Volume, mass and length

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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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