Applications of Integration494 15. Applications of Integration Example 15.1.5 Derive the formula for the volume of a cap of height h of a sphere of radius a, cf. Figure 15.10. Axis

Module 15

Applications of Integration

Prerequisites.The minimum prerequisites for Module 15, “Applications of Integration” are

• Definite integrals such as presented for example in Module 12.

• Indefinite integrals such as presented for example in Module 13. For the bulkof Sections 15.1 and 15.2 indefinite integrals as in Sections 13.1 and 13.2 aresufficient. In the remaining sections a significant number of exercises weredesigned to review techniques of integration.

• The Fundamental Theorem of Calculus such as presented for example inModule 14.

Each section in this module can be read independent of the others. In particular,parts of this module can be omitted, rearranged or read separately.

Figure 15.1: The launch of the Voyager 2 probe.In Sections 15.2 and 15.3 we consider the workneeded to launch a mass into space. Use of thispicture is consistent with [31]. Endorsement byNASA is neither claimed nor implied.

Figure 15.2: Definite integrals are used to com-pute volumes and centers of mass of rotationallysymmetric objects (cf. Sections 15.1 and 15.5).

Learning Objectives.After successfully completing this module the student will be able to

1. Compute volumes of three dimensional objects.

2. Compute the work required for certain physical tasks.

3. Compute improper integrals over infinite intervals and over singularities.

4. Determine if a given improper integral converges or diverges.

5. Compute the mass and the center of mass of a linearly distributed density.

6. Explain the modeling for volumes, work, mass and centers of mass as a sum-mation of small increments for which physical laws have simpler form.

IntroductionDefinite integrals can arise whenever large numbers of small quantities are

added. Module 12 has illustrated this idea through the introduction of Riemannsums. The Fundamental Theorem of Calculus provides a faster way to computedefinite integrals, but the modeling process of summing small increments remainsthe same. We revisit this modeling process in this module. In Sections 15.1 and15.2 we revisit volumes and work, respectively. Computation of work can involve

491

492 15. Applications of Integration

infinite intervals or singularities, so we investigate integrals over infinite intervalsor across singularities in Sections 15.3 and 15.4. We conclude with the computationof centers of mass of linear densities in Section 15.5. Since we had some exposureto modeling with integrals in Sections 12 and 14, this module will also highlightthe use of parameters.

15.1 Volumes

To compute volumes of arbitrary objects we reduce the computation to a sum ofvolumes of cylinders. The volume of a cylinder with base areaB and heighth isV = Bh (cf. Figure 15.3). This formula is independent of the shape of the base.

B

h

base area B, height hvolume V=Bh

Figure 15.3: The volume of any cylinder,inde-pendent of the shape of the base, is the productof the area of the base with the height.

To reduce an arbitrary geometric figure to a set of cylinders, slice it, say, hori-zontally, inton slices of thickness1y. Each slice is approximately a cylinder (cf.Figure 15.4). The height is the thickness1y and the base area isA(yk), the area

of the cross section at positionyk. Summing the volumes yieldsV ≈n∑

k=1

A(yk)1y.

As the numbern of slices goes to infinity, the error made by replacing non-cylindricalslices with cylinders goes to zero. Therefore the volume is

V = limn→∞

n∑k=1

A(yk)1y =∫ t

bA(y) dy,

Figure 15.4: A volume and an approximation ofit with cylindrical slices.

whereb is the object’s lower bound iny and t is its upper bound. We havederived the following theorem.

Theorem 15.1.1The volume of a given solid is the integral of the cross sec-

tional areas from bottom to top. Symbolically, V=∫ t

bA(y) dy.

It is also possible to slice vertically for the volume as an integral along thex-

axis V =∫ r

lA(x) dx, wherel is the object’s left bound andr is its right bound.

Theorem 15.1.1 suggests the following procedure (also cf. Examples 12.1.6 and14.1.5).

6

y = b

y = t

cross sectional areaA(y)

Figure 15.5: The volume of an object thatstretches fromy = b to y = t and which has

cross sectional areaA(y) is∫ t

bA(y) dy.

-x = l x = r

A(x)

cross sectional area

Figure 15.6: The volume of an object thatstretches fromx = l to x = r and which has

cross sectional areaA(x) is∫ r

lA(x) dx.

Algorithm 15.1.2 Volume computation with integrals (cf. Figures 15.5 and15.6).

1. Choose an axis through the object. The area of cross sections perpen-dicular to the axis should be easy to compute.

2. Make the chosen axis either the y-axis (Fig. 15.5) or the x-axis (Fig.15.6).

3. Compute the area A(y) (or A(x)) of the cross section at y (or x).

4. Obtain the lowest and highest projections b and t of the object onto they-axis (or the leftmost and rightmost projections l and r onto the x-axis).

5. Set up and compute the integral∫ t

bA(y) dy or

∫ r

lA(x) dx.

15.1. Volumes 493

The alignment of the axis does not influence the volume or the computation.Some objects are simply more easy to visualize sliced one way or the other.

Solids of rotation are obtained by rotating the graph of a functionf about thex- or y-axis, usually thex-axis. Their cross sections are circles and the radius atx is f (x). This makes volumes of solids of rotation easy to set up (cf. Example14.1.5 for an early example).

Figure 15.7: The function in Example 15.1.3.

Example 15.1.3Compute the volume of a board game token that is made by rotat-

ing the function f(x) =1

5

√−6x6+ 10x4− 5x2+ 1 on the interval[0,1] about

the x-axis (all dimensions in inches, cf. Figures 15.7 and 15.8).

Figure 15.8: Turned components have rotationalsymmetry. The miniature bowling pins weremade on a lathe. The cutting tool was shapedsimilar to the function in Figure 15.7.

Axis and coordinate system: The function is to be rotated about thex-axis andwe use the usual coordinate system.

Shape of cross section: Each cross section is a circle of radiusf (x).Area of cross section:πr 2

= π ( f (x))2.Bounds:l = 0, r = 1.The above leads to the integral∫ r

lA(x) dx =

∫ 1

0π

(1

5

√−6x6+ 10x4− 5x2+ 1

)2

dx

=π

25

∫ 1

0−6x6

+ 10x4− 5x2

+ 1 dx

=π

25

[−

6

7x7+ 2x5

−5

3x3+ x

]1

0

=π

25

[−

6

7+ 2−

5

3+ 1

]=

2

105π

Figure 15.9: In a lathe, metal bars are turnedat high velocities. By moving the cutting tool(front center) along the spinning bar, a mechaniccan produce any solid of revolution.

Discussion 15.1.4Whenever symmetry about a central axis is needed, solids ofrotation arise naturally. For example, a rotating shaft in an engine should be sym-metric about its central axis. Any significant deviation from symmetry about thecentral axis would make the shaft run less smoothly, leading to energy losses andwear and tear.

Solids of rotation can be produced on lathes (cf. Figure 15.9) and are alsocalled turned components. In a lathe, a metal rod is rotated around its centralaxis. A cutting tool proceeds along the outline of the functionf that defines thesolid. Usually, especially for transmission shafts, this function is made up of severalstraight lines. For more sophisticated shapes the cutting tool usually is computerguided, or a specific template that outlines the whole functionf is used. The cuttingtool for the board game tokens in Figure 15.8 was specifically made to look similarto the function in Figure 15.7.

In design, the volume of the components is used to calculate the mass.

The volume of a cap of a sphere can be used to compute the volume of liquidin a spherical tank from the more easily measured depth of the liquid. Exercise 6shows that, for example, losing half the height of liquid can mean losing signifi-cantly more than half the volume. In Exercise 11, a similar formula is derived for acylinder lying on its side, another common shape for tanks that store liquids.


Example 15.1.5Derive the formula for the volume of a cap of height h of a sphereof radius a, cf. Figure 15.10.

Axis and coordinate system: Since a sphere is symmetric in any direction, wecan choose any axis. To stay with the idea of a tank to be filled, let us choose they-axis and place the center at the origin.

Shape of cross section: Each cross section is a circle.

Figure 15.10: The (bottom) cap of a sphere, dot-ted lines indicate the rest of the whole sphere.

Area of cross section: Figure 15.11 shows how to obtain the radius of the cross

section at heighty. The area then isπr 2= π

(√a2− y2

)2= π

(a2− y2

).

6

-

y

x

−a

−a+ h

x =√

a2 − y2

x = −√

a2 − y2

Figure 15.11: Vertical cross section of the cap ofa sphere, used to determine the bounds and theradius of each horizontal cross section.

Bounds: The bottom of the cap is atb = −a and with a height ofh the top isat t = −a+ h (cf. Figure 15.11).

The above leads to the integral

∫ t

bA(y) dy =

∫−a+h

−aπ(a2− y2

)dy

= π

[a2y−

1

3y3

]−a+h

−a

= π

[a2(−a+ h)−

1

3(−a+ h)3−

(a2(−a)−

1

3(−a)3

)]= π

[−a3+ a2h+

1

3a3− a2h+ ah2

−1

3h3+ a3−

1

3a3

]= π

[ah2−

1

3h3

]= πh2

[a−

h

3

]

\Sanity check": For h=2a weshould obtain the volume of afull sphere of radius a and forh=a we should obtain halfthat volume. Explain toyourself why this should beand verify that the formulasare correct.h=2a:

h=a:

We can also investigate objects whose cross sections are not round.

Example 15.1.6Derive the formula for the volume of a frustrum of a square pyra-mid such that the side length of the base is a, the side length at the top is b and theheight is h (cf. Figure 15.12).

Axis and coordinate system: Embed the figure into a coordinate system withthe y-axis vertically through the center of the pyramid and the base aty = 0.

Shape of cross section: All cross sections are squares.Area of cross section: To obtain the side lengths of the square at heighty, con-

sider a vertical cross section as in Figure 15.13. The bottom left and right corners

of the cross section are at(−

a

2,0)

and(a

2,0)

and the top left and right corners

are at

(−

b

2, h

)and

(b

2, h

). Since the sides are straight lines, the equations of

the sides arex = −b− a

2hy−

a

2andx =

b− a

2hy+

a

2(note thatb−a is negative).

The equations are set up as functions ofy because we are integrating with respect toy. The side length of the cross section at heighty is the distance between the two

boundaries, which isb− a

2hy+

a

2−

(−

b− a

2hy−

a

2

)= a+

b− a

hy. Therefore

the area of the cross section at heighty is l 2=

(a+

b− a

hy

)2

.

Bounds: The bounds are 0 andh.The above leads to the integral

V =

∫ h

0

(a+

b− a

hy

)2

dy

15.1. Volumes 495

=

∫ h

0a2+ 2a

b− a

hy+

(b− a

h

)2

y2 dy

=

[a2y+ a

b− a

hy2+

1

3

(b− a

h

)2

y3

]h

0

= a2h+ ab− a

hh2+

1

3

(b− a)2

h2h3

= a2h+ abh− a2h+1

3b2h−

2

3abh+

1

3a2h

=1

3a2h+

1

3abh+

1

3b2h =

1

3h(a2+ ab+ b2

)�� C

CCCCCC

��

a

b

h

?

66

� -

-�

Figure 15.12: The frustrum of a square pyramidin Example 15.1.6.

-

y

�� A

AAAAAAAA

xa

2−

a

2 −b

2

b

2

h

6

Figure 15.13: Vertical cross section of the frus-trum in Example 15.1.6.

\Sanity check": Check that forb=0 we obtain the formula forthe volume of a squarepyramid with base length aand for b=a we obtain thevolume of an a× a× h box.b=0:

b=a:

One can also lay the frustrum on its side and then integrate in thex-direction.Exercise 17 shows that this gives the same result. In Exercise 18 we use classicalgeometry to derive the volume yet another way.

Example 15.1.7For 0 ≤ x ≤ 1, the vertical cross section at x of an object is anequilateral triangle with side length s(x) = x− x2. Compute the object’s volume.

Axis and coordinate system: As given, we should integrate along thex-axis.Shape of cross section: The cross sections are equilateral triangles.

TTTTTTTTTTTTT�

��

a a

a

2

a

2

h =

√a2 −

( a

2

)2=

√3

2a

Figure 15.14: Deriving/recalling the area for-mula for an equilateral triangle.

Area of cross section: The area of an equilateral triangle with side lengths is

A(s) =1

2sh=

1

2s

√3

2s=

√3

4s2 Figure 15.14 shows how this formula can quickly

be (re-)derived if not known or forgotten.Bounds: We integrate fromx = 0 to x = 1.The above leads to the integral

V =

∫ 1

0

√3

4(s(x))2 dx

=

∫ 1

0

√3

4

(x − x2

)2dx

=

∫ 1

0

√3

4

(x2− 2x3

+ x4)

dx

=

√3

4

[1

3x3−

1

2x4+

1

5x5

]1

0

=

√3

4

[1

3−

1

2+

1

5

]=

√3

120

Finally, we can combine volume combinations to obtain more complex shapes.

Example 15.1.8A hole of radius1

2has been drilled through a sphere of radius1.

What is the volume of the object thus obtained?Axis and coordinate system: Since the sphere is symmetric with respect to any

axis through its center, use thex-axis as central axis and assume that the hole isdrilled horizontally along thex-axis. Place the center of the sphere at the origin.

Shape of cross section: Each cross section is an annulus (a ring).

496 Module 15: Applications of Integration

6

-

y

x

y = ±

√1− x2

1

2

−1

2

√3

2−

√3

2

Figure 15.15: Finding the bounds for the integralin Example 15.1.8.

Area of cross section: The inner radius of each annulus is1

2and the outer radius is

√1− x2. This leads to an area of

A(x) = π(√

1− x2)2− π

(1

2

)2

=3π

4− πx2.

Bounds: The object starts and ends at the intersection points of

f (x) =√

1− x2 andg(x) =1

2, which are (also cf. Figure 15.15)

√1− x2 =

1

2

1− x2=

1

4

x = ±

√3

2.

Overall, the above leads to the following integral.

V =

∫ √3

2

−

√3

2

3π

4− πx2 dx =

[3π

4x − π

1

3x3

]√32

−

√3

2

=

3π

4

√3

2− π

1

3

(√3

2

)3

−

3π

4

(−

√3

2

)− π

1

3

(−

√3

2

)3

=3√

3π

4−

√3π

4=

√3π

2

Exercises

1. Common household items, such as bottles and glasses, have ro-tational symmetry. For each of the rotationally symmetric ob-jects below, sketch the function that is rotated about the axis ofsymmetry to obtain the object. To get a functionf (x), you’llneed to think of each object as lying on its side.

(a) The bottle on the left in Figure 15.16.

(b) The bottle in the middle in Figure 15.16.

(c) The bottle on the right in Figure 15.16.

Figure 15.16: Bottles for Exercise 1.

(d) The glass on the left in Figure 15.17.

(e) The glass in the middle in Figure 15.17.

(f) The glass on the right in Figure 15.17.

(g) Name at least two other objects that you encounter indaily life which have rotational symmetry.

Figure 15.17: Glasses with rotational symmetry for Exercise 1.The author has experimental evidence that alcohol impairs yourjudgment. Don’t drink and derive. And don’t even think aboutoperating machinery!

Section 15.1: Exercises 497

2. Find the volume obtained by rotating the function on the giveninterval about thex-axis.

(a) f (x) = x2− 1, x = −1, x = 1

(b) f (x) = x + 1, x = 0, x = 1

(c) f (x) = ex , x = 0, x = 1

(d) f (x) = xex , x = 0, x = 1

(e) f (x) = sin(x), x = 0, x = π

(f) f (x) = ln(x), x = 1, x = e

(g) f (x) = x2− 1, x = −2, x = 2

(h) y =1

4√

1− x2, 0≤ x ≤

1

2

3. Find the volume obtained by rotating the area between the func-tions (on the given interval) about thex-axis.

(a) f (x) = x2, g(x) =√

x

(b) f (x) = x2, g(x) =√

x, a = 0, b = 2

Hint. Be careful about which function is larger where.

(c) f (x) = e−x , g(x) = e−2x , a = 0, b = 1

(d) f (x) = cos(x), g(x) = sin(x), a = 0, b =π

4

4. Find the volume of the solid that is between the given positionson thex-axis and which has the given cross sections.

(a) x = 0, x = 4, cross sections are squares of side length2x

(b) x = 0, x = π , cross sections are squares of side lengthsin(x)

(c) x = −1, x = 1, cross sections are equilateral trianglesof side lengthxex

(d) x = 0, x = 2, cross sections are semicircles of radiusex

(e) x = 0, x = 3, cross sections are triangles with basex+1and height 2x + 5

(f) x = 0, x =π

2, cross sections are triangles with base

sin(x) and height cos(x)

(g) x = 1, x = 3, cross sections are equilateral trapezoidswith bottom width 3x − 1, top widthx2

+ 1 and height2x

5. Dr. Smith ’s house is heated by an oil-burning furnace. The oilis stored in a spherical underground tank of radius 1m. Winter(and thus heating season) usually lasts 124 days, from Decem-ber 1st to April 3rd. On December 1st the level of oil in thetank was 1.4m. By January 31st, the level of oil in the tank hasdropped to 0.7m. For how many more days will the remainingoil last? Assume that each day the same amount of oil is burned.

6. An oil tank is spherical with radiusr .

(a) Show that if the height of oil in the tank drops from3

4r

to3

8r , then more than half the volume of oil in the tank

has been pumped out.

Hint. Use the formula from Example 15.1.5.

(b) Is there any situation in which we can pump out less thanhalf the remaining oil and the fill height drops to half theoriginal height?Give a formal argument with the formula from Example15.1.5 or a geometric explanation.

7. Derive the formula for the volume of a right circular cone ofheighth and with base radiusr .

8. Derive the formula for the volume of a square pyramid of heightH and with base side lengtha.

9. Derive the formula for the volume of a pyramid whose base isan equilateral triangle of side lengtha and which has heighth.

10. Derive the formula for the volume of a truncated right circularcone (cf. Figure 15.18) of heighth with base radiusR and withtop radiusr .

��

LLLLLLLLL

Figure 15.18: The truncated cone in Exercise 10.

11. A right circular cylinder of radiusr and lengthl is lying on itsside. Find the volume of the part that remains if the cylinder istruncated at heighth. (For motivation, consider Exercise 6 inSection 15.2.)

12. A hole of radiusr has been drilled through a sphere of radiusR> r .

(a) Find the volume of the remaining part of the sphere.

(b) Find the volume of the material that was drilled out.

13. A torus (cf. Figure 15.19) is essentially the shape of an innertube with tube radiusr for which the radius of the circle in thecenter of the tube isR > r . Derive the formula for the volumeof a torus of main radiusR and “tube radius”r .Hint. Think of an easy shape that can be rotated to obtain thetorus.

Figure 15.19: A torus.

14. Find the volume common to two circular cylinders that have thesame radius and whose axes intersect at a right angle (cf. Figure15.20).


15. Two spheres of radiusr are positioned such that the center ofeach sphere is on the boundary of the other sphere.

(a) Find the volume that is common to the spheres.

(b) Find the total volume of the figure.

(c) Find the volume that is left when the intersection of thespheres is taken away from one of the spheres.

Figure 15.20: The intersection of two cylinders.

16. An ellipsoid is a three dimensional analogue of an ellipse (cf.

Figure 15.21). Its general equation isx2

a2+

y2

b2+

z2

c2= 1.

(a) Derive the formula for the volume of the general ellip-soid. (Use the result of Exercise 10a in Section 14.2.)

(b) Use your result from part 16a to obtain the formula forthe volume of a sphere of radiusr .

Figure 15.21: An ellipsoid is like a sphere with different radiiin thex-, y- andz-directions.

17. Derive the formula for the volume of a frustrum of a squarepyramid such that base length isa, the side length at the top isb and the height ish (cf. Example 15.1.6). Do so by laying thepyramid on its side and using thex-axis as the axis through themiddle of the pyramid.

18. Derive the formula for the volume of a frustrum of a squarepyramid such that base length isa, the side length at the top isband the height ish (cf. Example 15.1.6). Do this by subtracting

the volume of the missing top part from the volume of the wholepyramid that would contain the frustrum.

Hint. You will need to determine the height of the whole pyra-mid.

6HHH

HHHHHH

��

��−a

2

a

2

b

2

−b

2

h -

y

x

Figure 15.22: Vertical cross section of the pyramid that lies onits side in Exercise 17.

19. Solids with given cross sections.

(a) Prove that the solid betweenx = a andx = b obtainedby rotating the functionf (x) about thex-axis has vol-

umeV =∫ b

aπ ( f (x))2 dx.

(b) Prove that the solid betweenx = a and x = b ob-tained by rotating the area between the functionsf (x)andg(x) (with f (x) ≤ g(x)) about thex-axis has vol-

umeV =∫ b

aπ ( f (x))2− π (g(x))2 dx.

(c) Give a formula for the volume of the solid betweenx = aandx = b such that the cross section atx is a square withbase lengthf (x).

(d) Give a formula for the volume of the solid betweenx = aandx = b such that the cross section atx is an equilateraltriangle with base lengthf (x).

(e) Give a formula for the volume of the solid betweenx = aandx = b such that the cross section atx is an isoscelesright triangle with hypothenuse of lengthf (x).

(f) Give a formula for the volume of the solid betweenx = aand x = b such that the cross section atx is a regularpentagon such that each side has lengthf (x).

(g) After completing some of the above tasks, decide if it isuseful to memorize formulas such as the above. Explainwhy or why not.

15.2. Work 499

15.2 Work

The definition of work as force times distance,W = Fd, applies as long as theacting forceF is parallel to the distance movedd and F is constant. However,the force can vary over the distance (cf. Section 12.1.2 and Examples 12.2.5,12.2.16) and the computation can also depend on a geometric argument as in Exam-ple 12.1.7. Therefore integral formulas and algorithms similar to Theorem 15.1.1and Algorithm 15.1.2 do not exist for work. Instead, each situation needs to beanalyzed separately.

Our first example deals with the problem of moving a certain payload to orbit.

Example 15.2.1Compute the minimum amount of work that is needed to move apayload of mass m= 1kg from the surface of the earth to the international spacestation (cf. Figure 15.23).

Distance over which the force acts.Split the total distance into short steps oflength1r . For each1r , the force acting over the distance can be considered to beconstant. The total distance starts on the surface of the earth and ends in the orbitof the international space station.

Figure 15.23: The international space station(NASA photograph). In Example 15.2.1 wecompute the minimum amount of energy neededto move a 1kg mass from earth to the interna-tional space station. Use of this picture is consis-tent with [31]. Endorsement by NASA is neitherclaimed nor implied.

Force needed.The force that is needed to move the mass from a heightr to

a heightr + 1r is given by Newton’s law of gravitation asF(r ) = GmM

r 2, if we

ignore the slight decrease in the gravitational force over the short interval.Data. All needed quantities can be looked up. The mean radius of earth is

about 6,370,949m≈ 6,371km1, so the integral will start atre = 6,371km. Theaverage orbit height of the international space station is 354km, so the integral willend atr iss = 6,725km. The mass of earth isM ≈ 5.9763× 1024kg and the

gravitational constant isG ≈ 6.6720× 10−11Nm2

kg2 . We solve the integral with the

symbolic parameters and substitute numbers at the end.

W = limn→∞

n∑k=1

F(rk)1r =∫

F dr =∫ r iss

re

GmM

r 2dr = −G

mM

r

∣∣∣∣r iss

re

≈ −6.6720×10−11Nm2

kg2 ·1kg·5.9763×1024kg·

(1

6,725km−

1

6,371km

)≈ 3.2945× 106Nm

When a computation involvescomplicated numbers it isusually easier to solve thecomputation with symbolicparameters and onlysubstitute the numbers at theend.

In the last step we canceled meters (m) against kilometers (km), which leads toan extra factor 10−3.

With 1Nm= 1J= 1Ws we note that

3.2945× 106Nm= 3.2945× 106Ws= 915.1Wh,

which is about the energy used by nine 100W light bulbs in an hour. This a rel-atively small amount of energy. We must consider, however, that Example 15.2.1just gives the difference in gravitational potential energy between the surface ofthe earth and the international space station’s orbit. This is the absolute minimumamount of energy needed to lift the payload. The actual amount of energy is sig-nificantly larger, because the lifting mechanism (usually a rocket or a shuttle) itselfneeds to be lifted, requiring further energy. Also, friction is not taken into account.

1It’s a little more on the equator where the mean radius is 6,378,077m, but we will work withthe rounded mean value.


In Exercise 9 in Section 15.3 we compute the velocity to which the mass wouldneed to be accelerated on the surface of the earth to shoot it directly to the orbit.

A company that manufactures tanks may be interested in the following.

Example 15.2.2A full spherical oil tank of radius a is emptied by pumping the oilout through the top of the sphere. Determine the work needed to empty the tank thisway.

In this situation, the force on each oil molecule is equal to the force of gravitythroughout the motion. However, oil molecules at different depths will need totravel different distances to reach the top of the sphere. Thus we envision the oilbeing removed in horizontal layers. All molecules in a given layer will travel thesame height. Our analysis is done for a “typical” layer at heighty.

-

6

x

y

1h Base area:π(a2− y2

)?

6 � -√a2 − y2

y

a

a− y

?

6

Figure 15.24: Visualization of the setup in Ex-ample 15.2.2. The sphere is sliced into flat layersthat are almost cylinders. These layers are liftedto the top. One sample layer is given.

Distance over which the force acts.As shown in Figure 15.24, the layer atheighty must be lifted a distance ofd(y) = a− y.

Force needed.For each layer, the force isg1m, gravitational acceleration

g ≈ 9.81m

s2times the mass1m that is lifted. The mass1m of the layer isρ1V , the

product of the density of the oil with the volume1V of the layer. The volume1Vis a cylinder of height1y with circular base of radiusr =

√a2− y2 (cf. Figure

15.24). That is,1V = π(a2− y2

)1y. Overall, the force needed to lift the layer at

height y to the top of the tank is1F(y) = g1m= gρ1V = gρπ(a2− y2

)1y.

Since the radius is not specified, the result will be in terms ofa and we will notsubstitute numbers for any of the parameters.

The above leads to the following integral.

W =

∑1F(y)d(y) =

∑gρπ

(a2− y2

)1y(a− y)

=

∫ a

−agρπ

(a2− y2

)(a− y) dy= gρπ

∫ a

−a

(a2− y2

)(a− y) dy

= gρπ∫ a

−aa3− ay2

− a2y+ y3 dy= gρπ

[a3y−

1

3ay3−

1

2a2y2+

1

4y4

]a

−a

= gρπ

[a4−

1

3a4−

1

2a4+

1

4a4−

(−a4+

1

3a4−

1

2a4+

1

4a4

)]=

4

3gρπa4


Exercises

1. An object is moved fromx = a to x = b. The force used atpositionx is F(x). Compute the total amount of work done.

(a) F(x) = 3− 3x, a = 0, b = 1

(b) F(x) = x√

1− x2, a = 0, b = 1

(c) F(x) = xe−x, a = 0, b = 3

(d) F(x) = x2e−x , a = 0, b = 3

2. Solve the following earlier problems using integrals and the fun-damental theorem of calculus.

(a) Example 12.1.4

(b) Example 12.1.7

(c) Example 12.2.5

(d) Exercise 16 in Section 12.1

(e) Exercise 18 in Section 12.1

(f) Exercise 19 in Section 12.1

3. Emptying feed troughs. (Compare with Example 12.1.7 in Sec-tion 12.1.5.) A feed trough with the given cross section, depthand length is cemented to the ground and filled with water. Findthe amount of work that is needed to pump all the water out ofthe trough.

(a) Cross section is an equilateral triangle with the tip point-

ing downwards. The trough is1

2m deep and and 2m long.

(b) Cross section is a trapeze with top width 1m and bottom

width1

2m. The trough is

1

4m deep and and 3m long.

(c) Cross section is an trapeze with top widtht and bottomwidth b < t . The trough ish meters deep and andlmeters long.

(d) Cross section is a semicircle with radius 0.2m and thetrough is 1m long.

(e) Cross section is a semicircle with radiusr and the troughis l meters long.

4. (Compare with Exercise 10 in Section 14.1.) An elevator ofmassm is attached to a cable of linear densityρ. When thecable is pulled up all the way, the center of mass of the elevatoris at height 0. Compute the amount of work that it takes to pullthe elevator from heighta up to heightb > a. (Botha andb arenegative.)

5. A spherical oil tank of radiusr is filled to a height of1

2r .

(a) Determine the work that is needed to pump the oil throughthe top of the sphere.

(b) Determine the work that is needed to pump out the oil

that is left after the fill height has dropped to1

4r .

(c) Is your answer in part 5a twice the answer in part 5b?Explain why or why not.

6. A certain oil tank is shaped like a right circular cylinder that islying on its side. The tank has a radius of 1.5m and a length of5m.

(a) Find the amount of work that it takes to pump out thewhole contents through the top of the tank.

Hint. The integral can be solved using geometry andTheorem 12.2.12.

(b) Find the amount of work that it takes to empty the tankthrough the top if the tank is half full.

(c) Find the amount of work that it takes to empty the tankthrough the top if the tank is three quarters full.

Hint. Use a CAS is the integral is getting too tedious.

(d) Why is the amount in part 6b not half the amount in part6a? Why is the amount in part 6c not three quarters theamount in part 6a? Why is the amount in part 6b not twothirds the amount in part 6c? Explain.

7. A certain oil tank is shaped like a right circular cylinder that islying on its side. The tank has a radius ofr and a length ofl .

(a) Find the amount of work that it takes to pump out thewhole contents through the top of the tank.

(b) Find the amount of work that it takes to empty the tankthrough the top if the tank is half full.

(c) Find the amount of work that it takes to empty the tankthrough the top if the tank is three quarters full.

8. Compare the result of Example 15.2.1 with the amount of workcomputed with the formulaW = mgh for potential energy thatwe use for such computations on Earth. Theg denotes the accel-eration of Earth’s gravitational field on the surface of the Earth.

It is g ≈ 9.81m

s2. Then answer the following questions.

(a) Why are the two numbers not equal?

(b) Is the difference significant?

(c) Which approach should be used to compute the energyneeded to lift a payload to orbit? The approach of Exam-ple 15.2.1 or the approach usingW = mgh?

9. Compute the minimum amount of work that is needed to lift a1kg payload from the surface of the earth to geostationary orbit(radius≈ 42,164km, which is≈ 35,787km above the surfaceof the earth) in two ways.

(a) First use the approach of Example 15.2.1.

(b) Then compute the work using the formulaW = mghforpotential energy that we use for such computations on

earth. Recallg ≈ 9.81m

s2.

(c) Explain why the two numbers are not equal.

(d) Is the difference between the two numbers significant?

(e) Which approach should be used to compute the energyneeded to lift a payload to orbit? The approach of Exam-ple 15.2.1 or the approach usingW = mgh?

10. A water tank is in the shape of a right circular cylinder (standingupright) of height 10 meters and radius 5 meters. To empty thetank, the water is pumped to the top of the tank. Find the totalamount of work needed to empty the tank.


15.3 Improper Integrals I: Infinite Intervals

Integrals over infinite intervals arise naturally in physics and in statistics.

Example 15.3.1A deep space probe (also cf. Figures 15.1 and 15.25) must haveenough energy to (theoretically) move to any point in the universe. Otherwise, itwould eventually fall back to earth. In other words, the probe must have enoughenergy to move arbitrarily far away from Earth. To compute this energy, we needto compute the amount of work needed to move the probe infinitely far away fromthe surface of the Earth.

Figure 15.25: Saturn (NASA photograph, takenby the Voyager 2 probe). In Example 15.3.1 westart the discussion how much energy is neededto completely escape the gravitational pull ofearth. Voyager 2 needed even more than thatamount of energy to reach Saturn and go beyond.(The Voyager probes have since left the solarsystem.) Use of this picture is consistent with[31]. Endorsement by NASA is neither claimednor implied.

The work needed to move the probe a distancedr against the gravitational

force F = GmM

r 2of Earth isFdr . M denotes the mass of Earth,m is the mass

of the probe,r is the distance between the centers of mass andG is the gravita-tional constant. Summing up the differential amounts of work leads to the integral∫

F dr =∫

GmM

r 2dr . The lower bound should beR, the radius of the Earth,

while the upper bound would need to be infinity.Thus to complete this example we will need to compute an integral over an

infinite interval. This is done in Example 15.3.7.

Example 15.3.2 In probability theory/statistics, the probabilityP([a,b]) that theoutcome of an experiment lies in a certain interval [a,b] is given by the integral of

the probability density functionfX over the interval. That is,P([a,b]) =∫ b

afX dx.

This idea is expressed by saying thatarea represents probability(cf. Figure 15.26).-

a b

area representsprobability

P(a ≤ X<b)

Figure 15.26: In probability theory, area is inter-preted as probability. The greater the area overthe interval [a,b], the greater the probability ofan outcome in the interval.

To compute the probability that an outcome exceeds a numbera, we wouldneed to compute the integral offX from a to∞. Again we encounter an integralover an infinite interval. Statistics will be discussed in more detail in Modules 17and 18.

To define integrals over intervals [a,∞), we approximate the infinite intervalwith finite intervals [a, t ] and let the upper limit grow beyond all bounds (t →∞).

Definition 15.3.3 Let f be a continuous function on the interval[a,∞). If∫ t

af (x) dx exists for every t in[a,b) and lim

t→∞

∫ t

af (x) dx exists, then we

define theimproper integral of f from a to∞ to be∫∞

af (x) dx = lim

t→∞

∫ t

af (x) dx.

In this case the improper integral is calledconvergent. If the limit does notexist, the improper integral is calleddivergent.Improper integrals over intervals(−∞,a] are defined analogously.Finally, if f is defined for all real numbers, we define∫

∞

−∞

f (x) dx =∫ a

−∞

f (x) dx+∫∞

af (x) dx

provided both integrals on the right side exist.

6y

-x

a

��

��

Figure 15.27: The improper integral exists if andonly if the area underf (marked) froma to∞is finite.

Geometrically, an improper integral exists if and only if the area to the rightof a is finite, even though it has an infinitely long base (cf. Figure 15.27). This

15.3. Improper Integrals I: Infinite Intervals 503

means that the function decays rapidly. Otherwise the infinite width would makethe total area infinite. Symbolically, as long as we can compute the integrals andlimits involved, integration over infinite intervals is not a problem. However, unlikefor finite intervals, the improper integral of a continuous function may not exist.

Example 15.3.4Compute the improper integral∫∞

0e−2x dx or show that it does

not exist.

Figure 15.28: The area underf (x) = e−2x from0 to∞ is finite. That is,e−2x decays fast enoughthat the infinite base of the area is not enough tomake the total area infinite.

We find the antiderivative using substitution and then we compute the limit asthe upper bound goes to infinity.

∫∞

0e−2x dx = −

1

2e−2x

∣∣∣∣t→∞0

= limt→∞−

1

2e−2t−

(−

1

2e−2·0

)=

1

2.

The improper integral converges to1

2.

Example 15.3.5Compute the improper integral∫∞

1

1

xdx or show that it does not

exist.

Figure 15.29: The area underf (x) =1

xfrom 1

to∞ is infinite. That is,1

xdoes not decay fast

enough to make the total area finite.

∫∞

1

1

xdx = ln |x|

∣∣∣t→∞1= ∞− 0.

Since the limit is infinite, this improper integral diverges.

Integrals with lower bound−∞ are handled similarly.

Example 15.3.6Compute the improper integral∫ 0

−∞

xex dx or show that it does

not exist.

Figure 15.30: The area underf (x) = xex from−∞ to 0 is finite. The analysis is similar to thatfor intervals that go to∞. This figure also showsthat graphical predictions for improper integralsare impossible. The area in this figure looks asif it could be a significant portion of the area inFigure 15.29, yet it is finite.

∫ 0

−∞

xex dx = tet− et

∣∣0t→−∞

= 0− 1−

(lim

t→−∞tet− lim

t→−∞et

)= −1− lim

t→−∞

t

e−t+ 0= −1− lim

t→−∞

1

−e−t= −1

So the improper integral converges to 1.

The notion of an improper integral froma to ∞ allows us to conclude ourphysics example.

Example 15.3.7 (Conclusion of Example 15.3.1.) Compute the escape velocityfrom the surface of the earth.

First we compute the energy required to free a massm from the gravitationalpull of earth. From Example 15.3.1 we recall this energy as∫

∞

RG

mM

r 2dr = −G

mM

r

∣∣∣∣r→∞R

= GmM

R.

With R≈ 6.37× 106m, M ≈ 5.98× 1024kg andG = 6.67× 10−11Nm2

kg2 we obtain

mGM

R≈ m

6.67× 10−11Nm2

kg2 5.98× 1024kg

6.37× 106m≈ m× 6.26× 107Nm

kg.


The energy used to escape the Earth’s pull would have to be kinetic energy.

Kinetic energy is given byEkin =1

2mv2. Setting this equal to the energy above

allows us to cancel the mass and obtain a number for the velocityv.

1

2mv2

= mGM

R

v =

√2GM

R≈

√2 · 6.26× 107

kgms2 m

kg≈ 11,191

m

s

or about 11.2kms , which is 40,320km

h or over 25,000mih . This velocity is called

theescape velocityfrom the surface of the Earth. It is the minimum velocity thatan object on the surface of the Earth must have to escape the planet’s gravitationalpull. (Note that other factors such as air drag and how to achieve this velocity mustbe accounted for separately.)

15.3.1 Convergence Tests

For functions such asf (x) =1√

2πe−

x22 , which is important in statistics, there is no

antiderivative in closed form. In this situation we can try to estimate the improperintegral if it exists. Thus we first need an existence criterion for improper integrals.

Existence of an improper integral can be established by comparing it to a knownintegral, because an infinite area cannot fit inside a finite area.

Theorem 15.3.8 (The comparison test for improper integrals over infinite in-tervals.) Let f and g be continuous functions with f(x) ≥ g(x) ≥ 0 forx ≥ a.

1. If∫∞

af (x) dx is convergent, then

∫∞

ag(x) dx is convergent.

2. Equivalently, if∫∞

ag(x) dx is divergent, then

∫∞

af (x) dx is diver-

gent.

6y

-x

a

f

g

Figure 15.31: Visualization of the comparisontest (cf. Theorem 15.3.8). If the area underf isfinite, then the area underg must be, too. Simi-larly, if the area underg is infinite, then the areaunder f must also be infinite.

Proof. Deferred to Section B.9.

It must be emphasized that the comparison test doesnotapply if the lower func-tion g has a finite integral or if the upper functionf has an infinite integral. In eithercase the other function can still have an infinite or a finite integral, respectively.

Example 15.3.9Determine if the integral∫∞

1e−

x22 dx converges or diverges.

The function f (x) = e−x22 decays faster than an exponential function with

negative exponent. We have seen such a function in Example 15.3.4, so we expect

the integral to converge. Thus, to find a function that is larger thanf (x) = e−x22

and whose improper integral still converges, we should look for an exponential

function with an easier exponent that is larger than−x2

2for x ≥ 1. The expo-

nent−x

2will work, because forx ≥ 1 we havex2

≥ x, then−x2≤ −x and


−x2

2≤ −

x

2, and finallye−

x22 ≤ e−

x2 . Now

∫∞

1e−

x2 dx = −2e−

x2

∣∣∣t→∞1= 2e−

12 ,

so∫∞

1e−

x2 dx converges. By the comparison test,

∫∞

1e−

x22 dx converges.

The functione−x22 is normally considered on the interval(−∞,∞). To es-

tablish convergence of the improper integral on(−∞,∞) we can break up theimproper integral into the integrals on(−∞,0] and [0,∞), respectively. However,the comparison test works much better on the intervals(−∞,−1] and [1,∞). Thisis not a problem, because improper integrals can be pieced together just like regulardefinite integrals.

Theorem 15.3.10Let f be a continuous function on[a,∞) and let a< b.

Then∫∞

af (x) dx converges if and only if

∫∞

bf (x) dx converges. Moreover,

if the integrals exist, we have∫∞

af (x) dx =

∫ b

af (x) dx+

∫∞

bf (x) dx.

Proof. By Theorem 12.2.11 we have∫ t

af (x) dx =

∫ b

af (x) dx+

∫ t

bf (x) dx

for any t . The limit of the left side exists if and only if the limit of the right side

does. This proves the first part. The equation∫∞

af (x) dx =

∫ b

af (x) dx+

∫∞

bf (x) dx

now follows by taking limits on both sides.


−∞

e−x22 dx converges or diverges.

To show that∫∞

−∞

e−x22 dx converges, we need to show that

∫ 0

−∞

e−x22 dx and∫

∞

0e−

x22 dx converge. Example 15.3.9 shows that

∫∞

1e−

x22 dx converges. By

Theorem 15.3.10 we infer that∫∞

0e−

x22 dx converges. The inequality works for

x ≤ -1, but not for -1 < x < 0,so we need to use Theorem15.3.10.

To show that∫ 0

−∞

e−x22 dx converges, we use the comparison test once more.

Note that−x2≤ −|x| for x ≤ −1. This means that−

x2

2≤ −|x|

2for x ≤ −1 and

hencee−x22 ≤ e−

|x|2 for x ≤ −1. Since∫−1

−∞

e−|x|2 dx =

∫−1

−∞

ex2 dx = 2e

x2

∣∣∣−1

t→−∞= 2e−

12 ,

we have that∫−1

−∞

e−x22 dx converges by the comparison test and

∫ 0

−∞

e−x22 dx

converges by Theorem 15.3.10.

To be able to tackle a wider variety of integrals, we need more functions tocompare to integrands. The most natural family to use are power functions.

Figure 15.32: Illustration of Theorem 15.3.12.

The behavior of the functions1

x p changes at

p = 1. For p > 1 the functions decay fast

enough to give a finite integral∫∞

1

1

x p dx. For

p ≤ 1 they do not decay fast enough.

Theorem 15.3.12The p-integral test for integrals over infinite intervals (also

cf. Figure 15.32.) The improper integral∫∞

1

1

xpdx converges for p> 1 and

it diverges for0≤ p ≤ 1.


Proof. The proof is a direct computation of the integrals. Forp 6= 1 we obtain∫∞

1

1

xpdx =

∫∞

1x−p dx =

1

−p+ 1x−p+1

∣∣∣∣x→∞1

=

{1

1−p; for p > 1,∞; for p < 1,

which means that the improper integral converges forp > 1 and it diverges for

p < 1. For p = 1 we obtain∫∞

1

1

xdx = ln |x|

∣∣∣∣x→∞1

= ∞.

To apply the comparison test in conjunction with Theorem 15.3.12, we willnormally focus on the largest powers in numerator and denominator.


4

1√

x − 1dx converges or diverges.

This function is close to1√

x= x−

12 , whose improper integral diverges at∞.

We should thus try to show that1

√x − 1

≥1√

x. This is true forx > 1.

1√

x − 1

?≥

1√

x√

x − 1?≤√

x

−1 ≤ 0√

Thus by comparison test the integral∫∞

4

1√

x − 1dx diverges.


2

2x + 1

x4− x2dx converges or diverges.

Focusing on the highest powers only, we see that we should compare to the

function2

x3. Since

∫∞

2

2

x3dx converges we should try to show that

2

x3is an upper

bound of2x + 1

x4− x2. Forx > 1 and therefore certainly forx > 2 we have

2x + 1

x4− x2≤

2x

x4− x2≤

2x

x4=

2

x3,

so by comparison test, the integral∫∞

2

2x + 1

x4− x2dx converges.

15.3.2 Numerical Estimates

If the actual value of an improper integral is not accessible, an approximation is the

next best thing. Iff (x) ≥ g(x) ≥ 0 for x ≥ a, then∫∞

af (x) dx ≥

∫∞

ag(x) dx.

This inequality is the key to computing approximate values of improper integrals.

Example 15.3.15For the improper integral∫∞

0e−

x22 dx find a point a such that∫

∞

ae−

x22 dx < 10−5. Then use a CAS to compute

∫∞

0e−

x22 dx to four accurate

digits behind the decimal point.


For a ≥ 1 we have that∫∞

ae−

x22 dx ≤

∫∞

ae−

x2 dx. The integral

∫∞

ae−

x2 dx

can be computed. Moreover, if∫∞

ae−

x2 dx < 10−5, then

∫∞

ae−

x22 dx < 10−5.∫

∞

ae−

x2 dx

!< 10−5

2e−a2 < 10−5

−a

2< ln

(10−5

2

)a > −2 ln

(10−5

2

)≈ 24.4121

We can conclude that∫∞

24.5e−

x22 dx < 10−5. With

∫ 24.5

0e−

x22 dx ≈ 1.25331

(obtained with a CAS) we see that∫∞

0e−

x22 dx ≈ 1.25331.

15.3.3 Absolute Convergence

The comparison test cannot be applied to functions that assume negative values.We could try to apply it to the absolute value. Theorem 15.3.17 shows that if theabsolute value of a function can be integrated over the interval [a,∞), then so canthe function itself.

Definition 15.3.16 The improper integral∫∞

af (x) dx is calledabsolutely

convergentif and only if∫∞

a| f (x)| dx converges.

Theorem 15.3.17If the improper integral∫∞

af (x) dx converges absolutely,

then it converges.

Proof. Since∫∞

a| f (x)| dx converges, by Exercise 11 the integral

∫∞

a2| f (x)| dx

converges. Hence∫∞

af (x)+ | f (x)| dx converges by comparison test (note that

0≤ f (x)+ | f (x)| ≤ 2| f (x)|). But then, again by Exercise 11, we have that∫∞

af (x) dx =

∫∞

af (x)+ | f (x)| dx−

∫∞

a| f (x)| dx

converges also.


1

sin(x)

x2dx converges.

If it converges, find the value up to an error of at most5× 10−5.

Because

∣∣∣∣sin(x)

x2

∣∣∣∣ ≤ 1

x2we know that

∫∞

1

sin(x)

x2dx converges absolutely by

comparison test. In particular, it converges. Moreover,

508 Module 15: Applications of Integration∣∣∣∣∫ ∞a

sin(x)

x2dx

∣∣∣∣ ≤ ∫ ∞a

∣∣∣∣sin(x)

x2

∣∣∣∣ dx ≤∫∞

a

1

x2dx =

1

a.

Therefore, to force

∣∣∣∣∫ ∞a

sin(x)

x2dx

∣∣∣∣ ≤ 5× 10−5, we solve1

a≤ 5× 10−5,

which leads to a ≥1

5× 10−5= 20,000. With a CAS we ob-

tain∫ 20,000

1

sin(x)

x2dx ≈ 0.50292, which leads to the estimate∫

∞

1

sin(x)

x2dx ≈ 0.50292.

Example 15.3.19Absolute convergence and convergence are not equiva-

lent. The function f (x) =

1; for

2n∑k=1

1

k≤ x ≤

2n+1∑k=1

1

k

−1; for2n+1∑k=1

1

k≤ x ≤

2n+2∑k=1

1

k

is such that

∫∞

0f (x) dx converges, but it does not converge absolutely.

Exercises

1. Compute the improper integral or show that it diverges.

(a)∫∞

0e−3x dx

(b)∫∞

0xe−x dx

(c)∫∞

0x2e−5x3

dx

(d)∫∞

4

1

x ln(x)dx

(e)∫ 0

−∞

9

1+ x2dx

(f)∫∞

0

x2

1+ x3dx

(g)∫ 0

−∞

ex dx

(h)∫ 0

−∞

x2ex dx

(i)∫∞

0e−2x cos(3x) dx

(j)∫∞

0sin(2x) dx

(k)∫∞

1

√1+ 1

x

x2dx

(l)∫∞

0

1√

1+ xdx

(m)∫∞

0x2e−3x dx

(n)∫∞

1

x

1+ x2dx

(o)∫∞

0

dx

1+ x2

(p)∫∞

0

x

ex

(q)∫∞

2

1

x (ln x)2dx

(r)∫∞

1

x dx

1+ x2

(s)∫∞

0

dx

1+ x2

(t)∫∞

0

dx

ex

2. Find the area under the curvey = xe−x for 1≤ x <∞.

3. Determine if the improper integral converges or diverges. If theintegral converges, find a pointa such that the integral over theinfinite interval starting ata is less than 10−4.

(a)∫∞

0e−x3

dx

(b)∫∞

0

3

x3+ 5dx

(c)∫∞

5

1

x13 − 1

dx

4. Determine if the improper integral converges or diverges. If theintegral converges, find its approximate value to an accuracy of10−2.

(a)∫∞

0e−x2

dx

(b)∫∞

−∞

1√

2πe−

x22 dx

(c)∫∞

−∞

cos(x)

1+ x2dx

(d)∫∞

0e−x sin

(x2)

dx


(e)∫∞

1

ln(x)

x3dx

(f)∫∞

0x2e−x2

dx

(g)∫∞

2

1

x − 1dx

5. The velocity function of the mass in a given spring-mass-systemis given bys(t) = e−2t cos(3t). Prove that the total distance themass travels after timet = 0 is finite.

6. Compute the escape velocity from the surface of the given ce-lestial body.

(a) The moon (Mass:≈ 7.35× 1022kg, Radius:≈ 1.74×

106m)

(b) Jupiter (Mass:≈ 1.90×1027kg, Radius:≈ 6.98×107m,note: the surface is not solid)

(c) Pluto (Mass:≈ 1.4× 1022kg, Radius:≈ 1.5× 106m)

(d) Europa (Mass:≈ 4.8×1022kg, Radius:≈ 1.57×106m)

(e) From the outer layers of the atmosphere (say, 50km abovethe surface of the earth).

7. For the following questions we only consider the need to over-come gravitational attraction. (No need to consider friction,etc.)

(a) How far away from Earth could a space probe get if itstarts on the surface of the Earth with an initial velocityof exactly 11km

s ?

(b) How far away from Earth could a space probe get if itstarts on the surface of the Earth with an initial velocityof exactly 12km

s ?

8. TheSchwartzschild radiusof a mass is the radius of a spher-ical region from which not even light could escape if the masswas compressed into this sphere. This effect actually occurs forblack holes, where the Schwartzschild radius is also called theevent horizon.

(a) If not even light can escape from the inside of the Schwartzschildradius, then the escape velocity exactly at the boundarywould be the speed of lightc. Use this idea and the for-mula for the escape velocity to derive a formula for theSchwartzschild radius.

(b) Compute the Schwartzschild radius for the following ob-jects.

(i) Our sun. (Mass≈ 1.99× 1030kg.)

ii. A star whose mass is 30 times that of the sun.

iii. A 70kg human being. (This will just verify thatpeople are not capable of turning into black holes.)

iv. The Milky Way galaxy. Assume an estimated massof 600 billion solar masses.

9. Compute the velocity to which the 1kg mass in Example 15.2.1has to be accelerated on the surface of the Earth to be shot di-rectly to the orbit of the international space station. Since weassume there is no friction, this is an estimate of the minimumvelocity that might work.

10. Unlike Theorem 19.1.7 for series, there is no limit test for im-proper integrals. Give an example of a functionf such that∫∞

0f (x) dx exists and lim

x→∞f (x) 6= 0.

Hint. Use a sequence of triangles of height 1, alternatinglyabove and below the axis and let the areas shrink rapidly enough.

Remark: Example 15.3.5 shows that if the functionf goes to 0at∞, the improper integral need not converge. So for improperintegrals there is no connection at all between the limit of thefunction at∞ and convergence of the improper integral.

11. Prove the following (compare with Theorem 12.2.11)

Let f and g be functions on [a,∞) and let c be a

real number. If the improper integrals∫∞

af (x) dx

and∫∞

ag(x) dx converge, then so do the integrals∫

∞

af (x)± g(x) dx and

∫∞

ac f (x) dx and we have

∫∞

af (x)± g(x) dx =

∫∞

af (x) dx±

∫∞

ag(x) dx∫

∞

ac f (x) dx = c

∫∞

af (x) dx.

12. Analyzing Theorem 15.3.8.

(a) State the converse of Theorem 15.3.8.

(b) Provide an example that shows that the converse is nottrue.

13. State Theorem 15.3.12 as a biconditional (an “if and only if”statement).

14. Analyzing Theorem 15.3.17.

(a) State the contrapositive of Theorem 15.3.17.

(b) State the converse of Theorem 15.3.17.

(c) Provide an example that shows that the converse is nottrue.

Hint. There is one in this section.


15.4 Improper Integrals II: Singularities

Another type of improper integral arises when investigating the behavior of func-tions near a singularity.

Example 15.4.1 If the electrostatic force was the only acting force, how much en-ergy would be needed to bring two separate electrons from a distance of10−10m(about the diameter of a hydrogen atom) into exactly the same point in space?

The repulsive electrostatic force between two electrons is given byF =q1q2

4πε0

1

r 2,

wherer is the distance between the electrons. Thus the energy required to changethe distance bydr is F · dr and the total amount of energy needed to change the

distance froma to b is∫ b

a

q1q2

4πε0

1

r 2dr . q1 andq2 are both the charge of an electron

andε0 = 8.8542· 10−12 As

V mis the permittivity constant. We start ata = 10−10m

and because we want the electrons to ultimately occupy the same point, we wantb = 0m. To obtain an integral with the lower bound smaller than the upper bound,we can reversea andb.

We encounter a formal problem. The Riemann integral does not work for un-bounded functions as can be seen in Exercise 10. After introducing improper inte-grals over singularities we will finish this example in Example 15.4.4.

To circumvent formal problems on the boundary of an integral we make onebound of the integral a parameter and let it go towards the singularity.

Definition 15.4.2 Let f be a continuous function on the interval[a,b). If∫ t

af (x) dx exists for every t in[a,b) and lim

t↑b

∫ t

af (x) dx exists, then we

define theimproper integral of f from a to b to be∫ b

af (x) dx = lim

t↑b

∫ t

af (x) dx.

In this case the improper integral is calledconvergent. If the limit does notexist, the improper integral is calleddivergent.Improper integrals with a singularity on the left boundary are defined analo-gously (cf. Exercise 5).

Similar to integrals over infinite intervals, improper integrals that involve sin-gularities are a combination of integration and limit computation.

Example 15.4.3Compute the improper integral∫ 2

0

1√

2− xdx.

With the singularity atx = 2 we obtain∫ 2

0

1√

2− xdx = lim

b↑2

∫ b

0(2− x)−

12 dx = lim

b↑2−2(2− x)

12

∣∣∣b0

= limb↑2−2(2− b)

12 + 2 · 2= 4.

15.4. Improper Integrals II: Singularities 511

Example 15.4.4 (Concluding Example 15.4.1.) Compute∫ 10−10

0

q1q2

4πε0

1

r 2dr.

With the definition of the improper integral we note that∫ 10−10

0

q1q2

4πε0

1

r 2dr = lim

a↓0

∫ 10−10

a

q1q2

4πε0

1

r 2dr =

q1q2

4πε0lima↓0

∫ 10−10

a

1

r 2dr

=q1q2

4πε0lima↓0

[−

1

r

]10−10

a

=q1q2

4πε0lima↓0

(1

a− 1010

)= ∞.

So it would take an infinite amount of energy to put two electrons in the samepoint in space.

Discussion 15.4.5Examples 15.4.1 and 15.4.4 show the limitations of the macro-scopic laws of electrodynamics. Consider the following.

1. Two electrons repel each other, but two opposite charges, say an electron ora proton attract each other. Thus,if the electromagnetic force and gravitywere the only fundamental forces of nature, electrons and protons would getarbitrarily close to each other and release and release an infinite amount ofenergy in the process.

2. In nuclear fusion, two previously separated positive charges, say, two hydro-gen nuclei, are brought very close together to form a new atomic nucleus, inthis case a helium nucleus. Because two positive charges repel each other,this process would actually cost energyif the electromagnetic force and grav-ity were the only fundamental forces of nature. Instead, fusion of hydrogeninto helium actually releases energy.

The above two observations arenot designed to find fault with the theory ofelectromagnetism or with our observations about how charges behave at the atomicand subatomic level. They show however that there must be other fundamentalforces at work on these small scales. These forces are the strong and weak interac-tions which are discussed in detail in advanced physics classes.

Discussion 15.4.6On a more practical level, Examples 15.4.1 and 15.4.4 showwhy large amounts of energy are needed to start the fusion of, say, hydrogen atoms,in particle colliders. The strong and weak interaction dominate the electromagneticforces only over very short distances. This means to get particles close enoughtogether, significant amounts of energy are needed.

Convergence tests and numerical estimates of improper integrals with a sin-gularity are similar to those for improper integrals over infinite intervals. Thep-integral test is described in Exercise 8. The proof is similar to that of Theorem15.3.12. (Careful here. The roles ofp > 1 and p < 1 are reversed.) The com-parison test is described in Exercise 9. The proof is similar to that of Theorem15.3.10.


Exercises

1. Compute the given improper integral or show that it diverges.

(a)∫ 2

1

1

x − 1dx

(b)∫ 1

0

1

1− ex dx

Hint. Substituteu = 1− ex.

(c)∫ 3

2

1√

x − 2dx

(d)∫ 1

−1

13√x

dx

(e)∫ 1

0

1

1− e−x dx

(f)∫ 2

0ln |x| dx

2. Determine if the improper integral converges or diverges. If theintegral converges, find a pointc such that the integral over theinterval starting atc is less than 10−4.

Hint. Use the results of Exercises 7 and 8.

(a)∫ 1

0

ex√

xdx

(b)∫ 2

1

1

ln |x|dx

3. Determine if the improper integral converges or diverges. If theintegral converges, find its approximate value to an accuracy of10−2.

Hint. Use the results of Exercises 7 and 8.

(a)∫ 1

0

√ln(x) dx

(b)∫ 1

0

ex

x2dx

(c)∫ 2

0(ln(x))2 dx

4. Consider the functionf (x) = Cx ln1

xon the interval [0,1] (C

is a positive constant).

(a) Find the limit of f asx tends to 0, and asx tends to 1.

(b) Is f (x) positive or negative on [0,1]?

(c) Find the value ofC for which∫ 1

0f (x) dx = 1.

5. Formulate the definition for improper integrals∫ b

af (x) dx with

a singularity ata instead of atb.

6. Improper integrals with a singularityc inside the interval [a,b]

are defined as the sum∫ c

af (x) dx+

∫ b

cf (x) dx. Use this

idea to compute and explain the following.

(a) Explain why∫ 1

−1x−

13 dx exists and is zero.

(b) Explain why∫ 1

−1

1

xdx does not exist.

7. Prove the following theorem.

(The comparison test for improper integrals of functionswith a singularity.) Let f and g be continuous functionson [a,b) with f (x) ≥ g(x) ≥ 0 for x ≥ a.

(a) If∫ b

af (x)dx is convergent, then

∫ b

ag(x)dx is

convergent.

(b) Equivalently, if∫∞

ag(x)dx is divergent, then∫

∞

af (x)dx is divergent.

8. The p-integral test for improper integrals with a singularity.

(a) Prove the following theorem.

(The p-integral test for integrals with a singularity.)

i. For p ≥ 1, the improper integral∫ 1

0

1

x p dx di-

verges.

ii. For p < 1, the improper integral∫ 1

0

1

x p dx con-

verges.

(b) Explain why the integral of1√

xfrom 0 to 1 is finite

rather than infinite.

(c) Explain, in comparison, why the integral of1

x2from 0 to

1 is infinite rather than finite.


Let f be a continuous function on [a,b) and leta < c <

b. Then∫ b

af (x) dx converges if and only if

∫ b

cf (x) dx

converges. Moreover, if the integrals exist, we have∫ b

af (x) dx =

∫ c

af (x) dx+

∫ b

cf (x) dx.

10. For unbounded functions the definition of the definite integral(cf. Definition 12.2.1) leads normally to the conclusion that thefunction is not integrable. As an example, consider an attempt

to define a Riemann integral for the functionf (x) =1√

xon

the interval(0,1]. For this function we know that the improper

integral∫ 1

0

1√

xdx = 2.

(a) Show that if we choosex∗1 =1

n3in any Riemann sum

with n rectangles, then the Riemann sums go to infinityasn→∞.

(b) Show thatn∑

k=1

√k ≤

∫ n+1

1

√x dx=

2

3

((n+ 1)

32 − 1

).

(c) Write down the Riemann sum forf on(0,1] with n rect-angles and evaluation at the right endpoint.

(d) Show that the sum is bounded by 1 for alln.

(e) Explain why the above shows thatf is not Riemann in-tegrable on the interval(0,1], even if we extended thedefinition to half open intervals.

15.5. Centers of Mass of Linear Densities 513

15.5 Centers of Mass of Linear Densities

To start our investigation, consider two children playing on a teeter-totter (also cf.Figure 15.33 for a schematic). The heavier child must be closer to the middle ofthe teeter-totter to achieve balance. This is because to achieve balance, the centerof mass must be exactly in the middle. If the center of mass is too far to one side,the teeter-totter will tip towards that side.

BB

BB

BB

��

��

��

r

r

r

� �� m1

m1

m1

m2

m2

m2

center of mass is supported

center of mass is supported

center of mass is not supported

m1 = m2: balance,

m1 < m2, adjusted positions: balance,

m1 < m2, same positions: imbalance,

j

j

��

��XXXXXXXXXXXXXX

XXXXXXXXXXXXXX

Figure 15.33: When positioning two pointmasses on a teeter-totter, balance is achievedwhen the center of mass (solid dot) is supported(top and bottom). If the center of mass is notsupported (center), the teeter-totter drops on theside of the center of mass.

For a description with fundamental physics, let us assume that the beam is inperfect balance. Under this assumption we can neglect the beam’s mass and we canassume that two point massesm1 andm2 are placed on the beam at positionsx1 and

x2, respectively. The center of mass of this system is atxC =m1x1+m2x2

m1+m2. By

taking this weighted average, we take into account that the center of mass shouldbe closer to the larger mass. Ifm2 > m1, then its positionx2 is multiplied with alarger factor, and the computation produces anxC that is closer tox2 than tox1. Theformula for the center of mass can be experimentally verified with scales, leversand also with teeter-totters. For more than two point masses lined up along the

x-axis, a similar formula applies:xC =

∑nk=1 mkxk∑n

k=1 mk. Each positionx j contributes

proportional to the ratiom j∑nk=1 mk

of the mass located atx j to the overall mass.

length dxposition x

mass ρ(x)dxFigure 15.34: A continuously distributed densitycan be turned into point masses by consideringsmall segments of lengthdx.

Macroscopic masses are modeled as continuously distributed with a certaindensityρ. For masses distributed along thex-axis, we therefore have a linear den-sity functionρ(x), that gives the density in kilograms per meter at every point (alsocf. Figure 15.34). If the density is constant, the overall mass is mass times length.For a density that varies withx, we compute the mass as follows.

The quantity ρ(x)dx=dm is adifferential mass. If we sumup the differential massesthat are distributed from ato b we obtain the overallmass. The integral plays therole of this \continuous sum."

Theorem 15.5.1Let a,b be real numbers and letρ(x) be a linear density

function. Then the total mass m is m=∫ b

adm=

∫ b

aρ(x) dx.

The formula for the mass is obtained by assuming (cf. Figure 15.34) that thetotal mass is made up of short segments of lengthdx. Each such differential seg-ment has massdm = ρ(x) dx and it behaves like a point mass. Summing, thatis, integrating the differential masses gives the total mass. To obtain the center of

mass, each segment contributes a summandx dm

m=

xρ(x) dx

m. Summing these

contributions leads to a formula analogous to that from the point mass model.

Theorem 15.5.2Let a,b be real numbers and letρ(x) be a linear densityfunction of a linearly distributed mass m. Then the x-coordinate of the center

of mass is xC =1

m

∫ b

ax dm=

1

m

∫ b

axρ(x) dx.

The quantity xρ(x)dx=xdm isanalogous to a quantity mkxkin the point mass model.\Continuous sums" are onceagain set up as integrals.

With the above results we can compute the center of mass of any continuouslinear mass density.

Example 15.5.3Compute the center of mass of the linear mass density given byρ(x) = e−x for 0≤ x ≤ 1.

To compute the mass we integrate the density over the interval.

m =

∫ 1

0e−x dx = −e−x

∣∣10 = −e−1

− (−1) = 1−1

e.


Once we have the mass, we can compute thex-coordinate of the center of mass.

xC =1

1− 1e

∫ 1

0xe−x dx =

e

e− 1

[−xe−x

− e−x]1

0

=e

e− 1

[−e−1

− e−1− (−1)

]=

e

e− 1

e− 2

e=

e− 2

e− 1≈ 0.4180.

Solve the integral withintegration by parts.

Figure 15.35 shows another way to visualize the center of mass.rAA��

0 1

e− 2

e− 1

Figure 15.35: We can also visualize the centerof mass of a linear density as the point where thegraph of the density balances on a scale. The fig-ure shows the center of mass in Example 15.5.3.

15.5.1 Using Symmetry to Work With Actual Objects

Any kind of mass distribution extends not just linearly, but also in the two remain-ing directions of space. For objects with uniform densityρ that are symmetricabout an axis, the ideas for linear mass densities can be used to compute the threedimensional center of mass.

To do this, we align the axis of symmetry with the verticalz-axis. Since theobject is symmetric with respect to thez-axis and its density does not fluctuate,the center of mass must lie on thez-axis. That leaves only thez-coordinate tobe determined. For thez-coordinate we can then treat the object like a linearlydistributed mass with linear densityρ(z) = ρA(z). Here A(z) is the area of ahorizontal cross section at heightz. The next example shows the process.

Example 15.5.4Find the center of mass of a hemisphere of radius a with homo-geneous densityρ.

-

6

x

y

z

Figure 15.36: A hemisphere, aligned so that thez-axis is the axis of symmetry.

Explain the equation for themass of a slice of thicknessdz.

The volume of a hemisphere of radiusa is V =1

2

4

3πa3, so its mass is

m= ρV = ρ1

2

4

3πa3.

If we assume that the axis of symmetry is thez-axis (cf. Figure 15.36), then thecenter of mass is on thez-axis and we only need to find thez-coordinate. The massof a slice of thicknessdzat heightz is

dm= ρ dV = ρA(z) dz= ρπ(√

a2− z2)2

dz.

This gives for thez-coordinate of the center of mass

zc =1

ρ 12

43πa3

∫ a

0zρπ

(√a2− z2

)2dz

=3

ρ2πa3ρπ

∫ a

0a2z− z3 dz=

3

2a3

[1

2a2z2−

1

4z4

]a

0

=3

2a3

[1

2a4−

1

4a4

]=

3

2a3

1

4a4=

3

8a.

Thus the center of mass is at

(0,0,

3

8a

).


Although linearly distributed masses are purely conceptual entities, wehave seen that the approach can be used to compute the center of massfor axisymmetric three dimensional objects. This in itself is already goodpreparation for modeling realistic three dimensional mass densities in Sec-tion IMV.A . Finally, it should be mentioned that in statistics, probability isoften likened to mass. With this analogy probability density functions (cf.Section 17.5.1) act like linear mass densities.

Exercises

1. For the following linear density functions compute the mass andthe center of mass.

(a) ρ(x) = x2, 0≤ x ≤ 1

(b) ρ(x) = sin(x), 0≤ x ≤ π

(c) ρ(x) =1

1+ x2, 0≤ x ≤ 5

(d) ρ(x) = ex, 1≤ x ≤ 2

(e) ρ(x) =1

x2, 1≤ x ≤ 3

2. A metal rod of length 1 has densityρ(x) = x ln(1/x), wherexis the distance from, say, the left end-point. Assume that the rodis lying between 0 and 1.

(a) Find limx↓0 ρ(x) and limx→1 ρ(x) (Use L’Hopital whenneeded.)

(b) Find the critical point ofρ(x) and graphy = ρ(x) for0≤ x ≤ 1

(c) Compute the center of mass of this rod

3. Compute the coordinates of the center of mass for each of thefollowing objects. Assume that thez-axis is the axis of symme-try.

(a) A right circular cylinder of radiusa and heighth.

(b) A right circular cone of heighth and base radiusa.

(c) A square pyramid of heighth whose base has side lengtha.

(d) A cap of heighth of a sphere of radiusa.

4. Explain the following theorem.

Consider an object of constant densityρ. Assume that thehorizontal cross section of the object at heightz is given byA(z) for a ≤ z≤ b. Then the mass of the object is

m=∫ b

aρA(z) dz

and thez-coordinate of the center of mass is

zC =1

m

∫ b

az A(z) dz.

5. Find the center of mass for each of the sets of point massesbelow.

(a) m1 = 3kg atx1 = 2, m2 = 5kg atx2 = 6

(b) m1 = 1kg atx1 = 0, m2 = 4kg atx2 = 2, m3 = 1kg atx3 = 3

(c) m1 = 2kg atx1 = −2, m2 = 1kg atx2 = 1, m3 = 4kgat x3 = 5

(d) m1 = 1kg atx1 = 1, m2 = 5kg atx2 = 3, m3 = 2kg atx3 = 5, m4 = 3kg atx4 = 7


15.6 Projects

15.6.1 Arc Length

The arc length of a function is most naturally discussed in the context of parametriccurves (cf. SectionVVF.B). The formula for functionsy = f (x) can be derivedearlier and it is a good example of the approach of considering small increments toeliminate variations that only occur on a larger scale.

1. Prove the following theorem. You may use Figure 15.37 as guidance.-x

6y

uu

x x +1x

f

1x

1y ≈ f ′(x)1x

Figure 15.37: Deriving the arc length formulafor functionsy = f (x). Hint: Use Pythagoras’law on the marked triangle and then form a Rie-mann sum.

Theorem 15.6.1Let f be a differentiable function on[a,b]. Then the length

of the curve given by the graph of f is L=∫ b

a

√1+ ( f ′(x))2 dx.

2. Compute the length of the following functions over the given intervals. Usea CAS for integrals that cannot be computed by hand.

(a) f (x) =1

2x2 on [−2,2]

(b) f (x) =ex+ e−x

2on [−1,1]

(c) f (x) =e2x+ e−2x

2on [−1,1]

(d) f (x) = x3 on [−2,2]

(e) fr (x) =√

r 2− x2 on [−r, r ]

3. The functions in parts 2b and 2c only differ insignificantly. Explain why theintegral of the function in part 2b can be solved and why the integral of thefunction in part 2c cannot be solved.

4. Analyze the formula in Theorem 15.6.1. Do you expect there to be manyfunctions for which the arc length integral can be solved or do you think thatmost arc lengths will need to be computed numerically?

5. What geometric quantity is computed in part 2e?

6. The computation of the surface area of a volume of rotation is similar to thecomputation of the arc length.

Prove the following theorem.

Theorem 15.6.2The surface area of the surface of revolution obtained byrotating the function f between a and b about the x-axis is

S=∫ b

a2π f (x)

√1+ ( f ′(x))2 dy.

Hint. Remember that the surface area of the walls of a cylinder is the perime-ter times the height and take into account that the walls of any slices we takeare curved.

15.6. Projects 517

7. Use the result in part 6 to compute the surface area of the given object.

(a) A sphere of radiusr .

(b) The cone obtained by rotating the functionf (x) = 2x on [0,2] aboutthe y-axis.

(c) The circular cone of heighth with base radiusa.

15.6.2 The0 Function

The0 function is often considered to be a continuous version of the factorial func-tion n!. It arises in numerous contexts. For example, in the distribution in 15.6.3,the0 function is used in the renormalizing constant. In this project we explorethe definition of the0 function and prove its possibly most important property:0(α + 1) = α0(α).

Definition 15.6.3 The function0(α) :=∫∞

0xα−1e−x dx defined forα > 0 is

called theGamma function.

1. Prove that the improper integral∫∞

0xα−1e−x dx converges forα ≥ 1.

2. Prove that the improper integral also converges for 0< α < 1.

Parts 1 and 2 establish that the0 function is well defined.

3. Prove that the improper integral diverges forα ≤ 0.

4. Prove that0(1) = 1.


Theorem 15.6.4For all α ≥ 1 we have0(α + 1) = α0(α).

6. Prove by induction that for every natural numbern we have0(n) = (n−1)!.For this reason the Gamma function is also referred to as thegeneralizedfactorial function .

Similar to the normal distribution, the Gamma function has been studied exten-sively and tabulated.

15.6.3 Student’st-Distribution

Student’st-distribution is a statistical distribution that arises in the estimation ofthe mean of random samples from a normal distribution (cf. Theorem 18.2.8).Formally (cf. Definition 15.6.5), the Studentt-distribution is a parameter dependentfamily of functions, just like the normal distribution. In this project we prove someof the Studentt-distribution’s salient features and we also sketch the graph.


Definition 15.6.5 Let k be a positive integer. The function

fk(x) :=0(

k+12

)0(

k2

) 1√

kπ

1(1+ x2

k

) k+12

is the density of aStudent’s t-distribution a with k degrees of freedom.

aThe t-distribution was discovered by W. Gossett in the early 1900s. His employer (abrewery) had regulations regarding trade secrets, which only allowed him to publish under apseudonym (Student).

1. Show that for allk we have∫∞

−∞

fk(x) dx = 1.

This shows that the total area under the function is 1, as is necessary for aprobability density function.

2. Show that for allk > 1 we have∫∞

−∞

x fk(x) dx = 0 and show that the im-

proper integral diverges fork = 1.

This shows that the mean of the distribution is 0 when it exists.

3. Show that for allk > 2 we have∫∞

0x2 fn(x) dx =

k

k− 2and show that the

improper integral diverges fork = 1,2.

This shows that the variance of the distribution isk

k− 2when it exists.

4. Sketch the graph of the parameter dependent family of functions as indicatedin Section 11.2.2. In particular,

(a) Show that for allk, the function fk has a local maximum atx = 0.

(b) Show that for allk the inflection points offk arex1,2 = ±

√k

k+ 2.

(c) Compute asymptotes, intercepts, etc. as appropriate.

(d) Sketch the graph of the functionsfk. (You may use Example 11.2.5 forguidance.)

Applications of Integration494 15. Applications of Integration Example 15.1.5 Derive the formula for the volume of a cap of height h of a sphere of radius a, cf. Figure 15.10. Axis

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