40
EVALUAR CADA UNA DE LAS INTEGRALES ITERADAS. 1.-Evaluar cada una de las integrales iteradas. 1 2 X X 3 XYdydx 1 2 [ X X 3 XYdy ] dx 1 2 [ X.Y 2 2 ] / X 3 X dx 1 2 [ X ( X 3) 2 2 X ( X ) 2 2 ] dx 1 2 [ X.X 2 .3 2 X.X 2 2 ] dx 1 2 [ 3 X 3 2 X 3 2 ] dx 1 2 [ 2 X 3 2 ] dx 1 2 X 3 dx [ X 4 4 ] / 2 1 [ ( 2) 4 4 ( 1 ) 4 4 ] [ 16 4 1 4 ]

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∫1

2

∫X

X √3

XYdydx

∫1

2 [∫X

X √3

XYdy ]dx∫1

2 [ X .Y 22 ] /X √3Xdx

∫1

2 [ X ( X √3 )2

2−X (X )2

2 ]dx∫1

2 [ X . X2 .32− X . X

2

2 ]dx∫1

2 [ 3 X32 − X3

2 ]dx∫1

2 [ 2 X32 ]dx∫1

2

X3dx

[ X44 ]/21[ (2 )4

4−

(1 )4

4 ][ 164 − 1

4 ]154

2.-

∫0

∫a sinθ

a

rdrdθ

∫0

2π [ ∫a sin θ

a

rdr ]dθ∫0

2π [ r22 ] / aa sin θ

∫0

2π [ (a )2

2−

(a sin θ )2

2 ] dθ∫0

2π [ a22 −a2 sin2θ2 ]dθ

∫0

2π [ a2(1−sin2θ)2 ]dθ∫0

2π [ a2 cos2θ2 ]dθa2

2 [ θ2 + sin 2θ4 ]/2 π0

a2

2 [( 2π2 + sin 4 π4 )−( 02 +sin o

4 )]a2

2 [(π+ 04 )−(0+ 04 )]π a2

2

3.-

∫0

a

∫xa

xX

X2+Y 2dydx

∫O

a

[∫xa

xX

X 2+Y 2dy ]dx

∫0

a

X [ 1X tan−1 YX ]/

∫0

a

X1X [ tan−1 Y

∫0

a [ tan−1 XX

−tan−1XaX ]dx

∫0

a

[ π4−tan−1 1a ]dx

[ π4 X−X tan−1 1a ] /a0

[(a π4−a . tan−1 1a )−(0 π4−0. tan−1 1

a )]a [ π4−tan−1 1

a ]

4.-

∫O

a

∫y−a

2 y

XYdxdy

∫0

a [∫y−a

2 y

XYdx ]dy∫0

a [Y X22 ]/ 2 yy−ady∫0

a [Y (2Y )2

2−Y (Y−a )2

2 ]dy∫O

a [ 4Y 32 −Y (Y 2−2aY +a2 )

2 ]dy∫O

a [ 4Y 32 −(Y 3−2aY 2+a2Y )

2 ]dy∫O

a [ 4Y 3−Y 3+2aY 2−a2Y2 ]dy∫O

a [ 3Y 3+2aY 2−a2Y2 ]dy

[ 3Y 48 + aY3

3−a

2Y 2

4 ]/ a0[ 9Y 4+8aY 3−6 a2Y 224 ]/a0[ 9 a4+8a a3−6 a2a224 ]

11a4

24

5.-

∫b2

b

∫0

π2

rdθdr

∫b2

b [∫o

π2

rdθ ]dr∫b2

b

[rθ ] /π2odr

∫b2

b

[ rπ2 −r 0]dr

∫b2

b

[ rπ2 ]dr

π2 [ r22 ] / bb

2

π2 [ b22 −

( b2 )2

2 ]π2 [ b22 −

b2

42 ]

π2 [ b22 −b

2

8 ]3π b2

16

6.-

∫0

1

∫0

x

√1−X2dydx

∫0

1 [∫0

x

√1−X2dy ]dx∫0

1

[Y √1−X2 ]/ x0dx

∫0

1

[ X √1−X2−0√1−X2 ]dx

∫0

1

X √1−X2dx

1−X2=u

−2 Xdx=du

−12∫0

1

√1−X2 (−2 Xdx )

−12∫0

1

√u (du )

−12 [ u

32

32

] /10−12 [ 2 (1−X2 )

32

3 ]/10−[ (1−12 )

32

3−

(1−02 )32

3 ]13

7.-

∫0

2

∫0

√4−Y22

√4−Y 2dxdy

∫0

2 [ ∫0

√4−Y22

√4−Y 2dx ]dy

∫0

2

[ 2 X

√4−Y 2 ]/√4−Y2

0dx

∫0

2 [ 2 X √4−Y 2√4−Y 2

−2 X0

√4−Y 2 ]dx∫0

2

[2 X ]dx

[ 2 X22 ]/20[22−02 ]

4

8.-

∫0

π4

∫0

cosθ

3 r 2sin θdrdθ

∫0

π4 [∫

0

cosθ

3 r2sin θdr ] dθ

∫0

π4 [[ 3 r3 sin θ3 ]] /cos θo dθ

∫0

π4

[ (cosθ )3 sin θ−(0 )3sin θ ]dθ

∫0

π4

[ (cosθ )3 sin θ ]dθ

cosθ=u

−sin θdθ=du

−∫0

π4

(cosθ )3 (−sinθdθ )

−∫0

π4

(u )3du

−[ u44 ] / π4O

−[ (cosθ )4

4 ]/ π4O

−[(cos π4 )4

4−

(cos0 )4

4 ]

[ (√22 )4

4−1

4

4 ][ 464−1

4 ]−316

9.-

∫1

4

∫1

x

2Y e−x dydx

∫1

4 [∫1

x

2Y e− xdy ]dx∫1

4 [ 2Y 2 e−x2 ] /X1 dx∫1

4

[ (X )2 e−x−(1 )2 e−x ]dx

∫1

4

[ (X )2 e−x ]dx−∫1

4

[e−x ] dx

X2=u–∫dv=∫ e−x−dx

2 Xdx=du v=−e− x

[−X 2e− x ] /41−∫

1

4

−e−x 2 Xdx−[ (−e− x) ]/41

[−X 2e− x+e−x ] /41+∫1

4

e−x 2 Xdx

2 X=u∫ dv=−∫e− x−dx

2dx=du v=−e−x

[−X 2e− x+e−x ] /41−[2 X e−x ] /4

1−∫

1

4

−e−x2dx

[−X 2e− x+e−x−2 X e−x ] /41−¿ [2e− x ] /4

1¿

[−X 2e− x−e−x−2 X e− x ] /41

[−42 e−4−e−4−2.4 e−4 ]− [−12e−1−e−1−2.1e−1 ]

e−4 [−16−1−8 ]−e−1 [−1−1−2 ]

e−4 [−25 ]−e−1 [−4 ]

e−1 [4−25e−3 ]

10.-

∫0

1

∫2 X

2

(X−Y )dydx

∫0

1 [∫2X

2

(X−Y )dy ]dx∫0

1 [ 2 XY−Y 2

2 ]/ 22 X dx

∫0

1 [( 2 . X .22 −(2 )2

2 )−( 2. X .22 −(2 )2

2 )]dx−¿

∫0

1

[2 X−2 ]dx

[ 2 X22 −2 X ] /10[ 2. (1 )2

2−2.1]=−1

11.-

∫0

4

∫0

Y

√9+Y 2dxdy

∫0

4 [∫0

Y

√9+Y 2dx]dy∫0

4

[Y √9+Y 2 ]/Y0dy

∫0

4 √9+Y 22

2Ydy

9+Y 2=A

dA=2Ydy

∫0

4 √A2d A

[ A12+1

232

] /40[ (9+Y 2 )

32

3 ] /40[ (9+42 )

32

3−

(9+02 )32

3 ][ (25 )

32

3−

(9 )32

3 ][ 125−273 ]=32 23

12.-

∫0

1

∫X 2

X

(1+2Y )dydx

∫0

1 [∫X2

X

(1+2Y )dy ]dx∫0

1 [Y−2 (Y )2

2 ]/ XX 2dx∫0

1

[ (X−X2 )−(X2−X4)]dx

∫0

1

[ (X−2 X 2+X4 ) ]dx

∫0

1

[ (X−2 X 2+X4 ) ]dx

[ X22 −2 X3

3+ X

5

5 ] /10[ 122 −2.1

3

3+1

5

5 ]= 130

II- EVALUAR LAS INTEGRALES EN LAS REGUIONES DADAS.

1.- ∬R

(4−X2−Y 2 )dxdy donde R es la región plana limitada por la recta X=0,

X=1, Y=0, Y=3/2. a) 0≤ X ≤1

0≤Y ≤3/2

b) X: X=0 hasta X=1

Y: Y=0 hasta Y=3/2

c)

∫0

1

∫0

3 /2

(4−X2−Y 2 )dxdy

∫0

1 [∫0

3/2

(4−X2−Y 2 )dx ]dy∫0

1 [4 X−Y 2 X− X3

3 ]/3/20 dy

∫0

1 [4 32−Y 2 32−( 32 )3

3 ]dy∫0

1 [6−3Y 22 −98 ]dy

[ 398 Y−Y3

2 ] /10[ 398 −1

3

2 ]=358

2.- ∬R

(1+X+Y )dxdy donde R es la región plana limitada por la recta Y=-X ,

X=√Y , Y=2 .

a) −Y ≤ X≤√Y

0≤Y ≤2

b) X: X=−Y hasta X=√Y

Y: Y=0 hasta Y=2

c)

∫0

2

∫−Y

√Y

(1+X+Y )dxdy

∫0

2 [∫−Y

√Y

(1+X+Y )dx ]dy∫0

2 [1X+ X2

2+XY ] / √Y−Y

dY

∫0

2 [(√Y +(√Y )2

2+√Y .Y )−((−Y )+ (−Y )2

2+Y . (−Y ))]dY

∫0

2 [√Y + 3Y2

+Y 3/2+ Y2

2 ]dY

[Y 3 /23 /2+3Y

2

4+Y

5/2

5 /2+ Y

3

6 ] /20[ 2 (2 )3 /2

3+3 (2 )2

4+2 (2 )5/2

5+

(2 )3

6−2 (0 )3 /2

3−3 (0 )2

4−2 (0 )5 /2

5−

(0 )3

6 ]( 80√2+180+96√2+8060 )=44√2+6515

3.- ∬R

❑ (eyx )dydx donde R es la región plana limitada por la recta Y=X, Y=0 ,

X=1 .

a) 0≤ X ≤1

0≤Y ≤ X

b) X: X=0 hasta X=1

Y: Y=0 hasta Y=X

c)

∫0

1

∫0

X (eyx )dydx

∫0

1 [∫0

X (eyx )dy ]dx

∫0

1 [X e yx ]/X0dx

∫0

1

[ X e1−X e0 ] dx

∫0

1

[Xe−X ]dx

[ e . X22 − X2

2 ]/10[( e .122 −1

2

2 )−( e .022 −02

2 )][ e−12 ]

4.- ∬R

Y 2dA R={( X ,Y ) :−1≤Y ≤1 ,−Y−2≤ X≤Y }

a) −Y−2≤ X ≤Y

−1≤Y ≤1

b) X: X=-Y-2 hasta X=Y

Y: Y=-1 hasta Y=1

c)

∫−1

1

∫−Y−2

Y

Y 2dxdy

∫−1

1 [ ∫−Y−2

Y

Y 2dx ]dy∫−1

1

[Y 2 . X ] / Y−Y−2

dy

∫−1

1

[Y 2 .Y−Y 2 .(−Y−2)]dy

∫−1

1

[2Y 3+2Y 2 ]dx

[ 2.Y 44 + 2.Y3

3 ]/ 1−1[( (1 )4

2+2 (1 )3

3 )−( (−1 )4

2+2 (−1 )3

3 )]43

III.- Utilizar una integral iterada para hallar la región R

1.- R={( X ,Y ) :0≤Y ≤4−X2;0≤ X ≤2}

Y=0 (recta) Y=4-X2 → Y=X2-0X-4 → Vértice = (-b/2a) → (0;4)

Y=0 → X=±2

X=0 → Y=+4

a) 0≤ X ≤2

0≤Y ≤ 4−X2

b) X: X=0 hasta X=2

Y: Y=0 hasta Y=4−X2

∫0

2 [ ∫0

4−X2

dy ]dx∫0

2

[Y ]/4−X2

0dx

∫0

2

[ 4−X2 ] dx

[4 X− X3

3 ]/20[4.(2)− (2 )3

3 ]=163

2.- R={( X ,Y ) :X+2≤Y ≤ 4−X2;−2≤ X≤1}

Y=X+2 (linial) Y=4-X2 → Y=X2-0X-4 → Vértice = (-b/2a) → (0;4)

Y=0 → X=±2

X=0 → Y=+4

a) −2≤ X≤1

X+2≤Y ≤4−X2

b) X: X=-2 hasta X=1

Y: Y=X+2 hasta Y=4−X2

∫−2

1 [ ∫X+2

4−X 2

dy ] dx∫−2

1

[Y ] /4−X2

X+2dx

∫−2

1

[4−X2−(X+2 ) ]dx

∫−2

1

[2−X2−X ]dx

[2 X− X3

3− X

2

2 ] / 1−2[(2 (1 )−

(1 )3

3−

(1 )2

2 )−(2 (−2 )−(−2 )3

3−

(−2 )2

2 )][(2−13−12 )−(−4+ 83−42 )]

[( 76 )−(−206 )]=276

3.- R={( X ,Y ) X2+Y 2≤4 ;0≤ X ≤2 }

a) 0≤ X ≤2

0≤Y ≤√4−X2

b) X: X=0 hasta X=2

Y: Y=0 hasta Y=4−X2

∫0

2 [ ∫0

√4−X2

dy ]dx∫0

2

[Y ]/√4−X2

0dx

∫0

2

[√4−X2 ]dx

[ X √4−X22

+4 sin−1

X2

2 ]/20[ (2 ) √4−(2 )2

2+4sin−1( (2 )

2 )2 ]−[ (0 ) √4− (0 )2

2+4 sin−1( (0 )

2 )2 ]

4sin−1 (1 )2

2.π2

2.- R={(X ,Y ): 0≤Y ≤ 1

√ x−1;2≤ X≤5}

Y=1

√x−1→√x−1≠0→x≠1

Y¿0 → X=±2

a) 2≤ X≤5

0≤Y ≤1

√x−1

b) X: X=2 hasta X=5

Y: Y=0 hasta Y=1

√x−1

∫2

5 [ ∫0

1

√x−1

dy ]dx∫2

5

[Y ]/1

√x−10

dx

∫2

5

[ 1√ x−1 ]dx

∫2

5

[ (x−1 )−1/2 ]dx

[ ( x−1 )−12

+1

−12

+1 ] /52[2 ( x−1 )

12 ]/52

[(2 (5−1 )12)−(2 (1−1 )

12)]

[4−2 ]=2

IV.- Proporcione una integral doble para hallar el volumen del solido limitado por las gráficas indicadas.

1.- Z=XY, Z=0, Y=X, X=1

a) 0≤ X ≤1

0≤Y ≤ X

b) X: X=0 hasta X=1

Y: Y=0 hasta Y=X

∫0

1 [∫0

X

XYdy] dx∫0

1

X [Y 22 ]/X0 dx∫0

1

X [ (X )2

2 ]dx∫0

1 [ (X )3

2 ]dx[ X48 ]/10

[ (1 )4

8−

(0 )4

8 ]=18

2.- Z=X, Z=0, Y=X, y=0, X=0, x=5

a) 0≤ X ≤5

0≤Y ≤ X

b) X: X=0 hasta X=5

Y: Y=0 hasta Y=X

∫0

5 [∫0

X

Xdy ]dx∫0

5

X [ y ]/X0dx

∫0

5

X [x−0 ]dx

∫0

5

[ (X )2 ]dx

[ X33 ] /50[ (5 )3

3−

(0 )3

3 ]=1253

3.- Z=0, Z= X2, Y=0, y=4, X=0, x=2

a) 0≤ X ≤2

0≤Y ≤ 4

b) X: X=0 hasta X=2

Y: Y=0 hasta Y=4

∫0

2 [∫0

4

(X 2−0)dy] dx∫0

2

X2 [ y ] /40dx

∫0

2

X2 [4−0 ]dx

∫0

2

4 [ (X )2 ]dx

4 [ X33 ] /204 [ (2 )3

3−

(0 )3

3 ]=323

5.- X2+Z2=1, Y2+Z2=1, primer octante. X=√1−Z2

Y=√1−Z2

1−Z2≥0→z=±1

a) 0≤ X ≤√1−Z2

0≤Y ≤√1−Z2

−1≤Z≤1

b) X: X=0 hasta X=√1−Z2

Y: Y=0 hasta Y=√1−Z2

∫−1

1 [ ∫0

√1−Z2

√1−Z2dy ] dz∫−1

1

√1−Z2 [Y ] /√1−Z20

dz

∫−1

1

(1−Z2 )dz

[ z− z33 ]/ 1−1[(1− (1 )3

3 )−(−1− (−1 )3

3 )][1−13+1−13 ]

43

5.- Y=4-X2, Z=4-X2, primer octante.

Y>0 → −2≤ X≤2

Z>0 → −2≤ X≤2

a) 0≤Y ≤ 4−x2

0≤Z ≤4−x2

−2≤ X≤2

b) X: X=-2 hasta X=2

Y: Y=0 hasta Y=4−x2

∫−2

2 [ ∫0

4−x2

4−x2dy ]dx∫−2

2

4−x2 [Y ] /4−x2

0dx

∫−2

2

(4−x2 )2dx

∫−2

2

[16−8 X2+X 4 ] dx

[16 X−8 X3

3+ X

5

5 ] / 2−2[(16.2−8 (2 )3

3+

(2 )5

5 )−(16 (−2 )−8(−2 )3

3+

(−2 )5

5 )][ 25615 + 256

15 ]5123

V.- Encuentre el volumen del solido dado debajo del plano X+2Y-Z=0 y arriba de la región acotado por x=y, y= x4.

Z=X+2Y

x4<Y<X

X=Y= x4

0=X(1-X3)→X=1

a) X: X=0 hasta X=1

Y: Y=X hasta Y=x4

b) ∫0

1 [∫X 4

X

(X+2Y )dy ]dx∫0

1

[ XY+Y 2 ] / XX4dx

∫0

1

[ (X . X 4−( X4 )2 )−(X .X−(X )2 ) ]dx

∫0

1

[ (X5−X8 ) ]dx

[ X66 − X9

9 ] /10[ 16−1

9 ]118

VI.- Hallar la masa y centro de masa de la lámina limitada por las gráficas de las funciones con la densidad dada.

2.- Y=√X ,Y=0 , X=4 , ρ=KXY

a) 0≤Y ≤√X

0≤ X ≤4

b) X: X=0hasta X=4

Y: Y=0 hasta Y=√X

m=∫0

4

∫0

√ X

KXYdydx

m=∫0

4

KX [∫0

√X

Ydy ]dxm=KX∫

0

4

[Y 2 ] /√X0dx

m=K∫0

4X2

2dx

m=K [ X36 ]/40m=K [ 436 −0

3

6 ]m=32

3K

M Y=∫0

4

∫0

√X

K X2Ydydx

M Y=∫0

4

K X2[∫0

√X

Ydy ]dxM Y=

K X2

2∫0

4

[Y 2 ] /√X0dx

M Y=K2∫0

4

X3dx

M Y=K2 [ X44 ]/ 40

M Y=K2 [ 444 −0

4

4 ]M Y=32K

M X=∫0

4

∫0

√X

KXY 2dydx

M X=∫0

4

KX [∫0

√ X

Y 2dy] dxM X=

KX3∫0

4

[Y 3 ] /√X0dx

M X=K∫0

4X5 /2

3dx

M X=K [ 2 X7 /27 ]/ 40M X=K [ 2.47 /27

−2.07 /2

7 ]M X=256K

X= mM Y

;Y= mM X

X=32 /3K32K

→13;Y=32/3K

256K→124

3.- Y=X3, Y=0, X=2, ρ=KX

a) 0≤Y ≤ x3

0≤ X ≤2

b) X: X=0 hasta X=2

Y: Y=0 hasta Y=x3

m=∫0

2 [∫0

x3

KXdy ]dxm=∫

0

2

KX [ y ]/ x3

0dx

m=K∫0

2

x4dx

m=K [ X55 ]/20m=K [ (2 )5

5−

(0 )5

5 ]=325 K

M Y=∫0

2

∫0

X3

K X2dydx

M Y=∫0

2

K X2 [Y ]/X3

0dx

M Y=∫0

2

K X5dx

M Y=[K X66 ] /20=323 K

M X=∫0

2

∫0

X3

KXY dydx

M X=∫0

2

KX [∫0

X 3

Y dy ]dxM X=∫

0

2

KX [Y 22 ] /X30 dxM X=∫

0

2

KX [ X62 ]dxM X=K [ X816 ] /20M X=K [ 2816− 08

16 ]M X=16K

X= mM Y

;Y= mM X

X=32 /5K32 /3K

;Y=32/5K16K

X=35;Y=2

5

VII.- Hallar Ix, Iy, I0 para la lámina limitada por las gráficas de las funciones con la densidad dada.

1.- Y=√X;Y=0 ; X=4 ; ρ=KXY

a) 0≤Y ≤√X

0≤ X ≤4

b) X: X=0 hasta X=4

Y: Y=0 hasta Y=√X

IX=∫0

4

∫0

√X

KXY 2Y dydx

IX=∫0

4

KX [∫0

√ X

Y 3dy ]dxIX=∫

0

4

KX [ Y 44 ]/√X0 dxIX=∫

0

4

KX [ (√X )4

4−

(0 )4

4 ]dxIX=∫

0

4

K [ X34 ]dxIX=K [ X 416 ] /40=16KI Y=∫

0

4

∫0

√X

KX X2Ydydx

IX=∫0

4

K X 3[∫0

√ X

Y dy ]dx

IX=∫0

4

K X 3[Y 22 ] /√X0 dxIX=∫

0

4

K X 3[√X22 ]dxIX=K [ X510 ]/ 40

IX=K [ (4 )5

10−

(0 )5

10 ]IX=

5125K

I 0=I X+ I Y

I 0=5125K+16K

I 0=5925K