Chapter 6Chapter 6

Unit 6Unit 6Unit 6Unit 6

定积分的物理应用定积分的物理应用定积分的物理应用定积分的物理应用

New WordsWork 功 Pressure 压力

The universal gravitational constant

万有引力常数

Horizontal component 水平分力

Well-proportioned 均匀的

Perpendicular bisector 中垂线

Orthogonal triangle 直角三角形

The definite integral has wide applications in

mathematics, the physical science and engineering.

Here, we will introduce some simple applications in

the physical science.

At first, we will use the definite integral to compute

the work done by a varying force that moves an

object along a straight line.

Then we use the definite integral to compute the

pressure of water and the force between two objects.

To calculate a quantities by setting up a definite integral, the general procedure may be outlined as follows:

Step 1: Chop up the desired quantity into very small parts.

Step 2: Within each small part, calculate an approximation to the desired quantity.

Step 3: Add up the results of all of the small parts approximation from step 2

Step 4: Obtain a definite integral by taking the limit of

the sum in step 3 as the part get smaller and smaller.

1. Work1. Work

In this subsection we compute the work done by a

varying force that moves an object along a straight

line.

sFW

W

s

F

or

Distance Forcek Wor

simply is edaccomplish work the

then force, theofdirection in the distance aover

operatesit and alueconstant v thehas force theIf

But, if we are faced with a problem in which the force

is variable, we cannot use the above formula. Instead,

the definite integral is required.

Example 1Example 1

A rocket has mass 2000kg. Find the work done in

launching the rockets from the earth’s surface to an

altitude of 100 km. (Use 6400km as the radius of the

earth and ignore the weight of the fuel burned during

the flight)

SolutionSolution

According to the Newton’s law of gravitation. The

force F(x) needed to overcome the force of gravity

is 2x

GMmxF

body a of mass theis earth, theofthecenter

from distance theis and constant, nalgravitatio

universal theis )kg

Nkm10674.6( earth,

theof mass theis kg)10993.5( where

2

217

24

m

x

G

M

6500,6400d,,d

elyapproximat

is d to from liftingin done work The

xxxxxF

xxx

J

xxGMmxxfW

9

2417

6500

6400

26500

6400

10923.1

6500

1

6400

1200010993.510674.6

dd

is work total theHence

Example 2Example 2

A spring is stretched 0.5 meter longer than its rest len

gth. The force required to keep it at that length is 3 ne

wtons. Find the total work accomplished in stretching

the spring 0.5 meter from its rest length

SolutionSolution

Since the force is proportional to x, it is of the form kx

for some constant k. We know that F=3 when x=0.5, s

o6,5.03, kkkxF

To estimate the work involved in stretching the spring

from x to x+dx

The distance dx is small. As the end of the spring is str

etched from x to x+dx, the force is almost constant.

The work accomplished in stretching the spring from x

to x+dx is then approximately

xxxkxW d6dd

The element of work

Hence, the total work is

joule 75.065.0

0 xdxW

Example 3Example 3

SolutionSolution

A cylinder with base radius 3 meters and height 5

meters is full of work. How much work is required to

pump out all water in the cylinder?

See figure

variableof integral the

as ]5,0[ Choosing x

x

ox

dxx

5

ely approximat is

d,on cylinder small

of water of mass The

xxx

.d2.88d

ely approximat is d to

fromout water pumpingin done work The

xxw

xx

x

xd38.9 2

The element of work

Hence, the total work is

Jx

xxw 34622

2.88d2.885

0

25

0

Example 4Example 4

Solution:Solution:

?

,,

20N/s, 3m/s,

N,2000 50N, N,400

m,30 figure See

, ,

耳的功问克服重力需作多少焦

升至井口现将抓起污泥的抓斗提从抓斗缝隙中漏掉

的速率污泥以在提升过程中提升速度为

抓斗抓起的污泥重缆绳每米重

抓斗自重已知井深污泥后提出井口

抓起用缆绳将抓斗放入井底为清除井底的污泥

See figure

x

x

x+dx

0

30

321 WWWW

口需作的功

井抓起污泥的抓斗提升至

由题意知作的功

为提升污泥所所作的功

为克服缆绳重力作的功

为克服抓斗自重所其中

.

;

;

3

2

1

W

W

W

12000304001 W

xxW d3050d 2

22500d305030

02 xxW

作的功为

克服缆绳重力所处处提升到将抓斗由 , d xxx

提升污泥所作的功为在时间间隔 d, ttt

ttW d2020003d 3

s103

30时间为将污泥提升至井口共需

57000d202000310

03 ttW

JW 91500570002250012000

is work total theThus,

2. The pressure force of water2. The pressure force of water

Physics tells us the intensity of pressure of water

with depth h is p=rh, where r is the weight of one

cubic unit of water.

If a board with area A is placed horizontally in

water at a depth h, then the force on this board is

F=pA. But, if the board is submerged vertically in

water, we can not use the above formula. Instead,

the definite integral is required.

Example 5Example 5

Solution:Solution:

A cylindrical water tank has radius R (See figure), and it is full of water by half. Find the force against one end of the tank, where r is the weight of one cubic unit water.

See figure

x

ox

dxx

.d2 22 xxR

],0[],[

for and variable,of integral

theas ],0[ Choosing

Rdxxx

Rx

The area of small strip is

xxRxP d2d

ely approximat is strip small for the force typicalThe

22

xxRxPR

d2

is force total theThus,

22

0

.3

2

3

2

)(d

3

0

322

22

0

22

RxR

xRxR

R

R

Example 6Example 6

A orthogonal triangle board with side a and 2a is

SolutionSolution

The typical area of the small piece is

,d)(2 xxa

submerged vertically in water (see figure). Find

the force against the board

xxaaxP d1)(2)2(d

ely approximat is piece

small for the force typicalThe

x

oa2

a2a

The element of force

.3

7d))(2(2 3

0axxaaxP

a

3. Gravitation3. Gravitation

constant nalgravitatio universla theis

, mass and mass of distance theis where

is mass of

another and mass of particle onebetween

attraction nalgravitatio the:us tellsPhysics

21

221

2

1

k

mmrr

mmkF

m

m

If we want to calculate the gravitational attraction

between a mass and a piece of stick,

the definite integral is required, because the

distance between them is not constant.

Example 7Example 7

particle theandstick ebetween th

attraction nalgravitatio theFind bisector.

lar perpendicu itsin stick thefrom distance

thepoint with at the located is masswith

particle a is thereand , is length stick with

edproportion- wellof piece a ofdensity The

a

mM

l

SolutionSolution See figure

2l

2l

x

y

oMa

ry

dyy

]2

,2

[]d,[ interval smallany takeand

variable,of integral theas ]2

,2

[ Choosing

llyyy

lly

y

dyyy

d masswith

particle theas ],[ piece small theThinking

,d

is them

between attraction nalgravitatio typicalThe

,

is particle

theand piece small ebetween th distance The

22

22

ya

ymkF

yar

,)(

dd

iscomponent horizontal typical theAnd

23

22 ya

yamkFx

,)4(

2

)(

d

iscomponent horizontal the

21

23

2

2 2222 laa

lkm

ya

yamkF

l

lx

.0

iscomponent

vertical thesymmetry, the toAccording

yF

This section presented various physical applications o

f the definite integral. We showed that work, pressure

force and gravitational attraction can be represented a

s definite integrals.