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1.A = = = = = = = = (radioext) (radioint) p (1 + y)2 (1 + y 2 )2 Z 1 p (1 + y)2 (1 + y 2 )2 dy Z
0 1 2 2

p 1+2 y+y

(1 + 2y 2 + y 4 ) dy 1 2 y3 3 2 3 2y 2 y 4 dy

y2 2 3 (20 ( )y 2 + 3 2 14 + 3 2 39 30

Z

0 1

p 1+2 y+y

0

y5 1 )j 5 0 1 5

2.5 7 XX j=1 j=1

f (x; y)

= A = 25j = [f (5; 5) + f (5; 10) + f (5; 15) + f (5; 20) + f (5; 25) + f (5; 30) + f (10; 5) + f(10; 10) + f (10; 15) + f (10; 20) + f (10; 25) + f (10; 30) + f (15; 5) + f (15; 10) + f (15; 15) + f (15; 20) + f (15; 25) + f (15; 30) + f (20; 5) + f (20; 10) + f (20; 15) + f (20; 20) + f (20;25) + f (20; 30)] A = 4325

3.Z = = = =
1

Z

0 1

Z

1

(4

2y)dxdy

0

4x
1

2yxj1 dy 0 2ydy

Z 4

0

4

0

4y

2y 2 1 j 2 0 1=3

1

4.-

Z Z Z Z Z
2 12

y 3 dA
2y 1

Z

D

y 3 dxdy

2 y

Z

1 2

xy 3 j2y y1 dy 2 [(2y 1) 1 (2 y)] y 3 dy

1 2

(2y
2

2 + y)y 3 dy

Z Z

1

(3y
2 4

3)y 3 dy Z
2

1

3y dy

3y3 dy

1

1

= = =

3 52 3 42 y j y j 5 1 4 1 3 3 (32 1) (16 1) 5 4 18:6 11:25 = 7:35

Evalue la siguiente integral

2

Z Z

dA p Acotada por 4 Z 2 Z 2 2 e r rdrd
2

e

(x2 +y2 )

y2

por el eje Y

0

u 2 Z
2

= r2 = = = = = = 1 2 1 2 1 2 1 2 Z Z Z Z
2

2

Z

2

e

r2

rdrd

0

2 2

1 du = 2rdr = du = dr 2 Z 2 eu dud
0

2 2

eu j2 d 0e
r2 2 j0 d

2 2

(0:0183
2

1)d

(e 4 2 14:95

1)

Use el teorema de Green para evaluar la integarl de linea 1.-

3

I
c

(x2 + y)dx + xy 2 dy;

c es la curva cerrada dadapor y 2

= x y y = x de (0,0) a (1,-1) Z Z d d 2 (xy 2 ) (x + y) dA dx dy R Z Z = y 2 1 dA = = = = Z
R 0

Z

1 0 1 0

Z

y

(y 2

1)dxdy = ( y))

y2

Z

0

1

h

y2 xi xjy2y dy

(y 2 ( y) y3 + y
1

(y 2 (y 2 )

y 2 ) dy

1 4 1 2 1 5 y + y y + 4 2 5 1 1 = [0] ( 1)4 + ( 4 2 7 = 60 .

Z

y 4 + y 2 ) dy 1 3 y 3 1)2 1 1 ( 1)5 + ( 1)3 5 3

4