Lecture 2 Calculus of Multi Variables 2011

8/3/2019 Lecture 2 Calculus of Multi Variables 2011

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Calculus Of Multivariables

1

Double Integral

1. Objectives

To compute the volume of a solid bounded by a surfacez =f(x,y) and a region in thex-y

plane, we can integrate in one direction to find the cross-sectional area of thin slices of thesolid, then integrate in the other direction to find the volume of the solid.

You should be able to compute the double integral of a function of two variables for various

bounded regions in thex-y plane.

2. The antiderivative of functions of 2 variables

Let ),( yxf be a function of two variables. The antiderivative of ),( yxf with respect tox is

denoted by dxyxf ),( . Similarly, the antiderivative of ),( yxf with respect toy is denoted

by dyyxf ),( .

[ ] ),(),( yxfdxyxfx

=

and [ ] ),(),( yxfdyyxfy =

Example 1 Antiderivative with respect to x

Find the antiderivative of22

332 xyxy ++ with respect tox.

Solution

+++=++ )(3)332(32222 yCxxyyxdxxyxy

Example 2 Antiderivative with respect to y

Find the antiderivative of22

yx + with respect to y.

Solution

++=+ )(31

)(3222

xCyyxdyyx

3. The definite integral of functions of 2 variables


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2

How to find the definite integral b

a

dyyxf ),( of a function of 2 variables?

In

b

adyyxf ),( , the limits of integration refer to limits fory.

b

a

dyyxf ),( = ),(),( axFbxF , where = dyyxfyxF ),(),( .

Example 3

Evaluate the definite integral +1

0

22)( dyyx .

Solution

1

0

1

0

3222

3

1)(

=

=

+=+

y

y

yyxdyyx

=

+

3

12x [0+0]

=x2 +3

1

4. Iterated Integrals

4.1 Successive integralsdxdyyxf

xx

xx

xyy

xyy =

=

=

=

2

1

2

1

)(

)(),( , obtained by two successive integrations, is called an

iterated integral.


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3

Example 4

Evaluate the iterated integral dxdyyx

+

2

1

1

0

22 )(

Solution

dxdyyx

+

2

1

1

0

22)( = +

2

1

2)

3

1( dxx

=

2

1

3

33

+

xx

=

+

+

3

1

3

1

3

2

3

8

=

3

8

4.2 Interchanging the order of integration

Example 5

Evaluate the iterated integral dydxyx

+

1

0

2

1

22 )(

Solution

dydxyx

+

1

0

2

1

22 )( = +1

0

2 )3

7( dyy

=

1

0

3

3

7

3

1

+y

=

+

+

3

0

3

0

3

7

3

1

=3

8

Important result:

dxdyyx

+

2

1

1

0

22 )( and dydxyx

+

1

0

2

1

22 )( have the same value!!


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4

4.3 Variable Limits of Integration

Example 6

Evaluate the iterated integral

1

0

2dxxydy

x

x

Solution

Explanation of the steps

1

0 2dxxydy

x

x= dxxy

xy

xy

=

=

1

0

2

22

1(find the antiderivative with respect to y)

= dxxx

1

0

53

22(substitute the limits of integration)

= 1

0

53)(

2

1dxxx (find the antiderivative with respect tox)

=24

1

5. Concept of double integrals

1.1 Definite integral for functions of a single variable

xsfdxxfn

i

i

b

a

n=

=)(lim)(

1

provided that the limit exists.

a bx

i1s

ix

i

x

y=f(x)

f(si)

y

x=xi-xi-1


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5

5.2 Definite integral for functions of two variables

yxtsfdAyxf j

n

i

i

m

jm

Rn

= = =

),(limlim),(1 1

provided that the limit exists.

Note: We have chosen a rectangular area dA = xy.

Let ),( yxfz = be the height above thex-y plane at a point ),( yx in a closed regionR.

Can you give a meaning to the value of the double integral R

dAyxf ),( ?

R

xy

si

ti

RdA

z = f(x,y)


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If ),( yxfz = is equal to the height at ),( yx , then the value of the double integral

R

dAyxf ),( represents the solid volume over the region R bounded above and below by the

surfaces ),( yxfz = and 0=z (thex-y plane) respectively.

5.3 Some properties of double integrals

1. =R R

dAyxfcdAyxcf ),(),( , where c is a constant.

2. +=+R R R

dAyxgdAyxfdAyxgyxf ),(),()],(),([

3.

+=

21 1 2

),(),(),(

RR R R

dAyxfdAyxfdAyxf if the areasR1 andR2 do not overlap.

2. Use of inequalities to describe a regionHow to describe the points in a regionR enclosed by the curves )(1 xgy = and )(2 xgy = ?

)(2 xgy =

)(1 xgy =

R

a b

c

d


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The points (x, y) inR can be described by a set of inequalities:

I. g1(x) y g2(x)a x b

or

II. g21

(y)x g1

1(y)

c y d

Example 8

Use a set of inequalities to describe the region R bounded by the line y = x and the curvey

= x3.

y = x

y = x3

Solution

The liney = x and the curvey = x3

meet at (0,0) and (1,1).

The points (x, y) in R can be described by either of the following sets of inequalities:

I. x3 y x0 x 1

or

II. y

x 31

y 0 y 1

R

(1,1)


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7. Use of an iterated integral to evaluate a double integral

How to evaluate the double integralR

dAyxf ),(

?

R

dAyxf ),(

=

b

a

xg

gdxdyyxf

x

)(2

1 )(

),(

=

d

c

yg

ygdydxyxf

)(

)(

11

12

),(

Example 9:

By integrating with respect to y first and x second, evaluate the double integral R

xydA ,

whereR is the region bounded by the curves y = x andy = x3.

Solution

R is the region bounded below and above by g1(x) = x3 and g2(x) = x, and on the left and right

byx = 0 andx = 1.

g2(x) = x

g1(x) = x3

(1,1)


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R

xydA =

ba

xg

gdxdyyxf

x

)(2

1 )(

),(

=

1

0

3 dxxydyx

x

= dxxy x

x3

1

0

2

2

= dxxx

1

0

73

2

1

2

1

=16

1

8. Finding volumes with double integrals

Double integrals can be used to find the volumes of solid regions in 3-dimensional space.

dAyxfV

R

= ),(

where Vis the volume of a solid region bounded below by the region R in thex-y plane and

above by a portion of the surface ).,( yxfz =

Example 10

Find the volume of a solid region bounded below by the regionR = {(x,y): 0


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Solution

( )

3

2

3

1

)(

1

0

2

1

0

1

0

22

22

=

+=

+=

+=

dyy

dydxyx

dAyxV

R

9. Other Applications of Double Integrals

9.1 AreaThe area of a closed regionR in thex-y plane is given byA =

R

dxdy .

9.2 MassA thin sheet of material of uniform thickness covers a region R in the x-y plane. Suppose the

sheet has varying density ),( yx (in kg/m2) at each point (x,y) in the regionR.

The total massMof the sheet is given by =R

dxdyyxM ),( .

9.3 Mean value

The mean value off(x,y) over a closed regionR is defined as R

dxdyyxfA

),(1

,

whereA is the area of the regionR.

R


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Example 11:

Find the average density of the triangular sheet which has uniform thickness and varying

density ),( yx =y in the region bounded by the lines 3=y , 1+=xy and 1+= xy .

(The average density is defined asareaTotalmassTotal )

Answer:

Area = R

dxdy

4

)2()2(

0

2

2

0

0

2

2

0

3

1

3

1

=

++=

+=

++

dxxdxx

dydxdydx

xx

3=y

1+= xy1+= xy

2 2


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12

=R

dxdyyxMass ),(

3

28

832

18

32

1

]82[21]82[

21

])1(9[2

1])1(9[

2

1

22

2

0

23

0

2

23

0

2

2

0

22

0

2

2

0

22

0

2

2

0

3

1

23

1

2

0

2

2

0

3

1

3

1

=

++

++=

++++=

+++=

+

=

+=

++

++

xxx

xxx

dxxxdxxx

dxxdxx

dxy

dxy

ydydxydydx

xx

xx

The average density =areaTotalmassTotal

37

4328

=

=

10.0 Double Integral in Polar Coordinate


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=2

z

r= r2

=j+1

=j

=1

r= r2

r= rj+1

r= rj

1

i j

r

r {r= r1

A

If the integrating region is given in polar coordinate (r, ), the whole region will bepartitioned into some small area ij (as shown in the shaded part in the figure). The area ofthis small partition ij is rr and thus the volume subscribed by the surface z=f(x, y) andthe small partition ij is dV=f(x, y)rr. Therefore, the total volume subscribed by thesurface z=f(x, y) and the region

ij, or the double integral, is given by :

rdrdrrfdxdyyxf )sin,cos(),( = Or, if the integrating function f is already given in polar coordinate, then the volume is given

by

= rdrdrfV ),(

Example: Note the following integral transformation:

( ) ( )21 1 / 2 1

3/ 2 3/ 22 2 2

0 0 0 0

x

x y dydx r rdrd

+ =

Three aspects of the integral had to be transformed:

1. The integrand was transformed using the transformations cosx r = and siny r = .2. The integration region was transformed from a quarter-circle in thex-y plane to a

rectangle in the r- plane:


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3. The differential area element was transformed from dxdy to rdrd by geometricallyanalyzing small area elements in thex-y plane. Partitions of thex-y plane had the

shape of annular sectors, while corresponding partitions of the r-plane are

rectangular:

Example :(i) Let { }R x y x y= + ( , ) | 2 2 4 be a region bounded by a circle. Find the double integral

x y dxdyR

2 2+ .

Solution

By using the polar coordinatesx

y

=

=

2

2

cos

sin

, we have

( )x y dxdy r rdr d r

d

R

2 2

0

2

0

23

0

2

0

2

3

16

3+ =

=

=

11.0 Change of Variables in Double Integrals: Jacobian


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Theorem: If a region S in the u-v plane is mapped onto the regionR in thex-y plane by the

one-to-one transformation Tdefined by ( ),x g u v= and ( ),y h u v= , where g and h have

continuous first derivatives on S, and iffis continuous onR and the Jacobian( )( )

,

,

x y

u v

is

nonzero on S, then

( ) ( ) ( )( )( )( )

,, , , ,

,R S

x yf x y dA f g u v h u v dudv

u v

=

.


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Practices:

1.

Describe the region of integration and evaluate.i)

ii).

.


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iii).

iv).

v).

2. Integrate over the triangular region with vertices (0, 0), (1, 1), (1, 2).

3. Find the volume of the following regions in space.i) The region beneath z = x

2+ y

2and above the square with vertices (1, 1), (1, 1), (1,

1), (1, 1)

ii) The tetrahedron cut from the first octant by the plane .

iii). The first octant section cut from the region inside the cylinder x2 + z

2 = 1 by the

planes y = 0, z = 0, x = y.

4. Find the volume enclosed by a sphere 2 2 2 2 x y z a


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5. Find the area enclosed by the cardioid r=1+sin . The cardioids is shown in the figure.The region is described by the inequalities 0 1 sin and 0 2r .

6. Find the volume of the solid bounded by the xy-plane, the cylinder x 2+y2=4 and theparaboloid z=2(x

2+y

2).

-2

-10

221

4


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Triple integral

1. ObjectiveA triple integral is an integral taken over a volume of space.You should be able to compute

triple integrals.

2. Triple integralThe triple integral is defined with a three variable function f(x, y, z), which is called the

integrating function and an integrating region R, which is in the three dimensional space. It

should be noted that the function f(x, y, z) cannot be plotted out over a three dimensional

domain. In order to define the triple integral, the integrating region R, for example say a

parallelepiped as shown in the figure are equally partitioned into small cubics Rijk, which is

located at (xi, yj, zk) with lengths ofx, y and z in the x, y and z directions respectively.

The volume of the small partition Rijk= =xyz and the triple integral of the function f(x, y,z) over the integrating region R is defined as :

==kji

ijkkji

RR

dVzyxfdVzyxfdxdydzzyxf

,,

),,(),,(),,(

2.1 Calculation of triple integral :Order of Integration is Interchangeable

If a functionf(x,y) is integrable on the box Q = {(x,y,z) | axb, cyd, rzs }, then

we can write the triple integral offoverQ as:

R

Partition Rijkat position of

(xi,yj,zk) having vol. of

dVijk=xyz

x

y

z


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( ) ( )

( )

( )

( )

( )

( )

, , , ,

, ,

, ,

, ,

, ,

, ,

s d b

Q r c a

d s b

c r a

b s d

a r c

s b d

r a c

d b s

c a r

b d s

a c r

f x y z dV f x y z dxdydz

f x y z dxdzdy

f x y z dydzdx

f x y z dydxdz

f x y z dzdxdy

f x y z dzdydx

=

=

=

=

=

=

.

N.B. The order of integration is not a matter.

Examples :

Let [ ] [ ] [ ]E= 0 1 0 1 0 1, , , , find the following triple integrals

(a) x y zdxdydzE

(b) ( )x y z x y z dx dy dzE

sin 2 2 2+ +

Solution

(a)

x y z dx dy dz xdx yzdydzy

dy z dz

zdz

E

=

=

= =

0

1

0

1

0

1

0

1

0

1

0

1

2

4

1

8


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( ) ( )

( )

( ) ( )[ ]

( sin sin

cos

cos cos

b)

x y z x y z dxdydz x x y z dx yzdydz

x y z yzdydz

y z y z ydy zdz

E

2 2 2 2 2 2

0

1

0

1

0

1

2 2 2

0

1

0

1

0

1

2 2 2 2

0

1

0

1

1

2

12

1

+ + = + +

= + +

= + + +

( ) ( )[ ]

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]

( )

= + + +

= + +

= + + + +

= +

1

41

1

42 1 2

1

82 1 2

1

83 3 2 3 1 1

2 2 2 2

0

1

0

1

2 2 2

0

1

2 2 2

0

1

sin sin

sin sin sin

cos cos cos

cos cos cos

y z y z zdz

z z z zdz

z z z

2.2 Triple integral with a more general integrating region

IfQ has the form Q = {(x,y,z) | (x,y) R (a bounded region in thexy plane) and g1(x,y) zg2(x,y)}, then

( ) ( )( )

( )2

1

,

,

, , , ,

g x y

Q R g x y

f x y z dV f x y z dzdA=

Example: Find the volume bounded by the planes 2x+y+z=2, x=0, y=0 and z=0.

Solution :We need to findR

V dxdydz= bounded by the above 4 planes


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1 2 2 2 2 1 2 2 12

0 0 0 0 0 0(2 2 ) (2 4 2) 2 / 3

x x y x

R

V dxdydz dx dy dz dx x y dy x x dx

= = = = + =

Practices:

1. Find the triple integral R yzdVxy cos where R={ (x, y, z) : 0x1, 0y1, 0z/2}.

2. With the region R={ (x,y,z): 0x1, x2yx, x-yzx+y}.

(a) Find R zdVyx23

2 .

(b) Find the volume of the region R.

Lecture 2 Calculus of Multi Variables 2011

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