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U
n
iv ersity o f
S a s k at che w
a
n
DEO ET
PAT- RIE
Areas of Surfaces of Revolution
Let a cone have height h and radius r , so that the circumference of the base is 2 π r. If
the cone is cut along a straight line (of length R =
r 2
- h 2 ) from the vertex to the base
and flattened out, we obtain a sector of the circle whose radius is R. The area of the
circle of radius R is π R
2 = π (r
2
2 ) , so the ratio of the area of the sector to that of
the circle is
2 π r
2 π R
r
R
, so the area of the sector is
r
R
π R
2 = π r
r
2
2 .
H
r
R
r
R
If we now take two cones, with one being a subset of the other, we can calculate the area
of the region between the bases of the two cones. This region is called a frustum.
h 1
r 1
r 2
Let the larger and smaller cones have heights and radii h 2 and r 2 and h 1 and r 1.
Let R 2 =
r
2 2
2 2 and R 1 =
r
2 1
2 1 , so that the areas of the larger and smaller cones
are
A 2 = π r 2
r
2 2 +^ h
2 2 and^ A^1 =^ π r^1
r
2 1 +^ h
2 1
U
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iv ersity o f
S a s k at che w
a
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DEO ET
PAT- The area of the frustum is thus RIE
A = A 2 − A 1 = π r 2
r
2 2
2 2 − π r 1
r
2 1
2 1
π
[
r 2
r
2 2 +^ h
2 2 −^ r^1
r
2 1 +^ h
2 1
]
= π (r 2 R 2 − r 1 R 1 )
Writing R = R 2 − R 1 and r =
r 1 + r 2
, and using similar triangles, we derive
A = 2 π r R.
We can then use this formula to derive a formula for the area of the surface obtained
by rotating the curve (x(t), y(t)) , t 1 ≤ t ≤ t 2 about the x - and y −axes respectively:
S x =
t 2
t 1
2 π y(t)
(x
(t))
y
(t)
dt and
S
y
t 2
t 1
2 π x(t)
(x
(t))
y
(t)
dt
If the curve is the graph of a function y = f (x) , a ≤ x ≤ b , then the area of the surface
obtained by revolving the curve about the x -axis is
S
x
b
a
2 π f (x)
1 + (f
(x))
dx
and the area of the surface obtained by revolving the curve about the y -axis is
S y =
b
a
2 π x
1 + (f
(x))
dx
If the curve is the graph of a function x = g(y) , c ≤ x ≤ d , then the area of the surface
obtained by revolving the curve about the x -axis is
Example:(#6, p.552, green Stewart)
Find the area of the surface obtained by
rotating the curve y
2 = 4 x + 4, 0 ≤ x ≤ 8,
about the x -axis.
x
Solution 1: We have y = 2 (x + 1 )
1 (^2) , so y ′ = 2
(x + 1 )
−
1 (^2) = (x + 1 ) −
1 (^2) ,
and Sx =
8
0
2 π ( 2 )(x + 1 )
1 2
(x + 1 )
−
1 2
2 dx =
4 π
8
0
(x + 1 )
1 2
x + 1
dx = 4 π
8
0
(x + 2 )
1 (^2) dx = 4 π
(x + 2 )
3 2
3 2
8
0
8 π
(x + 2 )
3 2
8
0
8 π
[
3 2 − ( 0 + 2 )
3 2
]
16 π
Solution 2: We have x =
y
2 − 4
y
2 − 1, so x
′
y
and S (^) x =
y = 6
y = 2
2 π y
1 + (x ′ )
2 dy = 2 π
y = 6
y = 2
y
y
2
dy =
π
y = 6
y = 2
y
4 + y 2 dy = (letting y = 2 tan θ) =
π
θ =arctan 3
θ =arctan 1
2 tan θ
4 + 4 tan
2 θ
2 2 sec
2 θdθ =
π
θ =arctan 3
θ =arctan 1
2 tan θ
4 sec 2 θ 2 sec
2 θdθ =
π
θ =arctan 3
θ =arctan 1
2 tan θ( 2 sec θ) 2 sec
2 θdθ =
8 π
θ =arctan 3
θ =
π 4
sec
2 θ sec θ tan θdθ = ( letting u = sec θ) =
8 π
u =
√ 10
u =
√ 2
u
2 du = 8 π
u
3
u =
√ 10
u =
√ 2
π
16 π
Example:(#8, p.552), green Stewart)
Find the area of the surface obtained by
rotating the curve y =
x
2
ln x
, 1 ≤ x ≤ 4,
about the x -axis.
0 x
Solution: We have y
′
x
2 x
and S (^) x =
4
1
2 π
x
2
ln x
x
2 x
dx =
2 π
4
1
x
2
ln x
x
2 x
dx =
2 π
4
1
x
2
ln x
x
2 x
dx =, etc....
Example:(#6, p.524), brown Stewart) Find the area of the surface ob-
tained by rotating the curve y = cos x , 0 ≤ x ≤
π
, about the x -axis.
Solution: We have y
′ = − sin x ,
and S (^) x =
x =
π 3
x = 0
2 π cos x
1 + ( − sin x)
2 dx = ( letting u = sin x) =
2 π
√ 3 2
0
1 + u 2 du = ( letting u = tan θ) = 2 π
arctan
√ 3 2
0
sec
3 θdθ =
2 π
sec θ tan θ + ln |sec θ + tan θ |
arctan
√ 3 2
0
π ( sec θ tan θ + ln |sec θ + tan θ | ) |
arctan
√ 3 2 0 =
π
= π
Example:(#8, p.524), brown Stewart) Find the area of the surface ob-
tained by rotating the curve y =
x
2 (^3) , 1 ≤ x ≤ 8, about the x -axis.
