What is the surface area of the figure
obtained by rotating y = x2 about the
x-axis on the interval [0, 1]?
Z
b
2π ∗ (radius) ∗ (arclength)
Z 1 √
x2 1 + 4x2 dx
= 2π
Z0 arctan 2
1
1
= 2π
tan2 u ∗ sec u ∗ sec2 u du
4
0
|2 {z }
dx
Z
π arctan 2
=
sec2 u − 1 sec3 u du
4 0
Z
π arctan 2
sec5 u − sec3 u du
=
4 0
arctan 2
Z
Z
3
π 1
3
3
3
tan u sec u +
sec u du − sec u du
=
4 4
4
0
arctan 2
Z
1
π 1
=
tan u sec3 u −
sec3 u du
4 4
4
0
arctan 2
π 1
1 1
1
3
=
tan u sec u −
sec u tan u + ln |sec u + tan u|
4 4
4 2
2
0
1
√
√
1
π 1
1 x(4x2 + 1)3/2 − x 4x2 + 1 − ln 4x2 + 1 + 2x
=
4 2
4
8
0


√
√
1
1
1
3/2
∗ 1 ∗ 5 − ∗ 1 ∗ 5 − ln 5 + 2

2
4
8
π

=  √

√
4
1
1
1

−
∗ 0 ∗ (0 + 1)3/2 − ∗ 0 ∗ 0 + 1 − ln 0 + 1 + 2 ∗ 0
2
4
8
"
! √
#
π
53/2
5 1 √
1
=
−
− ln 5 + 2 − 0 − 0 − ln |1|
4
2
4
8
8
"
#
√
π 53/2
5 1 √
−
− ln 5 + 2
=
4
2
4
8
S=
a
≈ 3.80973 . . .
√
Noting the identity sinh−1 x = ln 1 + x2 + x, we could also express our
answer as
"
#
√
3/2
π 5
5 1
−
− sinh−1 (2)
4
2
4
8
1
2x = tan u
u = arctan 2x
“An ice cream cone where you can eat the ice cream but
not the cone.”
Also known as Gabriel’s Horn or Torricelli’s Trumpet, consider the surface
obtained by rotating y = 1/x on [1, ∞) about the x-axis.
The volume (i.e., amount of ice cream) is given by:
b
Z
π(radius)2
V =
a
Z
∞
1
dx
x2
1
Z R
1
dx
= π lim
R→∞ 1 x2
R
−1 = π lim
R→∞ x 1
−1 −1
−
= π lim
R→∞
R
1
=π∗1
=π
=π
So we have a finite volume, and thus our “cone” holds a finite ammount of ice
cream. Now for surface area (i.e., amount of waffle cone):
Z
b
2π(radius) ∗ (arc length)
s
2
Z ∞ 1
−1
1+
dx
=
2π
x
x2
1
Z R r
1
1
= lim 2π
1 + 4 dx
R→∞
x
x
1
S=
a
Let’s tackle this by finding an indefinite integral first and then dealing with
bounds:
r
Z r
Z 1
1
1 ∗ x4
1
2π
1 + 4 dx = 2π
1 + 4 dx
4
x
x
x∗x
x
2
r
1
1
= 2π x 1 + 4 ∗
dx
x
x5
Z −1
−π
1 √
1+u∗
du
u = 1/x4
=
2
u
4
Z √
−π
1+u
=
du
2
u
√
Z √
−π
1+u
1+u
=
∗√
du
2
u
1+u
Z
√
s2
= −π
ds
s
=
1+u
s2 − 1
Z 1/2
1/2
1+
= −π
−
ds
division, part. frac.
s−1 s+1
1
1
= −π s + ln |s − 1| − ln |s + 1| + C
2
2
1 √
√
√
1 1 + u + ln 1 + u − 1 − ln 1 + u + 1 + C
= −π
2
2
r
r
#
"r
1
1 1
1
1 1 + 4 + ln 1 + 4 − 1 − ln 1 + 4 + 1 + C
= −π
2 x
2 x
x
r
r
#
" r
−π
1
1
1
=
2 1 + 4 + ln 1 + 4 − 1 − ln 1 + 4 + 1 + C
2
x
x
x
√
√
#
" √
1 + x4
1 + x4
−π 2 1 + x4
+ ln − 1 − ln + 1 + C
=
2
2
2
x
x
2
x
√
√
#
" √
2
2
4
4
4
−π 2 1 + x
1+x −x 1+x +x + ln =
− ln +C
2
2
2
x
x
x2
#
√
" √
1 + x4 − x2 −π 2 1 + x4
=
+ ln √
+C
2
4
2
1+x +x 2
x
√
√
2
4
4
−π 1 + x
π 1+x +x =
+ ln √
+C
2
x
2 1 + x4 − x2 Z
4
Now that we have an indefinite integral, let’s evalute our original definite
integral:
s
2
Z ∞ 1
−1
S=
2π
1+
dx
x
x2
1
Z R r
1
1
1 + 4 dx
= lim 2π
R→∞
x
x
1
√
#R
" √
2
4
4
−π 1 + x
π 1+x +x = lim
+ ln √
2
R→∞
x
2 1 + x4 − x2 1
3
√
√
√
!
!#
√
2
4
4
−π 1 +
π 1+R +R −π 1 + 1
π 1 + 14 + 12 = lim
+ ln √
+ ln √
−
R→∞
R2
2 1 + R4 − R2 12
2 1 + 14 − 12 √
√
#
" √
2
4
4
√
−π 1 + R
π 2 + 1 π 1+R +R = lim
+ ln √
+ π 2 − ln √
R→∞
R2
2 1 + R4 − R2 2 2 − 1
q


q
2
1
1
2
R
+
1
+
1
−πR
+
1
√ 
√
R4
π
π R4

= lim 
ln
2
−
ln(3
+
2
2)
+
+
π
q
R→∞
R2
2 R2
2
1
+1−1 R4
q


r
1
+ 1 + 1 √ 
√
R4
1
π
π 
= lim −π
ln
2
−
ln(3
+
2
2)
+
1
+
+
π
q
R→∞
R4
2 2
1
+
1
−
1
R4
√
√
π
π
= −π + ∞ + π 2 − ln(3 + 2 2)
2
2
=∞
"
R4
and thus we have an infinite surface area.
Are you adequately convinced that you really don’t want to do this on a
test? Surely there’s an easier way...let’s go back to our integral of interest:
Z ∞ r
1
1
1 + 4 dx
2π
S=
x
x
1
For 1 ≤ x < ∞,
√
1
1≤ 1+ 4 ≤ 2≤2
x
r
1
=⇒ 2π ≤2π 1 + 4 ≤ 2 ∗ 2π
x
r
2π 2π
1
4π
=⇒
≤
1+ 4 ≤
x
x
x
x
r
Therefore,
Z
1
∞
2π
x
r
1
1 + 4 dx ≥
x
∞
Z
1
2π
dx = 2π
x
Z
1
∞
1
dx.
x
R∞
R∞ q
Since we know that 1 x1 dx diverges to +∞, we have that 1 2π
1 + x14 dx
x
diverges to +∞ by the comparison theorem. Therefore, we have an infinite
surface area.
4