Ejemplos integrales dobles

7/30/2019 Ejemplos integrales dobles

1/6

17.3 THE EVALUATION OF DOUBLE INTEGRALS BY REPEATED INTEGRALS I 881

evident, but actually it is quite difficult to prove. The result follows from what is known

as Fubinis theorem. J

Computations

Example 1 Evaluate

(x4 2y) dxdy with as in Figure 17.3.4. x

y

y=x2

y=

x2

x = 1 x = 1

x

Figure 17.3.4

SOLUTION By projectingonto thex-axis,we obtain theinterval [1, 1]. Theregion

consists of all points (x, y) with

1 x 1 and x2 y x2.

This is a region of Type I. By (17.3.1),

(x4 2y) dxdy =

11

x2x2

[x4 2y] dy

dx

= 1

1x4y y2

x2

x2dx

=

11

(x6 x4) (x6 x4)

dx

=

11

2x6 dx =

27x7

11=

47. J

Example 2 Evaluate

(xy y3) dxdy with as in Figure 17.3.5.

y = 1

y = 0

x = 1

x = y

x = 1

x

y

y

Figure 17.3.5

SOLUTION By projecting

onto the y-axis, we obtain the interval [0,

1]. The region consists of all points (x, y) with

0 y 1 and 1 x y.

After the Italian mathematician Guido Fubini (18791943).


2/6

882 I CHAPTER 17 DOUBLE AND TRIPLE INTEGRALS

This is a region of Type II. By (17.3.2),

(xy y3) dxdy =

10

y1

(xy y3) dx

dy

=

10

12x2y x y3

y1

dy

=

10

12y3 y4

12y + y3

dy

=

10 1

2y3 y4 1

2y

dy

=

1

8y4 1

5y5 1

4y2

10= 23

40.

We can also project onto the x-axis and express as a region of Type I, but then the

lower boundary is defined piecewise (see the figure) and the calculations are somewhat

more complicated: setting

(x) =

0, 1 x 0

x, 0 x 1,

we have as the set of all points (x, y) with

1 x 1 and (x) y 1;

y

x

y = 1

1 1

y = x

1

1

(0, 0)

(1, 1)

y = x2 (x = y1/2)

y = x1/4

(x = y4)

x

y

x

y

Figure 17.3.6

thus

(xy y3) dxdy =

11

1(x)

(xy y3) dy

dx

= 0

1

1

(x)(xy y

3

) dydx+

1

0

1

(x)(x y y

3

) dydx

=

01

10

(xy y3) dy

dx+

10

1x

(xy y3) dy

dx

as you can check =

1

2

+ 3

40

= 23

40.

Repeated integrals

ba

2(x)1(x)

f(x, y) dydx and

dc

2(y)1(y)

f(x, y) dxdy

can be written in more compact form by omitting the large parentheses. From now on

we will simply writeba

2(x)1(x)

f(x, y) dydx and

dc

2(y)1(y)

f(x, y) dxdy.

Example 3 Evaluate

(x1/2 y2) dxdy with as in Figure 17.3.6.

SOLUTION The projection of onto the x-axis is the closed interval [0, 1], and

can be characterized as the set of all (x, y) with

0 x 1 and x2 y x1/4.


3/6

17.3 THE EVALUATION OF DOUBLE INTEGRALS BY REPEATED INTEGRALS I 883

Therefore

(x1/2 y2) dxdy =1

0

x1/4x2

(x1/2 y2) dydx

=

10

x1/2y 1

3y3

x1/4x2

dx

=

10

23x3/4 x5/2 + 1

3x6

dx

=

8

21x7/4 2

7x7/2 + 1

21x7

10=

821 2

7+

121=

17.

We can also integrate in the other order. The projection of onto the y-axis is theclosed interval [0, 1], and can be characterized as the set of all (x, y) with

0 y 1 and y4 x y1/2.This gives the same result:

(x1/2 y2) dxdy =1

0

y1/2y4

(x1/2 y2) dxdy

=

1

0

23x3/2 y2x

y

1/2

y4dy

=

10

23y3/4 y5/2 + 1

3y6

dy

=

8

21y7/4 2

7y7/2 + 1

21y7

10=

821 2

7+

121=

17. J

x = y1/2

x = y4

y

x

y

Example 4 Use double integration to calculate the area of the region encolsedby

y = x2 and x + y = 2.

SOLUTION The region is pictured in Figure 17.3.7. Its area is given by the double

integral

dxdy.

We project onto the x-axis and write the boundaries as functions of x:

y = x2, y = 2 x. is the set of all (x, y) with 2 x 1 and x2 y 2 x . Therefore

dxdy =

12

2xx2

dydx =

12

(2 x x2) dx =

2x 12x2 1

3x3

12

= (2 12 13 ) (4 2+ 83 ) = 92 .We can also project onto the y-axis and write the boundaries as functions of y, but

then the calculations become more complicated. As illustrated in Figure 17.3.8, is

the set of all (x, y) with

0 y 4 and y x (y)


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884 I CHAPTER 17 DOUBLE AND TRIPLE INTEGRALS

12 1

1

4

y = 2 x

y = x2

x

y

(1, 1)

(2, 4)

Figure 17.3.7

12 1

4

x = 2 y

x = y

x

y

(1, 1)

(2, 4)

1

x = y

Figure 17.3.8

where

(y) =

y, 0 y 1

2 y, 1 y 4.Therefore

dxdy =1

0

yy

dxdy+4

1

2yy

dxdy.

As you can check, the sum of the integrals is 92

.

Symmetry in Double Integration

First wego back to the one-variable case. (Section 5.8) Lets suppose that g is continuous

on an interval which is symmetric about the origin, say [

a,a].

Ifg is odd, then

aa

g(x) dx = 0.

Ifg is even, then

aa

g(x) dx = 2

a0

g(x) dx.

We have similar results for double integrals.

Suppose that is symmetric about the y-axis.

If f is odd in x [ f(x, y) = f(x, y)], then

f(x, y) dxdy = 0.

If f is even in x [ f(x, y) = f(x, y)], then

f(x, y) dxdy = 2

right half

of

f(x, y) dxdy.

Suppose that is symmetric about the x-axis.

If f is odd in y [ f(x,y) = f(x, y)], then

f(x, y) dxdy = 0.

If f is even in y [ f(x,y) = f(x, y)], then

f(x, y) dxdy = 2

upper half

of

f(x, y) dxdy.


5/6

878 C H A P T E R 15 M U L T I P L E I N T E G R A T I O N

E X A M P L E 4 Evaluate

42

91

yex d y

d x.

Solution First evaluate the inner integral, treating x as a constant:

S(x) =

91

yex d y = ex9

1

y d y = ex

1

2y2 9

y=1

= ex

81 1

2

= 40ex

Then integrate S(x) with respect to x:

42

91

yex d y

d x=

42

40ex d x= 40ex

4

2

= 40(e4 e2)

If an iterated integral is written with d x preceding d y, then we integrate first with

respect to x: dc

ba

f(x, y) dx dy =

dy=c

bx=a

f(x, y) d x

d y

Here, for clarity, we have included the variables in the limits of integration of the iterated

integral on the right.

We often omit the parentheses in the

notation for an iterated integral:

ba

dc

f(x, y) dy dx

The order of the variables in dy dx tells us

to integrate first with respect to y between

the limits y = c and y = d.

E X A M P L E 5 Evaluate

/2

0

/2

0

sin(2x+ y) dx dy.

Solution Compute the inner integral, treating y as a constant:

/2

0

sin(2x+ y) d x= 1

2cos(2x+ y)

/2x=0

= 1

2cos(+ y)+

1

2cos y

/2

0

/2

0

sin(2x+ y) dx dy = 1

2

/2

0cos(+ y) cos y d y

= 1

2

sin(+ y) sin(y)

/2y=0

= 1

2

sin

3

2 sin

2

+

1

2

sin sin0

= 1

E X A M P L E 6 Reversing the Order of Integration Verify that1

0

4

2

x2y3 dy dx=

4

2

1

0

x2y3 dx dy

Solution Both integrals have the value 20:

10

42

x2y3 dy dx=

10

1

4x2y4

4y=2

d x=

10

64x2 4x2

d x

=1

060x

2

d x= 20x31

0= 20


6/6

S E C T I O N 15.1 Integration in Several Variables 879

42

10

x2y3 dx dy =

42

1

3x3y3

1x=0

d y =

1

3

42

y3 d y

=1

12y

44

2

=1

12(

256

16) =

20

The previous example illustrates a general fact: The value of an iterated integral does

not depend on the order in which the integration is performed. This is part of Fubinis

Theorem. Even more important, Fubinis Theorem states that a double integral over a

rectangle can be evaluated as an iterated integral.

CAUTIONWhen you reverse the order of

integration in an iterated integral,

remember to interchange the limits of

integration (the inner limits become the

outer limits).

THEOREM3 Fubinis Theorem The double integral of a continuous function f(x, y)

over a rectangle R = [a, b] [c, d] is equal to the iterated integral (in either order):R

f(x, y) d A =

ba

dc

f(x, y) dy dx =

dc

ba

f(x, y) dx dy

Proof We sketch the proof. Consider a Riemann sum SN,M for a regular partition. Then

SN,M is the sum of values f(Pi j )xy, where

x =b a

Nand y =

d c

M

Fubinis Theorem ultimately stems from the elementary fact that we may sum these val-

ues in any order. Think of the values f(Pi j ) as listed in an N M array. For example,

the values for S3,3 are listed in the array in the margin. If we first sum the columns and

3 f(P13) f(P23) f(P33)

2 f(P12) f(P22) f(P32)

1 f(P11) f(P21) f(P31)

ji 1 2 3

then add up the column sums, we obtain

SN,M =

Ni =1

Mj =1

f(Pi j )xy

Sum in any order=

Ni=1

M

j =1

f(Pi j )y

x

First sum the columns,then add up the column sumsLet us choose sample points of the form Pi j = (xi , yj ), where {xi } are sample points

for the regular partition on [a, b] and {yj } are sample points for the regular partition of

[c, d]. We observe thatx andy approach zero as M and N tend to infinity. Therefore,

the limit of the sums SN,M as P 0 may be expressed as a limit as N, M:R

f(x, y) d A = limN,M

Ni =1

M

j =1

f(xi , yj )y

x

The inner sum on the right is a Riemann sum that approaches the single integraldc

f(xi , y) d y as M approaches:

limM

SN,M = limM

Mj =1

f(xi , yj ) =

dc

f(xi , y) d y

To complete the proof, we take two facts for granted. First is the fact that the function

S(x) = d

c f(x, y) d y is continuous for a x b. Second, the limit limN,M SN,Mmay be computed by first taking the limit with respect to M and then with respect to N.

Ejemplos integrales dobles

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