27 1 Intro Srfc Ints

8/13/2019 27 1 Intro Srfc Ints

1/19

Contentsontents Multiple Integration27.1 Introduction to Surface Integrals 2

27.2 Multiple Integrals over Non-rectangular Regions 20

27.3 Volume Integrals 41

27.4 Changing Coordinates 66

Learning

In this Workbook you will learn to integrate a function of two variables over various

rectangular and non-rectangular areas. You will learn how to do this for various other

coordinate systems. You will learn to integrate a function of three variables over a volume.

outcomes
http://www.ebookcraze.blogspot.com/


2/19

Introduction to

Surface Integrals

27.1Introduction

Often in Engineering it is necessary to find the sum of a quantity over an area or surface. This can beachieved by means of a surface integral also known as a double integral i.e. a function is integratedtwice, once with respect to one variable and subsequently with respect to another variable. ThisSection looks at the concept of the double integral and how to evaluate a double integral over arectangular area.

Prerequisites

Before starting this Section you should. . .

thoroughly understand the various techniquesof integration

be familiar with the concept of a function oftwo variables

Learning OutcomesOn completion you should be able to. . . understand the concept of a surface integral

integrate a function over a rectangular region

2 HELM (2005):Workbook 27: Multiple Integration


3/19

1. An example of a surface integralAn engineer involved with the construction of a dam to hold back the water in a reservoir needs to beable to calculate the total force the water exerts on the dam so that the dam is built with sufficientstrength.

In order to calculate this force, two results are required:

(a) The pressurepof the water is proportional to the depth. That is

p=kd (1)

wherek is a constant.

(b) The force on an area subjected to constant pressure is given by

force= pressure area (2)

The diagram shows the face of the dam. The depth of water ishandA is a small area in the faceof the dam with coordinates (x, y).

A

y

x

h y

y

x

h

Figure 1

Using (1), the pressure atA k(h y). Using (2), the force on an areaA k(h y)A.Both of these expressions are approximate asy is slightly different at the top ofA to the bottom.Now

Total force on dam = sum of forces on all areas A making up the face of the dam

all A

k(h y)A

For a better approximation letAbecome smaller, and for the exact result find the limit asA 0.Then

Total force on the dam = limA0

k(h y)A

=

A

k(h y)dA

where A

k(hy)dAstands for thesurface integralofk(hy)over the areaA. Surface integrals

are evaluated usingdouble integrals. The following Section shows a double integral being developedin the case of the volume under a surface.

HELM (2005):Section 27.1: Introduction to Surface Integrals

3


4/19

2. Single and double integrals

As has been seen in 14.3, the area under the curvey =f(x) betweenx =a andx =b is

given by ba f(x)dx(assuming that the curve lies above the axis for allx in the rangea x b).

This is illustrated by the figure below.

a b

y

x

area required

y=f(x)

Figure 2

In a similar manner, the volume under a surface (given by a function of two variables) and abovethexy plane can be found by integrating the function twice, once with respect tox and once withrespect toy.

x

y

f(x, y)

x=a x=b

y =c

y=d

Figure 3

The above figure shows the part of a surface given byf(x, y) which lies above the rectanglea x b,c y d. This rectangle is shaded and the volume above this rectangle but below thesurface can be seen.



5/19

x

y

f(x, y)

x=a x=b

y =c

y=d

x

Figure 4

Imagine a vertical slice taken through this volume at right angles to the x-axis (figure above). Thisslice has thicknessx and lies at positionx. Assuming thatx is small enough that the areas ofboth sides (left and right) of this slice are virtually the same, the area of each face of the slice isgiven by the integral y=d

y=c

f(x, y)dy (wherexmeasures the position of the slice)

and the volume of the slice will be given by

x

y=dy=c

f(x, y)dy

To find the total volume between the surface and thexyplane, this quantity should be summed overall possible such slices, each for a different value ofx. Thus

V i

y=dy=c

f(xi, y)dy x

Whenxbecomes infinitesmally small, it can be considered to be dxand the summation will change

into an integral. Hence

V =

x=bx=a

y=dy=c

f(x, y) dydx

Thus the volume is given by integrating the function twice, once with respect tox and once withrespect toy.

The procedure shown here considers the volume above a rectangular area and below the surface.The volume beneath the surface over a non-rectangular area can also be found by integrating twice(see Section 27.2).


5


6/19

Key Point 1

Volume Integral

The volume under the surfacez=f(x, y) and above a rectangular region in thexy plane (that isthe rectanglea x b, c y d) is given by the integral:

V =

bx=a

dy=c

f(x, y) dydx

3. Inner and Outer integrals

A typical double integral may be expressed as

I=

x=bx=a

y=dy=c

f(x, y)dy

dx

where the part in the centre i.e.

y=dy=c

f(x, y)dy

(known as the inner integral) is the integral of a function ofx andy with respect toy. As theintegration takes place with respect toy , the variablex may be regarded as a fixed quantity (aconstant) but for every different value ofx, the inner integral will take a different value. Thus, the

inner integral will be a function ofxe.g.g(x) =

y=dy=c

f(x, y)dy.

This inner integral, being a function ofx, once evaluated, can take its place within the outer integral

i.e. I =

x=bx=a

g(x) dx which can then be integrated with respect tox to give the value of the

double integral.The limits on the outer integral will be constants; the limits on the inner integral may be constants(in which case the integration takes place over a rectangular area) or may be functions of the variableused for the outer integral (in this casex). In this latter case, the integration takes place over anon-rectangular area (see Section 27.2). In the Examples quoted in this Section or in the early partsof the next Section, the limits include the name of the relevant variable; this can be omitted oncemore familiarity has been gained with the concept. It will be assumed that the limits on the innerintegral apply to the variable used to integrate the inner integral and the limits on the outer integralapply to the variable used to integrate this outer integral.



7/19

4. Integration over rectangular areas

Consider the double integral

I= 5x=0

1y=1 (2x+y) dydx

This represents an integral over the rectangle shown below.

x

y

1

1

5

Figure 5

Here, the inner integral is

g(x) =

1

1

(2x+y)dy

and the outer integral is

I= 5

x=0

g(x)dx

Looking in more detail at the inner integral

g(x) =

1

1

(2x+y)dy

the function(2x+y)can be integrated with respect toy(keepingxconstant) to give2xy+ 12y2+C

(whereC is a constant and can be omitted as the integral is a definite integral) i.e.

g(x) =

2xy+

1

2y21

1

=

2x+

1

2

2x+

1

2

= 2x+

1

2+ 2x

1

2= 4x.

This is a function ofxas expected. This inner integral can be placed into the outer integral to get

I=

5

x=0

4x dx

which becomes

I=

2x2

5

0

= 2 52 2 02 = 2 25 0 = 50

Hence the double integral

I= 5x=0

1y=1 (2x+y) dydx= 50


7


8/19

Key Point 2

Double Integral

When evaluating a double integral, evaluate the inner integral first and substitute the result into theouter integral.

Example 1

Evaluate the double integral I=

2

x=1

3

y=2

x2y dydx

This integral is evaluated over the area shown below.

x

y

2

1

3

2

Figure 6

Solution

Here, the inner integral is

g(x) =

3

y=2

x2y dy =

x2

y2

2

32

=9

2x2

4

2x2 =

5

2x2

and hence the outer integral is

I= 2

x=

1

5

2

x2 dx= 5

2

1

3

x32

1

=5

6

85

6

(1) =15

2



9/19

Example 2Use the above approach to evaluate the double integral

I= 5x=0

1y=1 x

2

cos

y

2 dydx

Note that the limits are the same as in a previous case but that the function itselfhas changed.

Solution

The inner integral is

1

y=1

x2 cosy

2

dy= 2

x2 siny

21

1

= 2

x212

x2 (1) = 4

x2

so the outer integral becomes

I=

5

x=0

4

x2 dx=

4

3x350

= 4

3125

4

30 =

500

3 53.1

Clearly, variables other thanxandy may be used.

Example 3Evaluate the double integral

I=

4

s=1

t=0

s sin t dtds

Solution

This integral becomes (dispensing with the step of formally writing the inner integral),

I =

4

s=1

s cos t

0

ds=

4

1

[s cos +s cos0] ds=

4

1

[s (1) +s (1)] ds

=

4

1

2s ds=

s2

41

= 16 1 = 15


9


10/19

Clearly, evaluating the integrals can involve further tools of integration, e.g. integration by parts orby substitution.


I=

2

1

3

2

xyex

y2 + 1 dydx

Here, the limits have not formally been linked with a variable name but the limits on the outer integralapply toxand the limits on the inner integral apply toy. As the integrations are more complicated,the inner integral will be evaluated explicitly.

Solution

Inner integral =

3

2

xyex

y2 + 1 dy

which can be evaluated by means of the substitutionU=y2 + 1.

If U=y2 + 1 then dU= 2y dy so y dy=1

2dU .

Also if y= 2 then U= 5 and if y= 3 then U= 10.So the inner integral becomes (remembering thatxmay be treated as a constant)

105

12

xex

U dU=xe

x

2

105

dUU

=xex

2

ln U

105

=xex

2 (ln10 ln 5) =xex ln 2

2

and so the double integral becomes

I=

2

1

xexln 2

2 dx=

ln 2

2

2

1

xex dx

which can be evaluated by integration by parts.

I = ln 2

2

xe

x

2

1

2

1

1 ex

dx

= ln 2

2 2e2 + (1) e1 +

2

1

ex dx=

ln 2

2

2e2 e1 +

ex

2

1

= ln 2

2

2e2 e1 e2 +e1

=

ln 2

2

3e2

0.14



11/19

TaskaskEvaluate the following double integral.

I=

1

1 2

0 x2y+ 3y2

dydx

Your solution

Answer

The inner integral =

2

0

x2y+ 3y2

dy =

1

2x2y2 +y3

2

0

=1

2 4x2 + 8 (0 + 0) = 2x2 + 8

This can be put in the outer integral to give

I=

1

1

2x2 + 8

dx=

2

3x3 + 8x

11

=2

3+ 8 (

2

3 8) =

4

3+ 16 =

52

3

Exercises

Evaluate the following double integrals over rectangular areas.

1.I=

1

x=0

2

y=0

xy dydx

2.I=

3

2

4

0

x2 +y2

dxdy

3.I=

0 1

1

ysin2 x dydx

4.I=

2

0

3

1

st3 dsdt

5.I=

3

0

1

0

5z2ww2 1

4

dwdz (Requires integration by substitution.)

6.I=

2

0

1

0

tysin t dydt (Requires integration by parts.)

Answers

1. 1, 2. 460/3, 3. 0, 4. 16, 5. 9/2, 6.


11


12/19

5. Special cases

If the integrand can be written as

f(x, y) =g(x) h (y)

then the double integral

I=

ba

dc

g(x) h (y) dydx

can be written as

I=

ba

g(x)dx

dc

h (y)dy

i.e. the product of the two individual integrals. For example, the integral

I= 2x=1

3y=2

x2

y dydx

which was evaluated earlier can be written as

I=

2

x=1

x2 dx

3

y=2

y dy=

x3

3

21

y2

2

32

=

8

3

(1)

3

9

2

4

2

= 3 5

2=

15

2

the same result as before.

Key Point 3

Double Integral as a Product

The integral

b

a

d

c

g(x)h(y) dydx can be written as b

a

g(x)dx d

c

h(y)dy

Imagine the integral

I=

1

1

1

0

xey2

dydx

Approached directly, this would involve evaluating the integral 1

0

xey2

dy which cannot be done

by algebraic means (i.e. it can only be determined numerically).



13/19

However, the integral can be re-written as

I=

1

1

x dx

1

0

ey2

dy =

1

2x21

1

1

0

ey2

dy= 0

1

0

ey2

dy= 0

and the result can be found without the need to evaluate the difficult integral.

If the integrand is independent of one of the variables and is simply a function of the other variable,then only one integration need be carried out.

The integralI1 =

ba

dc

h (y) dydx may be written asI1 = (b a)

dc

h (y)dy and the integral

I2=

ba

dc

g(x) dydxmay be written asI2= (d c)

ba

g(x)dxi.e. the integral in the variable

upon which the integrand depend s multiplied by the length of the range of integration for the othervariable.


I=

2

0

2

1

y2 dydx

Solution

As the integral inycan be multiplied by the range of integration inx, the double integral will equal

I= (2 0)

2

1

y2 dy= 2

y3

3

21

= 2

23

3

(1)3

3

= 6

Note that the two integrations can be carried out in either order as long as the limits are associatedwith the correct variable. For example

I= 1x=0

2y=1 x

4

y dydx= 1x=0

x4y22

21 dx=

1x=0

2x

4

1

2x

4dx

=

1

0

3

2x4 dx=

3

10x510

= 3

10 1

3

10 0 =

3

10

and

I=

2

y=1

1

x=0

x4y dxdy=

2

y=1

x5y

5

10

dy =

2

1

y5 0

dy

= 21

y

5dy =y2

1021

= 4

10 1

10= 3

10


13


14/19

TaskaskEvaluate the following integral:

I=

1

0 1

1

z(w+ 1) dwdz.

Your solution

Answer1

Exercises

1. Evaluate the following integrals:

(a)I=

/20

1

0

(ycos x) dydx

(b)I=

3

8

1

1

y2 dydx

(c)I=

1

0

5

0

(s+ 1)4 dtds

2. Evaluate the integrals

3

1

2

0

x3y dydxand

2

0

3

1

x3y dxdyand show that they are equal.

As explained in the text, the order in which these integrations are carried out does not matterfor integrations over rectangular areas.

Answers

1. (a) 1/2, (b) 22/3, (c) 31

2. 40



15/19

6. Applications of surface integration over rectangular areas

Force on a dam

At the beginning of this Section, the total force on a dam was given by the surface integralA

k(h y)dA

Imagine that the dam is rectangular in profile with a width of 100 m and a height h of 40 m. Theexpression dA is replaced by dxdy and the limits on the variablesx andy are 0 to 100 m and 0to 40 m respectively. The constantk may be assumed to be 104 kg m2 s2. The surface integralbecomes the double integral

40

0

100

0

k(h y) dxdy that is

40

0

100

0

104(40 y) dxdy

As the integral in this double integral does not containx, the integral may be written 40

0

100

0

104(40 y) dxdy = (100 0)

40

0

104(40 y)dy

= 100 104

40y y2

2

400

= 106[(40 40 402/2) 0]

= 106 800 = 8 108 N

that is the total force is 800 meganewtons.

Centre of pressure

We wish to find the centre of pressure (xp, yp) of a plane area immersed vertically in a fluid. Takethex axis to be in the surface of the fluid and the y axis to be vertically down, so that the planeOxy contains the area.

A

x

y

x

y

surfaceO

Figure 7

We require the following results:

(a) The pressurepis proportional to the depthh, so thatp=hwhere is a constant.(b) The forceFon an areaA subjected to constant pressurepis given byF =pA


15


16/19

Consider a small element of areaA at the position shown. The pressure atA isy. Then the

force acting onA isyA. Hence the total force acting on the areaAis

A

y dA=

A

y dA.

Moment of force onA aboutOy = xyA

Total moment of force onA aboutOy = A

xy dA

Moment of force onA aboutOx = y2A

Total moment of force onA aboutOx =

A

y2 dA

Taking moments aboutOy:

total force xp = total moment

A

y dA xp = A

xy dA

xp

A

y dA =

A

xy dA

Taking moments aboutOx:

total force yp = total moment

A

y dA

yp =

A

y2 dA

yp A

y dA = A

y2 dA

Hence

xp=

A

xy dAA

y dA

andyp=

A

y2 dAA

y dA

.



17/19

Example 6A rectangle of sidesaandb is immersed vertically in a fluid with one of its edgesin the surface as shown in Figure 8. Where is the centre of pressure?

x

a

y

surfaceO

b

Figure 8

Solution

To express the surface integral as double integrals we will use cartesian coordinates and verticalslices. We need the following integrals.

Ay dA =

b

0 a

0

y dydx=

b

o

1

2y2a0

dx=

1

2a2 dx=

1

2a2x

b0

=1

2a2b

A

xy dA =

b

0

a

0

xy dydx=

b

o

1

2xy2

a

0

dx=

1

2xa2 dx=

1

4x2a2

b

0

=1

4a2b2

A

y2 dA =

b0

a0

y2 dydx=

bo

1

3y3a0

dx=

1

3a3 dx=

1

3a3x

b0

=1

3a3b

Hence

xp =

A

y2 dA

A

y dA=

1

3a3b

1

2a2b

=2

3a and yp =

A

xy dA

A

y dA=

1

4a2b2

1

2a2b

=1

2b

The centre of pressure is (1

2b,

2

3a), so is at a depth of

2

3a.


17


18/19

Areas and moments

The surface integral

A

f(x, y) dAcan represent a number of physical quantities, depending on the

functionf(x, y) that is used.

Properties:

(a) Iff(x, y) = 1 then the integral represents the area ofA.

(b) Iff(x, y) =xthen the integral represents the first moment ofAabout they axis.

(c) Iff(x, y) =ythen the integral represents the first moment ofAabout thexaxis.

(d) Iff(x, y) =x2 then the integral represents the second moment ofAabout they axis.

(e) Iff(x, y) =y2 then the integral represents the second moment ofAabout thexaxis.

(f) Iff(x, y) =x2 +y2 then the integral represents the second moment ofA about thezaxis.

Example 7Given a rectangular lamina of length, widthb, thicknesst (small) and density(see Figure 9), find the second moment of area of this lamina (moment of inertia)about they-axis.

x

y

O

b

density:

thickness:t (small)

Figure 9

Solution

By property (e) above, the moment of inertia is given by b0

0

y2t dxdy = t( 0)

b0

y2 dy

= t

y3

3

b0

= tb3

3

As the mass of the lamina isM=bt, the moment of inertia simplifies to 13Mb2. Thetandare

included in the integral to make it a moment of inertia rather than simply a second moment.



19/19

TaskaskBy a similar method to that in Example 7, find the moment of inertia of the samelamina about thex-axis.

Your solution

AnswerFrom property (d) above, the moment of inertia (or second moment of area) is given by the integral

l0

b0

x2t dydx = t(b 0)

l0

x2 dx

= bt

x3

3

l0

= bt

l3

3

As the mass of the lamina isM=lbt, the moment of inertia simplifies to 13Ml2. Again, thetand

are included in the integral to make it a moment of inertia rather than simple a second moment.

Exercises

By making use of the form of the integrand, evaluate the following double integrals:

1.I= 0

10

ycos2 x dydx

2.I=

3

8

1

1

y2 dydx

3.I=

1

0

5

0

(s+ 1)4 dtds

Answers 1. 4, 2. 22

3 3. 31

HELM (2005): 19

Date post:	04-Jun-2018
Category:	Documents
Upload:	ebookcraze
View:	229 times
Download:	0 times

27 1 Intro Srfc Ints

Documents