Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
7.5
Area of a surface of revolution
The surface of a solid of revolution is the lateral boundary of the solid. It
is formed when a curve is rotated about a line. In this section, we want to
find the area of such a surface.
Example 7.5.1
Find the surface area of a circular cone with base radius r and slant height
l.
Solution.
Recall from Example 6.2.12 the area of a circular sector of radius l and
central angle θ is given by A = 12 l2 θ. Cutting the cone along the dashed line
and flattening out we obtain a circular sector with radius l and central angle
θ as shown in Figure 7.5.1.
Figure 7.5.1
The area of this circular sector which is the area of the cone is
1 2
1
2πr
l θ = l2
= πrl
2
2
l
1
Example 7.5.2
Find the surface area of a portion of a cone as shown in Figure 7.5.2. Such
a solid is called a frustum.
Figure 7.5.2
Solution.
Let S be the surface area of the portion of the cone. Then
S = πr2 (l1 + l) − πr1 l1 .
From similar triangles we have
l1
l1 + l
=
=⇒ r2 l1 = r1 l1 + r1 l.
r1
r2
Thus,
S = π(r1 + r2 )l
Now, consider the surface shown in Figure 7.5.3 which is obtained by rotating
the function f (x), a ≤ x ≤ b about the x−axis.
2
Figure 7.5.3
Divide the interval [a, b] into n subintervals each of length ∆x = b−a
n using
the partition points xi = a + i∆x, i = 0, 1, · · · , n. We approximate the
arc from the point Pi−1 (xi−1 , f (xi−1 )) to Pi (xi , f (xi )) with the line segment
from these two points. When we rotate this line segment about the x−axis
we obtain a portion of a cone as in Example 7.5.2. Thus, the surface area
from x = xi−1 to x = xi is estimated by
q
Si ≈ π(f (xi ) + f (xi−1 ))|Pi−1 Pi | = π(f (xi ) + f (xi−1 )) 1 + [f (x∗i )]2 ∆x.
For small ∆x, we have f (xi−1 ) ≈ f (x∗i ) and f (xi ) ≈ f (x∗i ). Hence,
q
Si ≈ 2πf (x∗i ) 1 + [f (x∗i )]2 ∆x.
Hence, the surface area of the solid of revolution is approximated by
S≈
n
X
2πf (x∗i )
q
1 + [f (x∗i )]2 ∆x.
i=1
This approximation appears to become better when n → ∞ giving
Z b
p
S=
2πf (x) 1 + [f 0 (x)]2 dx.
a
Note that in the above
p formula, 2πf (x) is the circumference of the circle
with radius f (x) and 1 + [f 0 (x)]2 is an element of arc length. In general,
if a region is rotated about the x−axis its surface area is
Z upper limit
S=
2πyds
lower limit
3
where
s
ds =
1+
dy
dx
2
dx
dy
2
dx
if y = f (x) and
s
1+
ds =
dy
if x = g(y). Likewise, if a region is rotated about the y−axis its surface area
is
Z
upper limit
2πxds
S=
lower limit
where
s
ds =
1+
dy
dx
2
dx
dy
2
dx
if y = f (x) and
s
ds =
1+
dy
if x = g(y).
Example 7.5.3
Find
the surface area of the solid of revolution generated by rotating y =
√
4x + 1, 1 ≤ x ≤ 5 about the x−axis.
Solution.
We have
Z
S=
s
√
5
2
1+ √
4x + 1
2π 4x + 1
1
Z
=2π
5√
2
dx
5 + 4xdx
1
25 √
Z
π
π 2 3 25
=
udu =
u2
2 9
2 3
9
3
3
π
98
= [25 2 − 9 2 ] = π
3
3
Example 7.5.4
Find the surface area of the solid of revolution generated by rotating y =
x2 , 1 ≤ x ≤ 2 about the y−axis.
4
Solution.
We have
Z
2
S=
p
2πx 1 + (2x)2 dx
1
Z
π 17 √
udu
=
4 5
π h 3 i17
= u2
6
5
√
√
π
= (17 17 − 5 5)
6
where u = 1 + 4x2
Example 7.5.5
2
Find the exact surface area of the region obtained by rotating y = e−x ,
where −1 ≤ x ≤ 1 about the y−axis.
Solution.
The surface area is
1
Z
S=
p
2πx 1 + 4x2 e−2x2 dx ≈ 3.9603
0
Example 7.5.6
Find the exact surface area of the region obtained by rotating x = ln (2y + 1), 0 ≤
y ≤ 1 about the x−axis.
Solution.
The surface area is
Z
S=
s
1
2πy
0
1+
4
dy ≈ 4.258
(2y + 1)2
Example 7.5.7
Find the exact surface area of the region obtained by rotating x = ln (2y + 1), 0 ≤
y ≤ 1 about the y−axis.
Solution.
The surface area is
s
Z 1
S=
2π ln (2y + 1) 1 +
0
5
4
dy ≈ 5.605
(2y + 1)2