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Chapter 5
Applications of thedeﬁnite integral tocalculating volume,mass, and length
5.1 Introduction
In this chapter, we consider applications of the deﬁnite integral to calculating geometricquantities such as volumes of geometric solids, masses, centers of mass, and lengths of curves.The idea behind the procedure described in this chapter is closely related to those wehave become familiar with in the context of computing areas. That is, we ﬁrst imagine anapproximationusing a ﬁnite numberof pieces to representa desired result. Then,a limitingprocess of reﬁnement leads to the desired result. The technology of the deﬁnite integral,developed in Chapters 2 and 3 applies directly. This means that we need not re-derivethe link between Riemann Sums and the deﬁnite integral, we can use these as we did inChapter 4.In the ﬁrst parts of this chapter, we will calculate the total mass of a continuousdensity distribution. In this context, we will also deﬁne the concept of a center of mass.We will ﬁrst motivate this concept for a discrete distribution made up of a number of ﬁnitemasses. Then we show how the same concept can be applied in the continuous case. Theconnection between the discrete and continuous representation will form an important linkwith our study of analogous concepts in probability, in Chapters 7 and 8.Inthe secondpartofthis chapter,we will considerhowtodissect certainthreedimen-sional solids into a set of simpler parts whose volumes are easy to compute. We will usefamiliar formulae for the volumes of disks and cylindrical shells, and carefully constructa summation to represent the desired volume. The volume of the entire object will thenbe obtained by summing up volumes of a stack of disks or a set of embedded shells, andconsideringthe limit as the thicknessof the dissectioncuts gets thinner. Thereare some im-portant differences between material in this chapter and in previous chapters. Calculatingvolumes will stretch our imagination, requiring us to visualize 3-dimensional (3D) objects,and how they can be subdivided into shells or slices. Most of our effort will be aimed atunderstanding how to set up the needed integral. We provide a number of examples of thisprocedure,but ﬁrst we review the basics of elementaryvolumes that will play the dominantrole in our calculations.
81
82Chapter5. Applications ofthedeﬁnite integral tocalculatingvolume, mass,andlength
5.2 Mass distributions in one dimension
We start ourdiscussionwith a numberofexampleofmass distributedalonga singledimen-sion. First, we consider a discrete collection of masses and then generalize to a continuousdensity. We will be interested in computing the total mass (by summation or integration)as well as other properties of the distribution.In considering the example of mass distributions, it becomes an easy step to developthe analogous concepts for continuous distributions. This allows us to recapitulate the linkbetween ﬁnite sums and deﬁnite integrals that we developed in earlier chapters. Examplesin this chapter also further reinforce the idea of density (in the context of mass density).Later, we will ﬁnd that these ideas are equally useful in the context of probability,exploredin Chapters 7 and 8.
5.2.1 A discrete distribution: total mass of beads on a wire
5
m
1
m
2
m
3
m
4
m
5
x
1
x
2
x
3
x
4
x
Figure 5.1.
A discrete distribution of masses along a (one dimensional) wire.
In Figure 5.1 we see a number of beads distributed along a thin wire. We will labeleach bead with an index,
i
= 1
...n
(there are ﬁve beads so that
n
= 5
). Each bead has acertain position (that we will think of as the value of
x
i
) and a mass that we will call
m
i
.We will think of this arrangement as a
discrete mass distribution
: both the masses of thebeads, and their positions are of interest to us. We would like to describe some propertiesof this distribution.The total mass of the beads,
M
, is just the sum of the individual masses, so that
M
=
n
i
=1
m
i
.
(5.1)
5.2.2 A continuous distribution: mass density and total mass
We now considera continuousmass distributionwhere the mass per unit length (“density”)changes gradually from one point to another. For example, the bar in Figure 5.2 has adensity that varies along its length.The portion at the left is made of lighter material, or has a lower density than theportions further to the right. We will denote that density by
ρ
(
x
)
and this carries units of mass per unit length. (The density of the material along the length of the bar is shown inthe graph.) How would we ﬁnd the total mass?Suppose the bar has length
L
and let
x
(0
≤
x
≤
L
)
denote position along that bar.Let us imagine dividing up the bar into small pieces of length
∆
x
as shown in Figure 5.2.
5.2. Mass distributions in one dimension 83
x
mass distribution
!
( )
x
...
x
m
1
m
2
m
n
x
1
x
2
x
n
...
Figure 5.2.
Top: A continuous mass distribution along a one dimensional bar,discussed in Example 5.3.3. The density of the bar (mass per unit length),
ρ
(
x
)
is shownon the graph. Bottom: the discretized approximation of this same distribution. Here wehave subdivided the bar into
n
smaller pieces, each of length
∆
x
. The mass of each pieceis approximately
m
k
=
ρ
(
x
k
)
∆
x
where
x
k
=
k
∆
x
. The total mass of the bar (“sum of all the pieces”) will be represented by an integral (5.2) as we let the size,
∆
x
, of the piecesbecome inﬁnitesimal.
The coordinates of the endpoints of those pieces are then
x
0
= 0
,...,x
k
=
k
∆
x, ..., x
N
=
L
and the corresponding masses of each of the pieces are approximately
m
k
=
ρ
(
x
k
)
∆
x.
(Observethat units are correct, that is mass
k
=(mass/length)
·
length. Note that
∆
x
has unitsof length.) The total mass is then a sum of masses of all the pieces, and, as we have seen inan earlier chapter, this sum will approach the integral
M
=
L
0
ρ
(
x
)
dx
(5.2)as we make the size of the pieces smaller.
84Chapter5. Applications ofthedeﬁnite integral tocalculatingvolume, mass,andlength
We can also deﬁne a
cumulative function
for the mass distribution as
M
(
x
) =
x
0
ρ
(
s
)
ds.
(5.3)Then
M
(
x
)
is the total mass in the part of the interval between the left end (assumed at0) and the position
x
. The idea of a cumulative function becomes useful in discussions of probability in Chapter 8.
5.2.3 Example: Actin density inside a cell
Biologists often describe the density of protein, receptors, or other molecules in cells. Oneexample is shown in Fig. 5.3. Here we show a keratocyte, which is a cell from the scaleof a ﬁsh. A band of actin ﬁlaments (protein responsible for structure and motion of the
2
nucleusactin cortexactin cortex
bcd
!
11
!
101
x
=1
!
x
!
Figure 5.3.
A cell (keratocyte) shown in (a) has a dense distribution of actinin a band called the actin cortex. In (b) we show a schematic sketch of the actin cortex(shaded). In (c) that band of actin is scaled and straightened out so that it occupies alength corresponding to the interval
−
1
≤
x
≤
1
. We are interested in the distributionof actin ﬁlaments across that band. That distribution is shown in (d). Note that actin isdensest in the middle of the band. (a) Credit to Alex Mogilner.
cell) are found at the edge of the cell in a band called the
actin cortex
. It has been foundexperimentallythatthedensityofactinis greatestinthemiddleoftheband,i.e. thepositioncorresponding to the midpoint of the edge of the cell shown in Fig. 5.3a. According toAlex Mogilner
19
, the density of actin across the cortex in ﬁlaments per edge
µ
m is wellapproximated by a distribution of the form
ρ
(
x
) =
α
(1
−
x
2
)
−
1
≤
x
≤
1
,
where
x
is the fraction of distance
20
from midpoint to the end of the band (Fig. 5.3c and d).Here
ρ
(
x
)
is an actin ﬁlament density in units of ﬁlaments per
µ
m. That is,
ρ
is the number
19
Alex Mogilner is a professor of mathematics who specializes in cell biology and the actin cytoskeleton
20
Note that 1
µ
m (read “ 1 micro-meter” or “micron”) is 10
−
6
meters, and is appropriate for measuring lengthsof small objects such as cells.

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