Embed Size (px)
DESCRIPTION
mate2
Citation preview
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 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
2π
∫a sinθ
a
rdrdθ
∫0
2π [ ∫a sin θ
a
rdr ]dθ∫0
2π [ r22 ] / aa sin θ
dθ
∫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 ]/
xxadx
∫0
a
X1X [ tan−1 Y
X ]/xxadx
∫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
=π
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
