Section 8.2 Area of a Surface of Revolution
Ruipeng Shen
Feb 12
1
Area of a Portion of a Circular Cone
As shown in the following figure, we can calculate the area by flattening the surface
l1
l
l1
r1
2𝛑r1
l
r2
2𝛑r2
1
1
· 2πr2 · (l1 + l) − · 2πr1 · l1 = π[(r2 − r1 )l1 + r2 l].
2
2
By using symmetric triangles we have
A=
l1
l
r1 l
=
=⇒ l1 =
.
r1
r2 − r1
r2 − r1
Thus we have
A = π (r2 − r1 ) ·
r1 l
r1 + r2
+ r2 l = 2π ·
· l = 2πrl.
r2 − r1
2
1
2
Area of a General Surface of Revolution
y
y=f(x)
ds
f(x)
a
b
x
Figure 1: Surface of Revolution
Let us consider the surface shown in figure 1, which is obtained by rotating the curve y =
f (x), a ≤ x ≤ b about the x-axis. In order to calculate the surface area, we divide the interval
[a, b] into n sub-intervals with endpoints x0 , x1 , · · · , xn and divide the surface into “bands”
accordingly. Each band can be approximated by a portion of circular cone as shown in the
previous section. Thus the area of the i-th band is roughly
q
2πf (x∗i ) · ds ≈ 2πf (x?i ) 1 + [f 0 (x∗i )]2 ∆x
Adding them up and taking the limit we have
Z
n
q
X
S = lim
2πf (x?i ) 1 + [f 0 (x∗i )]2 ∆x =
n→∞
b
p
2πf (x) 1 + [f 0 (x)]2 dx.
a
i=1
Using the Leibniz notation for derivatives, the formula becomes
s
2
Z b
dy
S=
2πy 1 +
dx.
dx
a
If the curve is described as x = g(y), c ≤ y ≤ d, then the formula for the surface area becomes
s
2
Z d
dx
S=
2πy 1 +
dy.
dy
c
2
In summary, we can use the notation for the arc length and rewrite
Z
S = 2πy ds.
For rotation about the y-axis, the surface area formula becomes
Z
S = 2πx ds.
For ds we can use either
s
s
2
2
dy
dx
ds = 1 +
dx
or
ds = 1 +
dy
dx
dy
√
Example 1. The curve y = 4 − x2 , −1 ≤ x ≤ 1, is an arc of the circle x2 + y 2 = 4. Find the
area of the surface obtained by rotating this arc about the x-axis.
Figure 2: Part of A Sphere
Solution
We can calculate
dy
−x
=√
;
dx
4 − x2
Thus the surface area is
s
r
2
Z 1p
Z 1
dy
x2
2
S=
2πy 1 +
dx = 2π
4−x 1+
dx
dx
4 − x2
−1
−1
Z 1
=2π
2 dx = 8π.
−1
Example 2. The arc of the parabola y = x2 from (1, 1) to (2, 4) is rotated about y-axis. Find
the area of the resulting surface.
3
Figure 3: Paraboloid
Solution
Using
y = x2
dy
= 2x,
dx
and
we have
Z
S=
s
2
Z
2πx 1 +
2πx ds =
1
Z
dy
dx
2
dx
2
hπ
p
3/2 i2
2πx 1 + 4x2 dx =
1 + 4x2
6
1
1
π 3/2
3/2
17 − 5
=
6
Example 3. Find the area of the surface generated by rotating the curve y = ex , 0 ≤ x ≤ 1,
about the x-axis.
=
Solution
Using
y = ex
dy
= ex ,
dx
and
we have
s
2
dy
dx
dx
0
Z 1
Z ep
p
x
2x
=
2πe
1 + e dx = 2π
1 + u2 du
0
1
p
e
p
u
1 2
2
1 + u + ln u + 1 + u
= 2π
2
2
1
h p
p
√
√ i
2
2
= π e 1 + e + ln(e + 1 + e ) − 2 − ln(1 + 2) .
Z
S=
Z
2πy ds =
1
2πy
1+
4