by Formula 5, the surface area is:
2
 dy 
S   2 y 1    dx
1
 dx 
1
 2 
1
 2 
1
1
1
1
4  x2
4 x
2
x2
1
dx
2
4 x
2
dx
2
4 x
 4  1 dx  4 (2)  8
1
SURFACE AREA
Example 2
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.
E. g. 2—Solution 1
SURFACE AREA
Using y = x2 and dy/dx = 2x,
from Formula 8, we have:
S   2 x ds

2
1
2
 dy 
2 x 1    dx
 dx 
2
 2  x 1  4 x 2 dx
1
E. g. 2—Solution 1
SURFACE AREA
Substituting u = 1 + 4x2, we have du = 8x dx.
Remembering to change the limits
of integration, we have:
S


17
4 5

6
u du 

3 2 17
 23 u 
5
4
(17 17  5 5)
SURFACE AREA
E. g. 2—Solution 2
Using
x = y and dx/dy =
1,
2 y
we have the following solution.
E. g. 2—Solution 2
SURFACE AREA
2
S   2 xds  
4
1
 dx 
2 x 1    dy
 dy 
 2 
4
1

4
1




4
17
5

6
1
y 1  dy
4y
4 y  1dy
udu
(17 17  5 5)
(where u  1  4 y )
SURFACE AREA
Example 3
Find the area of the surface generated
by rotating the curve y = ex, 0 ≤ x ≤ 1,
about the x-axis.
Example 3
SURFACE AREA
Using Formula 5 with y = ex and dy/dx = ex,
we have:
2
 dy 
S   2 y 1    dx
0
 dx 
1
1
 2  e 1  e dx
x
2x
0
 2 
e
1
1  u du
2
(where u  e )
x
SURFACE AREA

 2  sec  d
 4
3
Example 3
(where u  tan  and   tan e)
1

 2 sec  tan   ln sec   tan  
1
2
4
(E.g.8, Sec.7.2)
  sec  tan   ln  sec   tan    2  ln


2 1 

Example 3
SURFACE AREA
Since tan α = e , we have:
sec2α = 1 + tan α = 1 + e2
Thus,


2
2

S   e 1  e  ln e  1  e  2  ln



2 1 
