Texas College Mathematics Journal
Volume 4, Number 2, Pages 13–20
S applied for(XX)0000-0
Article electronically published on December 12, 2007
DISK VERSUS FRUSTUM
CONG X. KANG AND EUNJEONG YI
Abstract. This note provides a friendly exposition which fills a gap in typical,
current calculus texts. We ask and answer the question: “Why, in dealing with
a solid of revolution, do we use disks for volume, but frusta for surface area?”
We provide a rationale with explicit examples.
1. Introduction
At some point in the calculus sequence, we tackle the volume and surface area
of a solid of revolution of a plane curve. The infinitesimal element to be integrated
is a disk in the case of finding volume, and it is the frustum of a cone in the case of
finding surface area. But why do we use two distinct shapes? Until the completion
of the first draft of this note, we had not been able to find a standard calculus
text which explains this matter (for examples, [1], [4], [5], and [8]). Since then, we
became aware that this topic was discussed by George B. Thomas on pp.180-181
of his 1983 text (see [6]). However, it is noteworthy that this topic is omitted or
stricken out of later editions or variations of the 1983 text (see [7]). Thus, it seems
timely to reprise this matter in the present note, which contains explicit examples
and more details than provided in [6].
For simplicity, we assume throughout this article that f (x) ≥ 0 and f 0 (x) is
continuous for all x on a closed interval [a, b]. Let us recall the formulas for the
volume V and the surface area A of the solid obtained by revolving the curve
y = f (x), a ≤ x ≤ b, about the x−axis:
Z
b
V=
πy 2 dx,
a
Z
A=
s
b
2πy dl, where dl =
a
µ
1+
dy
dx
¶2
dx.
As is well known, πy 2 dx represents the volume of an infinitesimal disk, and 2πydl
represents the surface area of an infinitesimal frustum. See Figure 1 and Figure 2.
Intuitively, the frustum is a better approximation than the cylindrical disk both
in finding the volume and the surface area, since the frustum is to the disk as the
trapezoid is to the rectangle –for calculations of areas under curves. So, why does
the cylindrical disk suffice as the elemental solid in finding volume? Dually, one
Received by the editors May 21, 2007 and, in revised form, October 17, 2007.
1991 Mathematics Subject Classification. Primary 26B15; Secondary 26-01.
Key words and phrases. volume, surface area, disk, frustum.
c
°2007
(copyright holder)
13
14
CONG X. KANG AND EUNJEONG YI
can ask why the cylindrical band of a disk is not good enough for the purpose of
surface area computation.
2. Volume
We first show that disk and frustum methods yield the same result in the calculations of volumes. Let ∆xi = xi − xi−1 , ∆yi = yi − yi−1 , x0 = a, and xn = b.
The volume ∆Vi of the disk (see Figure 1) is
∆Vi = π(yi∗ )2 ∆xi , where yi∗ = f (x∗i ), x∗i ∈ [xi−1 , xi ].
Thus, the volume V using the disks is the following:
V = lim
n→∞
n
X
∆Vi = lim
n
X
n→∞
i=1
Z
π(yi∗ )2 ∆xi
=
b
πy 2 dx.
a
i=1
y = f (x)
a
b
xi−1
x
xi
Figure 1
On the other hand, to find the volume of the frustum of a cone (see Figure 2), we
first find the x−intercept of the line that passes through (xi−1 , yi−1 ) and (xi , yi ).
∆yi
We can write the equation of the line as y −yi = ∆x
(x−xi ), and it follows that the
i
∆xi
x−intercept is xi −yi ∆yi . Without loss of generality, we may assume yi ≥ yi−1 > 0.
The volume ∆V̄i of the frustum of a cone, obtained by subtracting the volume of
the little cone from the volume of the big cone, is
1 2 ∆xi
1 2
∆xi
∆V̄i =
πy (yi
) − πyi−1
(xi−1 − xi + yi
)
3 i
∆yi
3
∆yi
1 2 ∆xi
1 2
∆xi
=
πyi (yi
) − πyi−1
(−∆xi + yi
)
3
∆yi
3
∆yi
1
1
1
2
2
=
π∆xi (yi3
+ yi−1
− yi−1
yi
)
3
∆yi
∆yi
1
yi
2
2
=
π∆xi (
(y 2 − yi−1
) + yi−1
)
3
∆yi i
1
2
=
π∆xi (yi2 + yi−1 yi + yi−1
).
3
DISK VERSUS FRUSTUM
15
y = f (x)
a
b
xi−1
x
xi
i
xi − yi ∆x
∆yi
Figure 2
Thus, the volume V̄ using the frusta is
V̄
(2.1)
=
=
n
X
lim
n→∞
lim
n→∞
∆V̄i
i=1
n
X
1
2
π(yi2 + yi−1 yi + yi−1
)∆xi .
3
i=1
Though (2.1) is not a Riemann sum, Bliss’ theorem (see [2]) says that it still equals
Rb 2
πy dx. Thus, the disk method and the frustum method yield the same volume.
a
3. Surface Area
We begin by observing that disk and frustum methods do not yield the same
result in computing surface areas. The surface area ∆Ai of the frustum of a cone
(see Figures 2, 3) is
∆li
∆yi
∆xi
a
y = f (x)
xi−1 xi
b
x
Figure 3
(3.1)
∆Ai = 2πri ∆li = 2πri
p
(∆xi )2 + (∆yi )2 , where ri =
f (xi−1 ) + f (xi )
.
2
16
CONG X. KANG AND EUNJEONG YI
Thus, the surface area A using the frusta is the following:
A =
(3.2)
=
lim
n→∞
lim
n→∞
n
X
∆Ai = lim
n→∞
i=1
n
X
n
X
r
π(f (xi−1 ) + f (xi ))
1+(
i=1
π(f (xi−1 ) + f (xi ))
∆yi 2
) ∆xi
∆xi
p
1 + [f 0 (ci )]2 ∆xi
i=1
for some ci ∈ (xi−1 , xi ) by the Mean Value Theorem as applied to f 0 . By Bliss’
theorem, (3.2) becomes
s
µ ¶2
Z b
Z b
p
dy
2πf (x) 1 + (f 0 (x))2 dx =
2πy 1 +
dx,
(3.3)
dx
a
a
as ∆xi approaches zero.
On the other hand, the surface area ∆Āi of the cylindrical band of a disk (see
Figure 1) is
(3.4)
∆Āi = 2π r¯i ∆xi , where r¯i = f (x∗i ), x∗i ∈ [xi−1 , xi ].
Thus, the surface area Ā using the disks is
(3.5)
Ā = lim
n→∞
n
X
i=1
∆Āi = lim
n→∞
n
X
Z
2πf (x∗i )∆xi
i=1
=
b
2πf (x)dx.
a
Clearly ri of (3.1) and r¯i of (3.4) Ã
become equal as ∆x
!i approaches zero. However,
r
³ ´2
dy
(∆xi )2 + (∆yi )2 − ∆xi tends to
1 + dx
− 1 dx as ∆xi approaches zero,
p
and the latter is of course nonzero wherever the curve has a non-horizontal tangent.
As a first example, consider the cone given by revolving about the x-axis the
line y = f (x) = mx, m 6= 0, 0 ≤ x ≤ b. The surface area A using (3.3), the
√
√
Rb
correct formula, gives A = 0 2πmx 1 + m2 dx = πm 1 + m2 b2 . However, Ā =
Rb
2πmxdx = πmb2 using (3.5), which is clearly different from A for m 6= 0.
0
4. Further Rationale
To better understand the issue at hand, we consider a viewpoint which presumably dates back to Newton and Leibniz; that is, only differentials of the lowest order need to be kept. Consider A = xy and A∗ = (x + ∆x)(y + ∆y), we
note that ∆A = A∗ − A = x∆y + y∆x + ∆x∆y. This suggests, a priori, that
dA = xdy + ydx + dxdy. But, of course, dxdy does not actually appear in the
product (Leibniz) rule. By the same token, the first order difference in surface
area between the elemental frustum and the elemental disk is important; but the
difference in volume between the elemental frustum and the elemental disk, being
of second order, may be ignored. Towards making clear this notion, we furnish two
pieces of computation. First, we show that the frustum integral (3.3) follows from
the more formal definition of the area of a parametrized surface. Then, we show a
less trivial example (as compared to the cone example) where, of the same surface,
the difference between the area as computed via the frustum method and the area
DISK VERSUS FRUSTUM
17
as computed via the disk method is seen to be of first order.
Recall the definition of the area of a parametrized surface (see p. 910 of [4]):
Definition 4.1. If a smooth parametric surface S is given by the equation
r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, (u, v) ∈ D
and S is covered just once as (u, v) ranges throughout the parameter domain D,
then the surface area of S is
Z Z
|ru × rv | dudv,
(4.1)
A(S) =
D
where ru =
∂x
∂u i
∂y
∂u j
+
+
∂z
∂u k
and rv =
∂x
∂v i
+
∂y
∂v j
+
∂z
∂v k.
Note that the area so defined is independent of parametrization (see section 7.6
of [3]).
Consider the surface obtained by revolving y = f (x) about the x−axis. Then
the topphalf of the surface (y ≥ 0 portion) is parametrized by r : (u, v) 7−→ (x =
u, y = (f (u))2 − v 2 , z = v) over the (generally non-rectangular) domain defined
by u = a, u = b, v = −f (u), and v = f (u). Using the above definition, we will
show that
Z b Z f (u)
Z b
p
(4.2)
|ru × rv | dvdu = π
f (u) 1 + (f 0 (u))2 du.
a
−f (u)
a
Note that ru × rv = √f (u)f
0
(u)
(f (u))2 −v 2
i−j− √
v
k,
(f (u))2 −v 2
s
|ru × rv | =
and thus
s
(f (u)f 0 (u))2
v2
+1+
=
2
2
(f (u)) − v
(f (u))2 − v 2
Therefore,
Z f (u)
|ru × rv | dv
Z
f (u)
s
=
−f (u)
−f (u)
Z
=
π
2
−π
2
Z
π
2
(f (u))2 [1 + (f 0 (u))2 ]
.
(f (u))2 − v 2
s
s
(f (u))2 [1 + (f 0 (u))2 ]
dv, put v = f (u) sin θ,
(f (u))2 − v 2
(f (u))2 [1 + (f 0 (u))2 ]
f (u) cos θ dθ
(f (u))2 − (f (u))2 sin2 θ
(f (u))2 [1 + (f 0 (u))2 ]
f (u) cos θdθ
(f (u))2 cos2 θ
−π
2
¯ π2
p
p
¯
= f (u) 1 + (f 0 (u))2 θ¯ −π = πf (u) 1 + (f 0 (u))2 ,
=
2
and thus (4.2) follows. Remembering that we computed only the area of one of two
equal halves of the surface of revolution, the frustum integral (3.3) is thus verified.
Note that we can satisfy a more restrictive definition of a parametrized surface
found in some calculus texts (cf. p.1108 of [7]) by pre-composing r with a change
of coordinates map given by T : (u, v) 7−→ (u, f (u)(2v − 1)) from [a, b] × [0, 1] to
18
CONG X. KANG AND EUNJEONG YI
the non-rectangular parametrizing domain.
Now, consider the surface S (a thin horizontal slice of the unit sphere at φ = π4 ,
see Figure 4) defined by π4 ≤ φ ≤ π4 + δ and 0 ≤ θ ≤ 2π on the unit sphere given by
ρ = 1, where (ρ, θ, φ) are the spherical coordinates. We denote by A, A∗ , and A∗∗
the actual area (as defined in (4.1)), the area as given by the elemental frustum,
and the area as given by the elemental disk, respectively, for the surface S. We will
show that
A − A∗∗
A − A∗
= 0 and lim
6= 0.
lim
δ→0
δ→0
δ
δ
z
(ρ = 1, θ, φ = π4 )
(ρ = 1, θ, φ =
π
4
+ δ)
y
x
Figure 4
To compute A, we use the parametrization r(φ, θ) = (sin φ cos θ)i+(sin φ sin θ)j+
(cos φ)k.
Z
2π
Z
π
4 +δ
A =
0
=
√
Z
2π
sin φ dφ dθ =
π
4
2π(1 − cos δ + sin δ) =
π
[− cos φ] π4
0
√
2π(2 sin2
4
+δ
dθ = 2π(cos
π
π
− cos( + δ))
4
4
δ
+ sin δ).
2
Next, we compute A∗ (see Figure 5).
A∗
=
=
=
=
=
π(r1 + r2 )l
r
π
π
π
π
π
π
π(sin + sin( + δ)) (sin( + δ) − sin )2 + (cos( + δ) − cos )2
4
4
4
4
4
4
√
p
2
π(1 + cos δ + sin δ) 2(1 − cos δ)
2
r
√
2
δ
π(1 + cos δ + sin δ) 4 sin2
2
2
√
δ
2π(1 + cos δ + sin δ) sin .
2
DISK VERSUS FRUSTUM
z
19
z
r1
r1
(ρ = 1, θ, φ = π4 )
(sin π4 , cos π4 )
l
r2
(ρ = 1, θ, φ =
π
4
+ δ)
r2
y
y
(sin( π4
+
δ), cos( π4
+ δ))
x
the slice on the yz − plane when θ =
π
2
Figure 5
Now, we compute A∗∗ . For convenience, we take a disk of radius 12 (r1 + r2 ) (see
Figure 6). Then
A∗∗
sin π4 + sin( π4 + δ)
π
π
)(cos − cos( + δ))
2
4
4
π
π
(1 + cos δ + sin δ)(1 − cos δ + sin δ) = ((1 + sin δ)2 − cos2 δ)
=
2
2
= π(sin δ + sin2 δ).
= 2π(
z
r1 +r2
2
φ=
π
4
y
φ=
π
4
+δ
x
Figure 6
Therefore, using limδ→0
∗
A−A
lim
δ→0
δ
=
=
=
sin δ
δ
√
= 1, we have
2π(2 sin2
δ
2
√
2π(1 + cos δ + sin δ) sin 2δ
δ→0
δ
√
δ
1 1
1
2π[lim (sin ) + 1 − − lim (cos δ) − lim (sin δ)]
δ→0
2
2 2 δ→0
2 δ→0
√
1 1
2π(0 + 1 − − − 0) = 0,
2 2
lim
+ sin δ) −
20
CONG X. KANG AND EUNJEONG YI
and
A − A∗∗
δ→0
δ
lim
√
=
=
=
It follows that
+ sin δ) − π(sin δ + sin2 δ)
δ→0
δ
√
√
δ
π[ 2 lim (sin ) + 2 − 1 − lim (sin δ)]
δ→0
δ→0
2
√
π( 2 − 1) 6= 0.
lim
∗
∗∗
limδ→0 A −A
δ
2π(2 sin2
δ
2
6= 0, and this is in contrast to the fact that
∆Vi − ∆V̄i
π
2
[((yi∗ )2 − yi2 ) + ((yi∗ )2 − yi−1 yi ) + ((yi∗ )2 − yi−1
)] = 0,
= lim
∆xi →0
∆xi →0 3
∆xi
where ∆Vi and ∆V̄i are as in section 2.
lim
Acknowledgement. The authors would like to thank Ricardo Alfaro for calling their
attention to [6]. They would also like to thank the editor and the referee for corrections and valuable suggestions.
References
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Hall, 2008.
2. H. B. Fine, Note on a substitute for Duhamel’s theorem, The Annals of Mathematics, 19
(1918) 172-173.
3. J. E. Marsden and A. J. Tromba, Vector Calculus, 4th ed., W. H. Freeman and Company,
2001.
4. J. Stewart, Calculus - early vectors, Brooks/Cole publishing company, 1999.
5. E. Swokowski, Calculus with analytic geometry, 4th ed., PWS-KENT publishing company,
1988.
6. G. B. Thomas, Jr., Calculus and Analytic geometry, classic edition, Addison Wesley, 1983.
7. G. B. Thomas, Jr. and R. L. Finney, Calculus and Analytic geometry, alternate edition,
Addison Wesley, 2003.
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Hall, 2007.
Texas A&M University-Galveston, Galveston, TX 77553
E-mail address: [email protected]
Texas A&M University-Galveston, Galveston, TX 77553
E-mail address: [email protected]