Multiple Integrals

Multiple Integrals Double Integral Let

D

C

IR2

liD II

be a bounded measurable set. Let

diameter of the set

D, i.e.,

sup

IIDII

be the

{J(XI

-

X2)2

+ (Yl

- Y2)2}.

(Xl,Yl)ED (X2,Y2)ED

=

Let ~

i,j

{Di

I

n

i = 1, ...

= 1, ... ,n,

, n}

be a division of

D

such that:

i #j. IDM 2009 - p. 1123

Let

II~II

:= 2=1, . max IIDill, f : D ... ,n

---+ JR,

the sum

and choose

Ni

Di,

i = 1, ...

, n. Consider

n

SD.

If the limit

DEFINITION

lim

=

L i=l

f(Ni)m(Di). ~ 1M. ie

SD. exists, and has the same value for any division

IID..II-+O

- {D1,

E

...

,

Dn} of D, and for any points N'i E

double integral of

f

Di,

i = 1, ...

, n, then it is called

on D, and is denoted by

Hence,

n

f

).

n

If

f = 1, we obtain

SD..

=

L i=l

m(Di)

= A(D),

therefore

the area

A(D)

of the set

D is

given by

IDM 2009 - p. 2123

Similarly, if P : D the domain

If

r

--t

JR

is the density of a mass distributed on

D,

= xI + y} ,

then the position vector

r

G of the centroid

(centre of mass) is given by:

IDM 2009 - p. 3/23

Double Integrals over a Rectangle

Let D

= [a,

THEOREM

function

b] x [c, d] C JR2 be a rectangle and f: D

Ifthe integral : [a,b] lR.,

fif(X,

( x)

--t

JR.

y) dxdy exists and the

= jdc f (x,

y) dy, is

integrable, then

IDM 2009 - p. 4/23

Proof: We prove the theorem in the particular case when the function

f

is continuous on

{[a

=

D.

Consider the divisions:

Xo, Xl], [Xl, X2], ...

and the points ~i E [Xi-I,

, [Xn-1, Xn

i = 1, ...

Xi],

=

b]},

, n. By using the

mean value theorem, we obtain:

n

L

f(~i,

rlj) (Yj - Yj-1),

j=l

where 1]j

E [Yj, Yj-1].

10M 2009 - p. 5/23

We deduce: n

Li=l

n I(~i)(Xi-Xi-1)

hence, for

=

n

Li=l L

f(~i,

1]j)(Xi-Xi-1)(Yj-Yj-1),

j=l

~~x

1<'l,J
{Xi -

Xi-I,

Yj - Yj-1}

lb I(x) dx = f!vf(x,

-+

0, we obtain

y) dxdy.

10M 2009 - p. 6/23

The integral

REMARK

IS

(x,y)

denoted also by

Therefore, if

f

is continuous, we have:

fLf(x,

= lbdx ldf(x,

y) dxdy

=

ld dy lbf(x,

y) dy

y) dx. IDM 2009 - p. 7/23

Double Integrals over Arbitrary Domains Let g, h : [a,

b]

-+

[c, d] be continuous. Consider the set

D

= {(x,

y)

I

a

<

x

<

b,

g(x)

<

y

<

h(x)}.

y

d

g

c

a

b IDM 2009 - p. 8/23

D

If f:

THEOREM

is continuous, then

---t JR.

f

D

Proof: The set function f*

: [a,

b]

f

is closed and bounded, hence it is measurable.

x

Define the auxiliary

[c, d] ---t JR.,

(x, y) y) E ([a, b] x [c, dJ) \ D. (x, E D, Since f* is discontinuous at most on the graph of g and h (set of of null measure), it is

f*(x,

integrable on

D.

={

y)

f(x, 0, y),

We have:

rr JJD

y) dxdy

f(x,

= JJ[a,b] rr

= Jarb dx Jcrdf*(x,y)dy=

x [c,d] f*(x,

Jarb

y) dxdy

dx Jg(x) rh(X)f(x,y)dy.

IDM 2009 - p. 9/23

Similarly, if g, h : [c, d]

D

=

{(x, y)

I

c
---t JR.

< d,

are continuous, and

g(y)

< x < h(y)},

y

d :£ -

hey)

c

x

a

b

we have

h~

f(x,

D

y) dxdy

=

ldc

dy

lh(Y) g(y)

f(x,

y) dx. IDM 2009 - p. 10/23

EXAMPLE

d:1xly, where

rr

Calculate lID (:1~2

parabolas y

x2,

is the domain bounded by the

y2.

::r _

1

D(x2 !J~

y)

dxdy

= 111ft o dx

o (X5/2 + :::. 2 _ x4 = 11

x2

2 _ x4)

(x2

dx

=

+ y)

dy

7 (~X7/2

=

11[ 0

+ x24

+ Y22]Y=ft y=x2

x2y x

10 _ ~X5)

0 11

dx

140 = ~. 10M 2009 - p. 11/23

RegionPlot [y > X2 && y2 < X, {X, 0,1}, {y, 0,1}, BoundaryStyle ---+ Dashed, Plot8tyle ---+ Yellow, AspectRatio

---+

Automatic]

1.0 -,

•

I

T

I~I 1

0.8

0.6

::1 f 00

l·,1 __

0.0

I

0.4

11 11 Boole 1 3

.~~

__,_,

0.6

[y

0.8

1.0

> x2 && y2 < xJ

dx dy 10M 2009 - p. 12/23

Green's Formula Suppose that the functions Consider the set

gl, g2 : [a,

M defined M

y

b]

-+

hI, h2

JR.,

: [c,

d]

-+

JR.

are continuous.

by:

a <x<

{(x,y)

I

{(x,y)

Ie

b, gl(X)

< y < d,

h1(y)

< Y < g2(X)} < X
D

d

A

C c

B a

We have:

-

= gl (t) { x=t

Denote the path

by fr

=

g2

(a

+b -

, t

E [a,b].

t)

M. IDM 2009 - p. 13/23

lVh, ... , lVln are bounded by piecewice smooth paths such that: n

U

JIv! -

1, ... , n,

y

y

ABCDA

If the sets

i,.i -

x=a+b-t

ABC:

i },

Afi n 1\1j

1\1,i,

·i=1 and

c fr( JlvIi) n fr( 1\Ij

w is a differential form over

fr

/ w=ti=l 1\!I

fr

f

.IvI.[

1.\1,

),

then

w

A

- -

B

0

----

~ ACDA ABC frM1UM2 frM2 JiG CA .•..

/

/

D

w+ / w.

IDM 2009 - p. 14/23

Suppose that THEOREM

M c ]R2

is a finite union of sets bounded by piecewise smooth paths.

(Green-Riemann)

P,

If the functions

possess continuous partial derivatives

q : 1\1 -+

are continuous and

~~ and ~~;, then

M

Proof: It is sufficient to prove the theorem when

is bounded by a piecewice smooth path.

We have:

dxdy !J~ fJP M (-8) y

= /

= - 1b dx 192(X) a 91 (x)

.f

P(x, g,(X»dx

Pdx -

/

Pdx

ADO

7fBC

8 fJP dy = 1b P(x, Y

a

-1b P(x,g,(x» =

/

Pdx

ABCJ5A

y)

Y=91(X) Y=92(X)

dx

dx

= / Pdx. fr

M

IDM 2009 - p. 15/23

Similarly, we obtain

/L~~

dxdy=

/

Qdy,

frM

hence

/L (~~- ~:) dxdy= x If in Green's Formula we take

q=

2 P

/

frM

Pdx+Qdy.

Y

= -"2'

we get

/L dxdy = ~frM/ xdy

- ydx,

IDM 2009 - p. 16/23

Calculate the area of the domain !vI bounded by the ellipse

r:

We have:

x2

y2

-2 a/ +

b')"-' ==

x

r:

y

arie(M)

== -

11277" 2 0

== ==

1.

a cos t, b sin

t

E [0,21rJ,

t,

= ~frMJ

+ xdy)

(-ydx

(a cos t(b sin t)'-b sin t( a cos t)') dt ==

== -

11277" 2 0

ab dt

1rab.

IDM 2009 - p. 17/23

Let

D

C

Change of Variables in Double Integrals JR.2a compact domain and T : ~ -+ D a differentiable bijection T: { Yx =y(u,v) = x(u, v)

,

(u, v) E ~.

T v

y

u

x THEOREM

If

f :D

-+

lR

0 is

continuous, then

) IDM 2009 - p. 18/23

where

D - {(x, e

integral, ,fa Fie ~

=

e

Calculate the integral

EXAMPLE

{(p,

B)

E ]R2 I x2

y)

y2

< R2}.

(method 3). E ]R2 I 0 < p < R, 0

T:~-tD,

B

<

211"}

~i fie

= peas B, = p sin e.

x

T:

<

Find the Gauss

y

D(x, y)

We get:

= p,

D(p, e)

I = fLe-P2pdpde

= 121rde

lR

e-p2pdp

= Jr(1_e-R2). IDM 2009 - p. 19/23

For

R

-t

00, we obtain

f~2e-X2_y2 dxdy

= Jr,

1.8.,

= 100 -00 100 -00 e

_x2_y2

dxdy

= Jr,

hence

e

IDM 2009 - p. 20/23

EXAM PLE

Find the area of the domain

D bounded by the

parabolas:

where

o

x2

ay

x2

by


b,

and

Y Y

0

2

px

2

qx

q.

IDM 2009 - p. 21/23

y' =qx

y'

= px

IDM 2009 - p. 22/23

We look for a change of variable of the form

{ Let D.

x2

y2

= {(u, v) Ip < u < q,

We have

(x, y) ED{::=:::}

= vy = ux a

u E [p, q],

,

v E [a,

b].

< v < b}.

(u, v) E D.. ,

(u, v) E D..

u2/3 v1/3 {

We deduce

~ ar';a(D)

-- ffD dxdy

Xy

u1/3

v2/3

D(x, y)

1

D(u,v)

3

--dudv -- 11.6. DD((xu" yv))

=

-dudv = .6.3 31 /11

arieD.

=

1

-(b-a)(q-p)

3

IDM 2009 - p. 23/23

.