applications of the deﬁnite integral to volume, mass, and ... · 5.3 mass distribution and the...

Chapter 5

Applications of thedefinite integral tovolume, mass, andlength

5.1 IntroductionIn this chapter, we consider applications of the definite integral to calculating geometricquantities such as volumes of geometric solids, masses, centers of mass, and lengths ofcurves.

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 anapproximation using a finite number of pieces to represent a 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 Chapter 8.

In the second part of this chapter, we will consider how to dissect certain three dimen-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, andconsidering the limit as the thickness of the dissection cuts gets thinner. There are 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 elementary volumes that will play the dominantrole in our calculations.

119

120 Chapter 5. Applications of the definite integral to volume, mass, and length

5.2 Mass distributions in one dimensionWe start our discussion with a number of example of mass distributed along a single dimen-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 Chapter 8.

5.2.1 A discrete distribution: total mass of beads on a wire

5

m1

m2

m3

m4

m5

x1

x2

x3

x4

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 mi

.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 =

nX

i=1

mi

. (5.1)

5.2.2 A continuous distribution: mass density and total massWe now consider a continuous mass distribution where 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 ⇢(x) and this carries units ofmass 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 121

x

mass distribution

ρ( )x

...

∆ x

m1

m2

mn

x1

x2

xn

...

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), ⇢(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

= ⇢(xk

)�x where xk

= k�x. The total mass of the bar (“sum ofall 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

x0 = 0, . . . , xk

= k�x, . . . , xN

= L

and the corresponding masses of each of the pieces are approximately

mk

= ⇢(xk

)�x.

(Observe that units are correct, that is massk

=(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 =

ZL

0⇢(x)dx (5.2)

as we make the size of the pieces smaller.


We can also define a cumulative function for the mass distribution as

M(x) =

Zx

0⇢(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 ofprobability in Chapter 8.

5.2.3 Example: Actin density inside a cellBiologists 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

nucleus

actin cortex

actin cortex

b

c

d

−1 1

−1 0 1x

=1−xρ

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 foundexperimentally that the density of actin is greatest in the middle of the band, i.e. the positioncorresponding to the midpoint of the edge of the cell shown in Fig. 5.3a. According toAlex Mogilner20, the density of actin across the cortex in filaments per edge µm is wellapproximated by a distribution of the form

⇢(x) = ↵(1� x2) � 1 x 1,

where x is the fraction of distance21 from midpoint to the end of the band (Fig. 5.3c and d).Here ⇢(x) is an actin filament density in units of filaments per µm. That is, ⇢ is the number

20Alex Mogilner is a professor of mathematics who specializes in cell biology and the actin cytoskeleton21Note that 1µm (read “ 1 micro-meter” or “micron”) is 10�6meters, and is appropriate for measuring lengths

of small objects such as cells.

5.3. Mass distribution and the center of mass 123

of actin fibers per unit length.We can find the total number of actin filaments, N in the band by integration, i.e.

N =

Z 1

�1↵(1� x2

) dx = ↵

Z 1

�1(1� x2

) dx.

The integral above has already been computed (Integral 2.) in the Examples 3.6.2 of Chap-ter 3 and was found to be 4/3. Thus, we have that there are N = 4↵/3 actin filaments inthe band.

5.3 Mass distribution and the center of massIt is useful to describe several other properties of mass distributions. We first define the“center of mass”, x̄ which is akin to an average x coordinate.

5.3.1 Center of mass of a discrete distributionThe center of mass x̄ of a mass distribution is given by:

x̄ =

1

M

nX

i=1

xi

mi

.

This can also be written in the form

x̄ =

Pn

i=1 xi

miP

n

i=1 mi

.

5.3.2 Center of mass of a continuous distributionWe can generalize the concept of the center of mass for a continuous mass density. Ourusual approach of subdividing the interval 0 x L and computing a Riemann sum leadsto

x̄ =

1

M

nX

i=1

xi

⇢(xi

)�x.

As �x ! dx, this becomes an integral. Based on this, it makes sense to define the centerof mass of the continuous mass distribution as follows:

x̄ =

1

M

ZL

0x⇢(x)dx .

We can also write this in the form

x̄ =

RL

0 x⇢(x)dxRL

0 ⇢(x)dx.


5.3.3 Example: Center of mass vs average mass densityHere we distinguish between two (potentially confusing) quantities in the context of anexample.

A long thin bar of length L is made of material whose density varies along the lengthof the bar. Let x be distance from one end of the bar. Suppose that the mass density is givenby

⇢(x) = ax, 0 x L.

This type of mass density is shown in a panel in Fig. 5.2.

(a) Find the total mass of the bar.

(b) Find the average mass density along the bar.

(c) Find the center of mass of the bar.

(d) Where along the length of the bar should you cut to get two pieces of equal mass?

Solution

(a) From our previous discussion, the total mass of the bar is

M =

ZL

0ax dx =

ax2

2

��L

0

=

aL2

2

.

(b) The average mass density along the bar is computed just as one would compute theaverage value of a function: integrate the function over an interval and divide by thelength of the interval. An example of this type appeared in Section 4.5. Thus

⇢̄ =

1

L

ZL

0⇢(x) dx =

1

L

✓aL2

2

◆=

aL

2

A bar having a uniform density ⇢̄ = aL/2 would have the same total mass as the barin this example. (This is the physical interpretation of average mass density.)

(c) The center of mass of the bar is

x̄ =

RL

0 xp(x) dx

M=

1

M

ZL

0ax2 dx =

a

M

x3

3

��L

0

=

2a

aL2

L3

3

=

2

3

L.

Observe that the center of mass is an “average x coordinate”, which is not the same asthe average mass density.

(d) We can use the cumulative function defined in Eqn. (5.3) to figure out where half of themass is concentrated. Suppose we cut the bar at some position x = s. Then the massof this part of the bar is

M1 =

Zs

0⇢(x) dx =

as2

2

,

5.4. Miscellaneous examples and related problems 125

We ask for what values of s is it true that M1 is exactly half the total mass? Using theresult of part (a), we find that for this to be true, we must have

M1 =

M

2

, ) as2

2

=

1

2

aL2

2

Solving for s leads to

s =1p2

L =

p2

2

L.

Thus, cutting the bar at a distance (

p2/2)L from x = 0 results in two equal masses.

Remark: the position that subdivides the mass into two equal pieces is analogous tothe idea of a median. This concept will appear again in the context of probability inChapter 8.

5.4 Miscellaneous examples and related problemsThe idea of mass density can be extended to related problems of various kinds. We givesome examples in this section.

Up to now, we have seen examples of mass distributed in one dimension: beads on awire, actin density along the edge of a cell, (in Chapter 4), or a bar of varying density. Forthe continuous distributions, we determined the total mass by integration. Underlying theintegral we computed was the idea that the interval could be “dissected” into small parts(of width �x), and a sum of pieces transformed into an integral. In the next examples, weconsider similar ideas, but instead of dissecting the region into 1-dimensional intervals, wehave slightly more interesting geometries.

5.4.1 Example: A glucose density gradientA cylindrical test-tube of radius r, and height h, contains a solution of glucose which hasbeen prepared so that the concentration of glucose is greatest at the bottom and decreasesgradually towards the top of the tube. (This is called a density gradient). Suppose that theconcentration c as a function of the depth x is c(x) = 0.1 + 0.5x grams per centimeter3.(x = 0 at the top of the tube, and x = h at the bottom of the tube.) In Figure 5.4 we show aschematic version of what this gradient might look like. (In reality, the transition betweenhigh and low concentration would be smoother than shown in this figure.) Determine thetotal amount of glucose in the tube (in gm). Neglect the “rounded” lower portion of thetube: i.e. assume that it is a simple cylinder.

Solution

We assume a simple cylindrical tube and consider imaginary “slices” of this tube alongits vertical axis, here labeled as the “x” axis. Suppose that the thickness of a slice is �x.Then the volume of each of these (disk shaped) slices is ⇡r2�x. The amount of glucose inthe slice is approximately equal to the concentration c(x) multiplied by the volume of theslice, i.e. the small slice contains an amount ⇡r2�xc(x) of glucose. In order to sum up the


x=0

r

x=h

∆ x

Figure 5.4. A test-tube of radius r containing a gradient of glucose. A disk-shapedslice of the tube with small thickness �x has approximately constant density.

total amount over all slices, we use a definite integral. (As before, we imagine �x ! dxbecoming “infinitesimal” as the number of slices increases.) The integral we want is

G = ⇡r2Z

h

0c(x) dx.

Even though the geometry of the test-tube, at first glance, seems more complicated thanthe one-dimensional highway described in Chapter 4, we observe here that the integralthat computes the total amount is still a sum over a single spatial variable, x. (Note theresemblance between the integrals

I =

ZL

0C(x) dx and G = ⇡r2

Zh

0c(x) dx,

here and in the previous example.) We have neglected the complication of the rounded bot-tom portion of the test-tube, so that integration over its length (which is actually summationof disks shown in Figure 5.4) is a one-dimensional problem.

In this case the total amount of glucose in the tube is

G = ⇡r2Z

h

0(0.1 + 0.5x)dx = ⇡r2

✓0.1x+

0.5x2

2

◆ ��h

0

= ⇡r2✓0.1h+

0.5h2

2

◆.

Suppose that the height of the test-tube is h = 10 cm and its radius is r = 1 cm.Then the total mass of glucose is

G = ⇡

✓0.1(10) +

0.5(100)

2

◆= ⇡ (1 + 25) = 26⇡ gm.

In the next example, we consider a circular geometry, but the concept of dissectingand summing is the same. Our task is to determine how to set up the problem in termsof an integral, and, again, we must imagine which type of subdivision would lead to thesummation (integration) needed to compute total amount.

5.4. Miscellaneous examples and related problems 127

5.4.2 Example: A circular colony of bacteriaA circular colony of bacteria has radius of 1 cm. At distance r from the center of thecolony, the density of the bacteria, in units of one million cells per square centimeter, isobserved to be b(r) = 1�r2 (Note: r is distance from the center in cm, so that 0 r 1).What is the total number of bacteria in the colony?

(one ring)

∆ r

Side view Top−down view

r

b(r)

b(r)=1−r

r

2

Figure 5.5. A colony of bacteria with circular symmetry. A ring of small thickness�r has roughly constant density. The superimposed curve on the left is the bacterial densityb(r) as a function of the radius r.

Solution

Figure 5.5 shows a rough sketch of a flat surface with a colony of bacteria growing on it.We assume that this distribution is radially symmetric. The density as a function of distancefrom the center is given by b(r), as shown in Figure 5.5. Note that the function describingdensity, b(r) is smooth, but to accentuate the strategy of dissecting the region, we haveshown a top-down view of a ring of nearly constant density on the right in Figure 5.5. Wesee that this ring occupies the region between two circles, e.g. between a circle of radius rand a slightly bigger circle of radius r +�r. The area of that “ring”22 would then be thearea of the larger circle minus that of the smaller circle, namely

Aring = ⇡(r +�r)2 � ⇡r2 = ⇡(2r�r + (�r)2).

However, if we make the thickness of that ring really small (�r ! 0), then the quadraticterm is very very small so that

Aring ⇡ 2⇡r�r.

Consider all the bacteria that are found inside a “ring” of radius r and thickness �r(see Figure 5.5.) The total number within such a ring is the product of the density, b(r) andthe area of the ring, i.e.

b(r) · (2⇡r�r) = 2⇡r(1� r2)�r.

22Students commonly make the error of writing Aring = ⇡(r + �r � r)2 = ⇡(�r)2. This trap should beavoided! It is clear that the correct expression has additional terms, since we really are computing a difference oftwo circular areas.


To get the total number in the colony we sum up over all the rings from r = 0 to r = 1

and let the thickness, �r ! dr become very small. But, as with other examples, this isequivalent to calculating a definite integral, namely:

Btotal =

Z 1

0(1� r2)(2⇡r) dr = 2⇡

Z 1

0(1� r2)rdr = 2⇡

Z 1

0(r � r3)dr.

We calculate the result as follows:

Btotal = 2⇡

✓r2

2

� r4

4

◆ ��1

0

= (⇡r2 � ⇡r4

2

)

��1

0

= ⇡ � ⇡

2

=

⇡

2

.

Thus the total number of bacteria in the entire colony is ⇡/2 million which is approximately1.57 million cells.

5.5 Volumes of solids of revolutionWe now turn to the problem of calculating volumes of 3D solids. Here we restrict attentionto symmetric objects denoted solids of revolution. The outer surface of these objects isgenerated by revolving some curve around a coordinate axis. In Figure 5.7 we show onesuch curve, and the surface it forms when it is revolved about the y axis.

5.5.1 Volumes of cylinders and shellsBefore starting the calculation, let us recall the volumes of some of the geometric shapesthat are to be used as elementary pieces into which our shapes will be carved. See Fig-ure 5.6.

1. The volume of a cylinder of height h having circular base of radius r, is

Vcylinder = ⇡r2h.

2. The volume of a circular disk of thickness ⌧ , and radius r (shown on the left inFigure 5.6 ), is a special case of the above,

Vdisk = ⇡r2⌧.

3. The volume of a cylindrical shell of height h, with circular radius r and smallthickness ⌧ (shown on the right in Figure 5.6) is

Vshell = 2⇡rh⌧.

(This approximation holds for ⌧ ⌧ r.)

5.5. Volumes of solids of revolution 129

r

h

τ

r

τ

disk shell

Figure 5.6. The volumes of these simple 3D shapes are given by simple formulae.We use them as basic elements in computing more complicated volumes. Here we willpresent examples based on disks. At the end of this chapter we give an example based onshells.

y

x x x

y y

Figure 5.7. A solid of revolution is formed by revolving a region in the xy-planeabout the y-axis. We show how the region is approximated by rectangles of some givenwidth, and how these form a set of approximating disks for the 3D solid of revolution.

5.5.2 Computing the VolumesConsider the curve in Figure 5.7 and the surface it forms when it is revolved about they axis. In the same figure, we also show how a set of approximating rectangular stripsassociated with the planar region (grey rectangles) lead to a set of stacked disks (orangeshapes) that approximate the volume of the solid (greenish object in Fig. 5.7). The totalvolume of the disks is not the same as the volume of the object but if we make the thicknessof these disks very small, the approximation of the true volume is good. In the limit, asthe thickness of the disks becomes infinitesimal, we arrive at the true volume of the solid


of revolution. The reader should recognize a familiar theme. We used the same concept incomputing areas using Riemann sums based on rectangular strips in Chapter 2.

Fig. 5.8 similarly shows a volume of revolution obtained by revolving the graph ofthe function y = f(x) about the x axis. We note that if this surface is cut into slices, theradius of the cross-sections depend on the position of the cut. Let us imagine a stack ofdisks approximating this volume. One such disk has been pulled out and labeled for ourinspection. We note that its radius (in the y direction) is given by the height of the graphof the function, so that r = f(x). The thickness of the disk (in the x direction) is �x. Thevolume of this single disk is then v = ⇡[f(x)]2�x. Considering this disk to be based atthe k’th coordinate point in the stack, i.e. at x

k

, means that its volume is

vk

= ⇡[f(xk

)]

2�x.

Summing up the volumes of all such disks in the stack leads to the total volume of disks

Vdisks =X

k

⇡[f(xk

)]

2�x.

When we increase the number of disks, making each one thinner so that �x ! 0, we

disk thickness:

disk radius:

x∆

x

y

x

y=f(x)

r=f(x)

Figure 5.8. Here the solid of revolution is formed by revolving the curve y = f(x)about the y axis. A typical disk used to approximate the volume is shown. The radius of thedisk (parallel to the y axis) is r = y = f(x). The thickness of the disk (parallel to the xaxis) is �x. The volume of this disk is hence v = ⇡[f(x)]2�x

arrive at a definite integral,

V =

Zb

a

⇡[f(x)]2dx.

In most of the examples discussed in this chapter, the key step is to make careful observationof the way that the radius of a given disk depends on the function that generates the surface.


(By this we mean the function that specifies the curve that forms the surface of revolution.)We also pay attention to the dimension that forms the disk thickness, �x.

Some of our examples will involve surfaces revolved about the x axis, and otherswill be revolved about the y axis. In setting up these examples, a diagram is usually quitehelpful.

Example 1: Volume of a sphere

k

y

x

f(x )k

∆ x

∆ xx

Figure 5.9. When the semicircle (on the left) is rotated about the x axis, it gen-erates a sphere. On the right, we show one disk generated by the revolution of the shadedrectangle.

We can think of a sphere of radius R as a solid whose outer surface is formed byrotating a semi-circle about its long axis. A function that describe a semi-circle (i.e. thetop half of the circle, y2 + x2

= R2) is

y = f(x) =pR2 � x2.

In Figure 5.9, we show the sphere dissected into a set of disks, each of width �x. The disksare lined up along the x axis with coordinates x

k

, where �R xk

R. These are justinteger multiples of the slice thickness �x, so for example,

x0 = �R, x1 = �R+�x, . . . , xk

= �R+ k�x .

The radius of the disk depends on its position23. Indeed, the radius of a disk through the xaxis at a point x

k

is specified by the function rk

= f(xk

). The volume of the k’th disk is

Vk

= ⇡r2k

�x.

By the above remarks, using the fact that the function f(x) determines the radius, we have

Vk

= ⇡[f(xk

)]

2�x,

23Note that the radius is oriented along the y axis, so sometimes we may write this as rk = yk = f(xk)


Vk

= ⇡

qR2 � x2

k

�2�x = ⇡(R2 � x2

k

)�x.

The total volume of all the disks is

V =

X

k

Vk

=

X

k

⇡[f(xk

)]

2�x = ⇡

X

k

(R2 � x2k

)�x.

as �x ! 0, this sum becomes a definite integral, and represents the true volume. We startthe summation at x = �R and end at x

N

= R since the semi-circle extends from x = �Rto x = R. Thus

Vsphere =

ZR

-R⇡[f(x

k

)]

2 dx = ⇡

ZR

-R(R2 � x2

) dx.

We compute this integral using the Fundamental Theorem of calculus, obtaining

Vsphere = ⇡

✓R2x� x3

3

◆ ��R

�R

.

Observe that this is twice the volume obtained for the interval 0 < x < R,

Vsphere = 2⇡

✓R2x� x3

3

◆ ��R

0

= 2⇡

✓R3 � R3

3

◆.

We often use such symmetry properties to simplify computations. After simplification, wearrive at the familiar formula

Vsphere =4

3

⇡R3.

Example 2: Volume of a paraboloid

Consider the curvey = f(x) = 1� x2.

If we rotate this curve about the y axis, we will get a paraboloid, as shown in Figure 5.10.In this section we show how to compute the volume by dissecting into disks stacked upalong the y axis.

Solution

The object has the y axis as its axis of symmetry. Hence disks are stacked up along they axis to approximate this volume. This means that the width of each disk is �y. Thisaccounts for the dy in the integral below. The volume of each disk is

Vdisk = ⇡r2�y,

where the radius, r is now in the direction parallel to the x axis. Thus we must expressradius as

r = x = f�1(y),


2

y

x x

y

y=f(x)=1−x

Figure 5.10. The curve that generates the shape of a paraboloid (left) and theshape of the paraboloid (right).

i.e, we invert the relationship to obtain x as a function of y. From y = 1 � x2 we havex2

= 1� y so x =

p1� y. The radius of a disk at height y is therefore r = x =

p1� y.

The shape extends from a smallest value of y = 0 up to y = 1. Thus the volume is

V = ⇡

Z 1

0[f(y)]2 dy = ⇡

Z 1

0[

p1� y]2 dy.

It is helpful to note that once we have identified the thickness of the disks (�y), we areguided to write an integral in terms of the variable y, i.e. to reformulate the equationdescribing the curve. We compute

V = ⇡

Z 1

0(1� y) dy = ⇡

✓y � y2

2

◆ ��1

0

= ⇡

✓1� 1

2

◆=

⇡

2

.

The above example was set up using disks. However, there are other options. InSection 5.8.1 we show yet another method, comprised of cylindrical shells to compute thevolume of a cone. In some cases, one method is preferable to another, but here eithermethod works equally well.

Example 3

Find the volume of the surface formed by rotating the curve

y = f(x) =px, 0 x 1

(a) about the x axis. (b) about the y axis.

Solution

(a) If we rotate this curve about the x axis, we obtain a bowl shape; dissecting this surfaceleads to disks stacked along the x axis, with thickness �x ! dx, with radii in the ydirection, i.e. r = y = f(x), and with x in the range 0 x 1. The volume will thusbe

V = ⇡

Z 1

0[f(x)]2 dx = ⇡

Z 1

0[

px]2 dx = ⇡

Z 1

0x dx = ⇡

x2

2

��1

0

=

⇡

2

.


(b) When the curve is rotated about the y axis, it forms a surface with a sharp point at theorigin. The disks are stacked along the y axis, with thickness �y ! dy, and radii inthe x direction. We must rewrite the function in the form

x = g(y) = y2.

We now use the interval along the y axis, i.e. 0 < y < 1 The volume is then

V = ⇡

Z 1

0[f(y)]2 dy = ⇡

Z 1

0[y2]2 dy = ⇡

Z 1

0y4 dy = ⇡

y5

5

��1

0

=

⇡

5

.

5.6 Length of a curve: Arc lengthAnalytic geometry provides a simple way to compute the length of a straight line segment,based on the distance formula24. Recall that, given points P1 = (x1, y1) and P2 = (x2, y2),the length of the line joining those points is

d =

p(x2 � x1)

2+ (y2 � y1)2.

Things are more complicated for “curves” that are not straight lines, but in many cases, weare interested in calculating the length of such curves. In this section we describe how thiscan be done using the definite integral “technology”.

The idea of dissection also applies to the problem of determining the length of acurve. In Figure 5.11, we see the general idea of subdividing a curve into many small“arcs”. Before we look in detail at this construction, we consider a simple example, shownin Figure 5.12. In the triangle shown, by the Pythagorean theorem we have the length ofthe sloped side related as follows to the side lengths �x,�y:

�`2 = �x2+�y2,

�` =p�x2

+�y2 =

r1 +

�y2

�x2

!�x =

s

1 +

✓�y

�x

◆2

�x.

We now consider a curve given by some function

y = f(x) a < x < b,

as shown in Figure 5.11(a). We will approximate this curve by a set of line segments, asshown in Figure 5.11(b). To obtain these, we have selected some step size �x along thex axis, and placed points on the curve at each of these x values. We connect the pointswith straight line segments, and determine the lengths of those segments. (The total lengthof the segments is only an approximation of the length of the curve, but as the subdivisiongets finer and finer, we will arrive at the true total length of the curve.)

We show one such segment enlarged in the circular inset in Figure 5.11. Its slope,shown at right is given by �y/�x. According to our remarks, above, the length of thissegment is given by

�` =

s

1 +

✓�y

�x

◆2

�x.

24The reader should recall that this formula is a simple application of Pythagorean theorem.

5.6. Length of a curve: Arc length 135

x

y

y=f(x)

x

y

y=f(x)

x

y

y=f(x)

x

y

∆

∆

Figure 5.11. Top: Given the graph of a function, y = f(x) (at left), we drawsecant lines connecting points on its graph at values of x that are multiples of �x (right).Bottom: a small part of this graph is shown, and then enlarged, to illustrate the relationshipbetween the arc length and the length of the secant line segment.

y

x

x

y

∆

∆∆ l

Figure 5.12. The basic idea of arclength is to add up lengths �l of small linesegments that approximate the curve.

As the step size is made smaller and smaller �x ! dx, �y ! dy and

�` !

s

1 +

✓dy

dx

◆2

dx.

We recognize the ratio inside the square root as as the derivative, dy/dx. If our curve is


given by a function y = f(x) then we can rewrite this as

d` =

q1 + (f 0

(x))2dx.

Thus, the length of the entire curve is obtained from summing (i.e. adding up) these smallpieces, i.e.

L =

Zb

a

q1 + (f 0

(x))2dx. (5.4)

Example 1

Find the length of a line whose slope is �2 given that the line extends from x = 1 to x = 5.

Solution

We could find the equation of the line and use the distance formula. But for the purpose ofthis example, we apply the method of Equation (5.4): we are given that the slope f 0

(x) is-2. The integral in question is

L =

Z 5

1

p1 + (f 0

(x))2 dx =

Z 5

1

p1 + (�2)

2 dx =

Z 1

5

p5 dx.

We get

L =

p5

Z 5

1dx =

p5x

��5

1

=

p5[5� 1] = 4

p5.

Example 2

Find an integral that represents the length of the curve that forms the graph of the function

y = f(x) = x3, 1 < x < 2.

Solution

We find thatdy

dx= f 0

(x) = 3x2.

Thus, the integral is

L =

Z 2

1

p1 + (3x2

)

2 dx =

Z 2

1

p1 + 9x4 dx.

At this point, we will not attempt to find the actual length, as we must first develop tech-niques for finding the anti-derivative for functions such as

p1 + 9x4.


y = f(x) =1-x^2

0.0 1.00.0

1.5

y = f(x) =1-x^2

cumulative length L

length increment

Arc Length

0.0 1.00.0

1.5

Figure 5.13. The spreadsheet can be used to compute approximate values ofintegrals, and hence to calculate arclength. Shown here is the graph of the function y =

f(x) = 1 � x2 for 0 x 1, together with the length increment and the cumulativearclength along that curve.


Using the spreadsheet to calculate arclength

Most integrals for arclength contain square roots and functions that are not easy to integrate,simply because their antiderivatives are difficult to determine. However, now that we knowthe idea behind determining the length of a curve, we can apply the ideas developed haveto approximate the length of a curve “numerically”. The spreadsheet is a simple tool fordoing the necessary summations.

As an example, we show here how to calculate the length of the curve

y = f(x) = 1� x2 for 0 x 1

using a simple numerical procedure implemented on the spreadsheet.Let us choose a step size of �x = 0.1 along the x axis, for the interval 0 x 1.

We calculate the function, the slopes of the little segments (change in y divided by changein x), and from this, compute the length of each segment

�` =p1 + (�y/�x)2 �x

and the accumulated length along the curve from left to right, L which is just a sum ofsuch values. The Table 5.6 shows steps in the calculation of the ratio �y/�x, the valueof �`, the cumulative sum, and, finally the total length L. The final value of L = 1.4782represents the total length of the curve over the entire interval 0 < x < 1.

x y = f(x) �y/�x �` L =

P�`

0. 0 1.0000 -0.1 0.1005 0.00000.1 0.9900 -0.3 0.1044 0.10050.2 0.9600 -0.5 0.1118 0.20490.3 0.9100 -0.7 0.1221 0.31670.4 0.8400 -0.9 0.1345 0.43880.5 0.7500 -1.1 0.1487 0.57330.6 0.6400 -1.3 0.1640 0.72200.7 0.5100 -1.5 0.1803 0.88600.8 0.3600 -1.7 0.1972 1.06630.9 0.1900 -1.9 0.2147 1.26351. 0 0.0000 -2.1 0.2326 1.4782

Table 5.1. For the function y = f(x) = 1� x2, and 0 x 1, we show how tocalculate an approximation to the arc-length using the spreadsheet.

In Figure 5.13(a) we show the actual curve y = 1 � x2. with points placed on itat each multiple of �x. In Figure 5.13(b), we show (in blue) how the lengths of the littlestraight-line segments connecting these points changes across the interval. (The segmentson the left along the original curve are nearly flat, so their length is very close to �x. Thesegments on the right part of the curve are much more sloped, and their lengths are thus


bigger.) We also show (in red) how the total accumulated length L depends on the positionx across the interval. This function represents the total arc-length of the curve y = 1� x2,from x = 0 up to a given x value. At x = 1 this function returns the value y = L, as it hasadded up the full length of the curve for 0 x 1.

5.6.1 How the alligator gets its smileThe American alligator, Alligator mississippiensis has a set of teeth best viewed at somedistance. The regular arrangement of these teeth, i.e. their spacing along the jaw is im-portant in giving the reptile its famous bite. We will concern ourselves here with how thatpattern of teeth is formed as the alligator develops from its embryonic stage to that of anadult. As is the case in humans, the teeth on an alligator do not form or sprout simultane-ously. In the development of the baby alligator, there is a sequence of initiation of teeth,one after the other, at well-defined positions along the jaw.

Paul Kulesa, a former student of James D Muray, set out to understand the pattern ofdevelopment of these teeth, based on data in the literature about what happens at distinctstages of embryonic growth. Of interest in his research were several questions, includingwhat determines the positions and timing of initiation of individual teeth, and what mecha-nisms lead to this pattern of initiation. One theory proposed by this group was that chemicalsignals that diffuse along the jaw at an early stage of development give rise to instructionsthat are interpreted by jaw cells: where the signal is at a high level, a tooth will start toinitiate.

While we will not address the details of the mechanism of development here, wewill find a simple application of the ideas of arclength in the developmental sequence ofteething. Shown in Figure 5.14 is a smiling baby alligator (no doubt thinking of somefuture tasty meal). A close up of its smile (at an earlier stage of development) reveals theshape of the jaw, together with the sites at which teeth are becoming evident. (One of thesesites, called primordia, is shown enlarged in an inset in this figure.)

Paul Kulesa found that the shape of the alligator’s jaw can be described remarkablywell by a parabola. A proper choice of coordinate system, and some experimentation leadsto the equation of the best fit parabola

y = f(x) = �ax2+ b

where a = 0.256, and b = 7.28 (in units not specified).We show this curve in Figure 5.15(a). Also shown in this curve is a set of points at

which teeth are found, labelled by order of appearance. In Figure 5.15(b) we see the samecurve, but we have here superimposed the function L(x) given by the arc length along thecurve from the front of the jaw (i.e. the top of the parabola), i.e.

L(x) =

Zx

0

p1 + [f 0

(s)]2 ds.

This curve measures distance along the jaw, from front to back. The distances of the teethfrom one another, or along the curve of the jaw can be determined using this curve if weknow the x coordinates of their positions.

The table below gives the original data, courtesy of Dr. Kulesa, showing the orderof the teeth, their (x, y) coordinates, and the value of L(x) obtained from the arclength


formula. We see from this table that the teeth do not appear randomly, nor do they fill inthe jaw in one sweep. Rather, they appear in several stages.

In Figure 5.15(c), we show the pattern of appearance: Plotting the distance along thejaw of successive teeth reveals that the teeth appear in waves of nearly equally-spaced sites.(By equally spaced, we refer to distance along the parabolic jaw.) The first wave (teeth 1, 2,3) are followed by a second wave (4, 5, 6, 7), and so on. Each wave forms a linear patternof distance from the front, and each successive wave fills in the gaps in a similar, equallyspaced pattern.

The true situation is a bit more complicated: the jaw grows as the teeth appear asshown in 5.15(c). This has not been taken into account in our simple treatment here, wherewe illustrate only the essential idea of arc length application.

Tooth number position distance along jawx y L(x)

1 1.95 6.35 2.14862 3.45 4.40 4.70003 4.54 2.05 7.11894 1.35 6.95 1.40005 2.60 5.50 3.20526 3.80 3.40 5.48847 5.00 1.00 8.42418 3.15 4.80 4.15009 4.25 2.20 6.3923

10 4.60 1.65 7.370511 0.60 7.15 0.607212 3.45 4.05 4.657213 5.30 0.45 9.2644

Table 5.2. Data for the appearance of teeth, in the order in which they appearas the alligator develops. We can use arc-length computations to determine the distancesbetween successive teeth.


Figure 5.14. Alligator mississippiensis and its teeth


1

2

3

4

5

6

7

8

9

10

11

12

13

Alligator teeth

-6.0 6.00.0

8.0

jaw y = f(x)

arc length L(x) along jaw

0.0 5.50.0

10.0

(a) (b)Distance along jaw

1

2

3

4

5

6

7

8

9

10

11

12

13

teeth in order of appearance

0.0 13.00.0

10.0

(c) (d)

Figure 5.15. (a) The parabolic shape of the jaw, showing positions of teeth andnumerical order of emergence. (b) Arc length along the jaw from front to back. (c) Distanceof successive teeth along the jaw. (d) Growth of the jaw.

5.7. Summary 143

5.6.2 References1. P.M. Kulesa and J.D. Murray (1995). Modelling the Wave-like Initiation of Teeth

Primordia in the Alligator. FORMA. Cover Article. Vol. 10, No. 3, 259-280.

2. J.D. Murray and P.M. Kulesa (1996). On A Dynamic Reaction-Diffusion Mechanismfor the Spatial Patterning of Teeth Primordia in the Alligator. Journal of ChemicalPhysics. J. Chem. Soc., Faraday Trans., 92 (16), 2927-2932.

3. P.M. Kulesa, G.C. Cruywagen, S.R. Lubkin, M.W.J. Ferguson and J.D. Murray (1996).Modelling the Spatial Patterning of Teeth Primordia in the Alligator. Acta Biotheo-retica, 44, 153-164.

5.7 SummaryHere are the main points of the chapter:

1. We introduced the idea of a spatially distributed mass density ⇢(x) in Section 5.2.2.Here the definite integral represents

Zb

a

⇢(x) dx = Total mass in the interval a x b.

2. In this chapter, we defined the center of mass of a (discrete) distribution of n massesby

¯X =

1

M

nX

i=0

xi

mi

. (5.5)

We developed the analogue of this for a continuous mass distribution (distributed inthe interval 0 x L). We defined the center of mass of a continuous distributionby the definite integral

x̄ =

1

M

ZL

0x⇢(x)dx . (5.6)

Importantly, the quantities mi

in the sum (5.5) carry units of mass, whereas theanalogous quantities in (5.6) are ⇢(x)dx. [Recall that ⇢(x) is a mass per unit lengthin the case of mass distributed along a bar or straight line.]

3. We defined a cumulative function. In the discrete case, this was defined as In thecontinuous case, it is

M(x) =

Zx

0⇢(s)ds.

4. The mean is an average x coordinate, whereas the median is the x coordinate thatsplits the distribution into two equal masses (Geometrically, the median subdividesthe graph of the distribution into two regions of equal areas). The mean and medianare the same only in symmetric distributions. They differ for any distribution that isasymmetric. The mean (but not the median) is influenced more strongly by distantportions of the distribution.


5. In the later parts of this chapter, we showed how to compute volumes of variousobjects that have radial symmetry (“solids of revolution”). We showed that if thesurface is generated by rotating the graph of a function y = f(x) about the x axis(for a x b), then its volume can be described by an integral of the form

V =

Zb

a

⇡[f(x)]2dx.

We used this idea to show that the volume of a sphere of radius R is Vsphere

=

(4/3)⇡R3

In the Chapter 8, we find applications of the ideas of density and center of mass tothe context of a probability distribution and its mean.

5.8 Optional Material5.8.1 The shell method for computing volumesEarlier, we used dissection into small disks to compute the volume of solids of revolution.Here we use an alternative dissection into shells.

Example: Volume of a cone using the shell method

x

y

y=f(x)=1−x

y

x

x

y

y=1−x

dx

Figure 5.16. Top: The curve that generates the cone (left) and the shape of thecone (right). Bottom: the cone showing one of the series of shells that are used in thisexample to calculate its volume.

5.8. Optional Material 145

We use the shell method to find the volume of the cone formed by rotating the curve

y = 1� x

about the y axis.

Solution

We show the cone and its generating curve in Figure 5.16, together with a representativeshell used in the calculation of total volume. The volume of a cylindrical shell of radius r,height h and thickness ⌧ is

Vshell = 2⇡rh⌧.

We will place these shells one inside the other so that their radii are parallel to the x axis(so r = x). The heights of the shells are determined by their y value (i.e. h = y = 1�x =

1� r). For the tallest shell r = 0, and for the flattest shell r = 1. The thickness of the shellis �r. Therefore, the volume of one shell is

Vshell = 2⇡r(1� r) �r.

The volume of the object is obtained by summing up these shell volumes. In the limit,as �r ! dr gets infinitesimally small, we recognize this as a process of integration. Weintegrate over 0 r 1, to obtain:

V = 2⇡

Z 1

0r(1� r) dr = 2⇡

Z 1

0(r � r2) dr.

We find that

V = 2⇡

✓r2

2

� r3

3

◆ ��1

0

= 2⇡

✓1

2

� 1

3

◆=

⇡

3

.


5.9 ExercisesExercise 5.1 Five beads are distributed along a thin 1 dimensional wire. Their massesand positions are: m1 = 5, x1 = 0;m2 = 2, x2 = 3;m3 = 1, x3 = 4;m4 = 2, x4 =

5;m5 = 10, x5 = 6. (a) Find the total mass M and (b) the center of mass x̄ of this discretemass distribution.

Exercise 5.2 Suppose that the function ⇢(x) represents the density of a bar for a x b,.Explain the distinction between:

(a) the average density of the bar, ⇢̄

(b) the mass of the bar, M , and

(c) the center of mass (or centroid) of the bar, x̄.

Exercise 5.3 The mass density of a bar is given by

⇢(x) = 1� x 0 < x < 1

(a) Sketch the density of the bar and the function

M(x) =

Zx

0p(s) ds.

Find the total mass of the bar.

(b) Find the average mass density ⇢̄ along the bar.

(c) Find the center of mass x̄ of the bar.


(e) What fraction of the mass of the bar is found between x = 0 and x = 0.5 ?

Exercise 5.4 The mass density of a bar is given by

⇢(x) = ax20 < x < L

(a) Find the total mass M of the bar.

(b) Find the average mass density ⇢̄ along the bar.

(c) Find the center of mass x̄ of the bar.


Exercise 5.5 The density of a beam is given by the function ⇢(x) = xm/n where 0 x 1.

5.9. Exercises 147

(a) Find the center of mass x̄.

(b) Explain what happens to the center of mass if m is very large (for n = 1).

(c) What happens to the center of mass if n is very large, (for m = 1)?

Exercise 5.6 Density of cars along a highway At rush hour, the density of cars along ahighway is given by

C(x) = 100x⇣1� x

10

⌘0 x L,

where x is distance in kilometers.

(a) What is the largest value of L for which this density makes sense?

(b) Where along the highway is the congestion greatest? What is the car density at thatlocation?

(c) What is the total number of cars along the road?

Exercise 5.7 The density of a band of protein along a one-dimensional strip of gel in anelectrophoresis experiment is given by p(x) = 2(x� 1)(2� x) for 1 x 2, where x isthe distance along the strip in cm and p(x) is the protein density (i.e. protein mass per cm)at distance x.

(a) Graph the density p as a function of x.

(b) Find the total mass of the protein in the band for 1 x 2.

(Hint: simplify the function first.)

Exercise 5.8 The air density h meters above the earth’s surface is

p(h) = Ae�ahkg/m3.

Find the mass of a cylindrical column of air of radius r = 2 meters and height H = 25000

meters. Let A = 1.28 (kg/m3), a = 0.000124 per meter. (Hint: set the integral up as aRiemann sum).

Exercise 5.9 A test-tube contains a solution of glucose which has been prepared so thatthe concentration of glucose is greatest at the bottom and decreases gradually towards thetop of the fluid. (This is called a density gradient). Suppose that the concentration c as afunction of the depth x is c(x) = x/10 (in units of g/cm3). The radius of the tube is 2 cmand the height of the glucose-containing solution is 10 cm. Determine the total amount ofglucose in the tube (in g).

Exercise 5.10 To investigate changes in the Earth’s weather, scientists examine the distri-bution of pollen grains in a 1 dimensional drilled “core sample”, i.e. a sample of the Earth’scrust that contains archaeological deposits of soil from many thousands of years. Assume


that the core sample is a cylinder of unit cross-sectional area and length L. Suppose thatpollen grain density p(x) at a point x in this sample is in a core sample of length L is givenby

p(x) = A sin(ax), 0 < x < L =

⇡

a.

where p(x) are the number of particles per unit volume at a distance x from one end of thesample.

(a) Where is the pollen grain most concentrated along this one dimensional sample?

(b) Find the average density of pollen grains along the length of the sample.

(c) Find the center of mass of the pollen grain distribution. (Note: you can use the fact thatthe density is distributed symmetrically to avoid having to integrate.)

Exercise 5.11 Population density The population density of inhabitants living along thebanks of the river Nile at a distance x km from its mouth is found to be n(x) = 1000e�kx.Twenty km from the river mouth, the population density is half as large as it is at the mouthof the river. Find the total population of people living along the Nile. (Hint: consider theNile being “very long”, i.e. let x ! 1 be the “length” for the purpose of the integration.)

Exercise 5.12 Find the volume of a cone whose height h is equal to its base radius r, byusing the disc method. We will place the cone on its side, as shown in the Figure 5.17, andlet x represent position along its axis.

(a) Using the diagram shown below (Figure 5.17), explain what kind of a curve in thex�y-plane we would use to generate the surface of the cone as a surface of revolution.

y

x

h

r

generating curve

Figure 5.17. For problem 5.12

(b) Using the proportions given in the problem, specify the exact function y = f(x) thatwe need to describe this “curve”.

(c) Now find the volume enclosed by this surface of revolution for 0 x 1.

5.9. Exercises 149

(d) Show that, in this particular case, we would have gotten the same geometric object,and also the same enclosed volume, if we had rotated the “curve” about the y axis.

Exercise 5.13 Find the volume of the cone generated by revolving the curve y = f(x) =1 � x (for 0 < x < 1) about the y axis. Use the disk method, with disks stacked up alongthe y axis.

Exercise 5.14 Find the volume of the “bowl” obtained by rotating the curve y = 4x2

about the y axis for 0 x 1.

Exercise 5.15 On his wedding day, Kepler wanted to calculate the amount of wine con-tained inside a wine barrel whose shape is shown below in Figure 5.18. Use the disk methodto compute this volume. You may assume that the function that generates the shape of thebarrel (as a surface of revolution) is y = f(x) = R� px2, for �1 < x < 1 where R is theradius of the widest part of the barrel. (R and p are both positive constants.)

x

y

y = R − px2

1


Exercise 5.16 Consider the curve

y = f(x) = 1� x20 < x < 1

rotated about the y-axis. Recall that this will form a shape called a paraboloid. Use thecylindrical shell method to calculate the volume of this shape. Note: this technique isoptional. See the Appendix of course notes for an example of the shell method.

Exercise 5.17 Find the volume of the solid obtained by rotating the region bounded bythe given curves f(x) and g(x) about the specified line.

(a) f(x) =px� 1, g(x) = 0, from x = 2 to x = 5, about the x-axis.

(b) f(x) =px, g(x) = x/2, about the y-axis.

(c) f(x) = 1/x, g(x) = x3, from x = 1/10 to x = 1, about the x-axis.

Exercise 5.18 Let R denote the region contained between y = sin(x) and y = cos(x) for0 x ⇡/2. Write down the expression for the volume obtained by rotating R about


(a) the x-axis

(b) the line y = �1.

Do not integrate.

Exercise 5.19 Suppose a lake has a depth of 40 meters at its deepest point and is bowl-shaped, with the surface of the bowl generated by rotating the curve z = x2/10 around thez-axis. Here z is the height in meters above the lowest point of the bowl. The distributionof sediment in the lake is stratified by height along the water column. In other words, thedensity of sediment (in mass per unit volume) is a function of the form s(z) = C(40� z),where z is again vertical height in meters from the point at the bottom of the lake. Findthe total mass of sediment in the lake (Your answer will have the constant C in it.). Thevolume of the lake is the volume above the curve z = x2/10 and below z = 40.

Exercise 5.20 In this problem you are asked to find the volume of a height h pyramidwith a square base of width w. (This is related to the Cheops pyramid problem, but we willuse calculus.) Let the variable z stand for distance down the axis of the pyramid with z = 0

at the top, and consider “slicing” the pyramid along this axis (into horizontal slices). Thiswill produce square “slices” (having area A(z) and some width �z). Calculate the volumeof the pyramid as an integral by figuring out how A(z) depends on z and integrating thisfunction.

z

w

h

0


Exercise 5.21 Set up the integral that represents the length of the following curves: Donot attempt to calculate the integral in any of these cases

(a) y = f(x) = sin(x) 0 < x < 2⇡.

(b) y = f(x) =px 0 < x < 1.

(c) y = f(x) = xn � 1 < x < 1.

Exercise 5.22 Compute the length of the line y = 2x + 1 for �1 < x < 1 using thearc-length formula. Check your work by using the simple distance formula (or Pythagoreantheorem).

5.9. Exercises 151

Exercise 5.23 Spreadsheet assignment You are told that the derivative of a certain func-tion is

f 0(x) = 1� 2 sin

2(x/3)

and that f(0) = 0. Use the spreadsheet to create one plot that contains all of the followinggraphs:

(a) The graph of the function y = f(x) (whose derivative is given to you). This should beplotted over the interval from 0 to 9.

(b) The graph of the “element of arc-length” dl =

p1 + (f 0

(x))2dx showing how thisvaries across the same interval.

(c) The graph of the cumulative length of the curve L(x). Briefly indicate what you didto find the function in part (a). (You might consider how the spreadsheet would helpyou calculate the values of the desired function, y = f(x), rather than trying to find anexpression for it.)

Exercise 5.24 Work of spring A spring has a natural length of 16 cm. When it is stretchedx cm beyond that, Hooke’s Law states that the spring pulls back with a restoring forceF = kx dyne, where the constant k is called the spring constant, and represents the stiffnessof the spring. For the given spring, 8 dyne of force are required to hold it stretched by2 cm. How much work (dyne-cm) is done in stretching this spring from its natural lengthto a length 24 cm? (Note: use integration to set up this problem.)

Exercise 5.25 Work of pump Calculate the work done in pumping water out of a paraboliccontainer up to the height h = 10 units. Assume that the container is a surface of revolutiongenerated by rotating the curve y = x2 about the y axis, that the height of the water in thecontainer is 10 units, that the density of water is 1 g/cm3 and that the force due to gravityis F = mg where m is mass and g = 9.8 m/s2.

5.10. Solutions 335

5.10 SolutionsSolution to 5.1

(a) M = 20 (b) x̄ = 4

Solution to 5.2

(a) ⇢̄ =

Zb

a

⇢(x)dx

b� a(b) M =

Zb

a

⇢(x) dx (c) x̄ =

1

M

Zb

a

x⇢(x) dx

Solution to 5.3

(a) M =

1

2

, ⇢(x) and M(x) are shown in Figure 5.3.

0

0.2

0.4

0.6

0.8

1

0.2 0.4 0.6 0.8 1x

Figure 5.3. Solution for problem 7.4

(b) ⇢̄ =

1

2

(c) x̄ =

1

3

(d) s = 1�p2

2

(e)3

4

Solution to 5.4

(a) M =

aL3

3

(b) ⇢̄ =

aL2

3

(c) x̄ =

3

4

L (d) s =1

2

1/3L

Solution to 5.5

(a) x̄ =

m+ n

m+ 2n(b) x̄ ! 1 (c) x̄ ! 1

2


Solution to 5.6

(a) L 10 (b) 250 cars/km at x = 5 km (c) 1666

Solution to 5.7

(a) see Figure 5.4 (b)1

3

3/2 x

p(x)

0

1/2

1 2

Figure 5.4. Solution for problem 5.7

Solution to 5.8 123873.71 kg

Solution to 5.9 20⇡ g

Solution to 5.10

(a) x =

⇡

2a(b) p̄ =

2A

⇡(c) x̄ =

⇡

2a

Solution to 5.11 N =

20000

ln 2

⇡ 28854

Solution to 5.12

(a) straight line (b) y = x (c)⇡

3

(d)⇡

3

Solution to 5.13 V =

⇡

3

Solution to 5.14 V = 2⇡

5.10. Solutions 337

Solution to 5.15 V = 2⇡

✓R2 � 2Rp

3

+

p2

5

◆


⇡

2

Solution to 5.17

(a) V = 7.5⇡ (b) V =

64

15

⇡ (c)⇡

7

(62 +

1

10

7)

Solution to 5.18

(a) V = 2

Z ⇡4

0⇡ cos(2x) dx.

(b) V = 2

Z ⇡4

0⇡�(1 + cos(x))2 � (1 + sin(x))2

�dx.

Solution to 5.19 16000⇡C�20� 40

3

�⇡ 335103C


w2h

3

Solution to 5.21

(a) L =

Z 2⇡

0

p1 + cos

2(x) dx.

(b) L =

Z 1

0

r1 +

1

4xdx.

(c) L =

Z 1

�1

p1 + n2x2n�2 dx.

Solution to 5.22 L = 2

p5

Solution to 5.23

Solution to 5.24 W = 128 dynes·cm

Solution to 5.25 W = 980⇡ · 103�12 � 1

3

�⇡ 5.131⇥ 10

5 g cm2/s2


the given f’(x) (optional)

f(x)dl (element of arc-length)

V|

L(x) - total length

Arc length

0.0 9.0-2.0

12.0

V

|

0.0 9.00.0

0.5

Figure 5.5. Solution to problem 5.23. The right panel is a magnification to showthat dl is not constant.

applications of the deﬁnite integral to volume, mass, and ... · 5.3 mass distribution and the...

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