CALCULO VECTORIAL
TITULO:
RECOPILACIÓN DE PRUEBAS
TUTOR:
ING. EDISON GUAMAN
ALUMNO:
BRYAN ZAMORA
SANGOLQUI, 04-03-2015
x2 + y 2 = 4 x =
− m2
y = 1 −
m
2
y = 1 − x2
1 − y + y 2 = 4 y2 − y − 3 = 0 y1 = 2, 3 y2 = −1, 3 2
2
y2
x + y = 4 x = 4 − y 2 = 4 − (−1, 3)2 = 1, 52 x = 2cost y = 2sent
1, 52 = 2cost
t = 0, 71 t1 = 2π − 0, 71 = 5, 57
´ 1 √
5π 2
´ 0
√
A = 21 5,57 4dθ − 0 1 − ydy − −1,3 [ 1 − y − (− 0,y87 )]dy A = 4, 57 − 0, 67 − 0, 69 A = 3, 21
´
5π 2
´ 1
´ 0
M y = 31 5,57 23 cosθdθ − 0 (1 − y )dy − −1,3 [(1 − y ) − (− 0,y87 )2 ]dy M y = 4, 41 − 0, 5 − 1, 18 = 2, 73
´
¯ = d = x
2,73 3,21
= 0 , 85
V = 2 π ∗ d ∗ A = 2π ∗ (0, 85) ∗ (3, 21) V = 17 , 14
r = 2(1 − cosθ ) r =
−6cosθ y = 4+x
2(1 − cosθ) = 2cosθ = −1 θ = 23π
−6cosθ
´ π
4(1 − cosθ)2 dθ A = 21 A = 7, 039 2π 3
´ π
8(1 − cosθ)3 dθ M x = 31 M x = 7, 29 2π 3
M y = M y =
1 3
´ π
8(1 − cosθ )3 cosθdθ −14, 979 2π 3
,979 = −1, 91 x = −14 7,86
y =
7,29 7,86
= 0, 93
x − y + 4 = 0
1,91−0,93+4 √ +by+c = −√ = 0 , 82 d = ax a +b 2
2
12 +(−1)2
V = 2 π ∗ d ∗ A = 2π ∗ (0, 82) ∗ (7, 039) V = 36 , 28
9ay 2 = x (x
y =
x(x−3a)2 9a
A = 2 A = A =
´ 3a
−
−
0 √ 2 a 3a
x(x−3a)2 dx = 9a
´ 3a
0 2 2, 77a u2
M y =
√
2 a 3a
−− √ M y = − 23
a a
x
3 2
− 3ax
´ 3a 0
2 72 x 7
a a
dx =
3 2
− 3ax − 65 ax |30 x x
M y = 3, 56a3 a x = My = 32,,56 = 1, 28a A 77a 3
1 2
√
− 23
5 2
a
1 2
´ 3a √ 0 √ 2 a 3a
−
dx =
2
V rev = 2π (1, 28a)(2, 77a2 ) = 22, 36a3
− 3a)2
x(x
− 3a)dx 2 2 |30 5 x − ax
−
5 2
√
2 a 3a
3 2
´ 3a 0
x
5 2
a
− 3ax
3 2
dx
r = 1 + cos(θ)
x2 = 1
− 2y
x = rcos(θ ) = (1 + cos(θ ))cos(θ) y = rsen(θ) = (1 + cos(θ))sen(θ)
x2
− 1 + 2y = ((1 + cos(θ))cos(θ))2 − 1 + 2(1 + cos(θ))sen(θ) = 0
θ1 = 5, 79 x1 = 1, 65 θ2 = 2, 058 x2 = 0, 24 LT = L L2 ´ 51, + 79 L1 = 2,058 ( sen(θ )2 + (1 + cos(θ ))2 dθ =
−
− √ | − √ √ √ | | √ | ´ 5,79
L1 = 2,058 2cos θ2 dθ = 4sen θ2 L1 = 3, 59
´ 1,65
2
´ 5,79
2,058
sen2 θ + 1 + 2cosθ + cos2 θdθ
5,79 2,058
´ 1,65
L2 = −0,24 1 + ( x) dx = −0,24 1 + x2 dx 1,65 L2 = x2 1 + x2 + 12 ln x + x2 + 1 −0,24 L2 = 2, 47 LT = 3, 59 + 2, 47 = 6, 061 ´ 5,79 ´ 5,79 M y1 = 2,058 (1 + cosθ ) cosθ 2cos θ2 dθ = 2,058 cosθ + cos2 θ M y1 =
´ 5,79
2,058
2cos θ2 + 2cos3 θ2 + cos θ2 + cos 32θ + cos 52θ dθ
M y1 = 4sen θ2 + 43 sen 32θ + 2sen θ2 + 23 sen 32θ + 25 sen 52θ M y1 = 1, 26
´ 1,65
M y2 = −0,24 x 1 + x2 dx = 3 1 + x2 M y2 = 2, 03 M yT = 1, 26 + 2, 03 = 3, 29 x = 0, 5428 S = = 2π (x + 2)(6, 061) = 96, 83
3 2
1,65
−0,24
5,79 2,058
2cos θ2 dθ
r = 2 + 2 cosθ y = x 2 + 1
rsenθ = r 2 cosθ + 1 r = 2 + 2cosθ f θ = 2senθ + sen2θ 4cos2 θ 8cos3 θ
−
−
− 4cos4θ − 1
θ1 θ1 = 1, 12 r1 = 2, 87 x1 = 1, 26 y1 = 2, 58 θ2 θ2 = 1, 93 r2 = 1, 29 x2 = 0, 457 y2 = 1, 21
−
´ 1,93
(2 + 2cosθ)2 dθ A = 21 1,12 (2 1) (2, 053x)]dx
−
}
− {´ − 0 0 457[(x2 + 1) − (−2, 645x)]dx + ´ 0 1 259[(x2 + ,
,
1,93 1 2 (4θ + 8 senθ + 2θ + sen2θ )1,12 2,053x2 1,259 )0 2 A = 21 (3, 73) 0, 51
A =
}
1 2
´ 1,259
2
´ 1,259
1 3
− (2, 053x) ]dx} − 0, 36
´ 1,93 1,12
2
y¯ =
= =
}
−
,
0,589 1,36 = 0, 43 2,37 1,36 = 1, 75
V = 2 πAd √ +by+c = (−0,94)(0,43)−1,75+3,77 = 1 , 17 d = ax a +b 2
−
− {´ − 0 0 457[(x2 + 1)x − (−2, 645x)x]dx +
(2 + 2 cosθ)3 cosθdθ
−
My A Mx A
+ x
,
2
[(x + 1)x (2, 053x)x]dx M y = 31 (1, 83) 0, 023 M y = 0, 589 ¯ = x
2,645x2 0 x3 ) + ( − 0 457 , 2 3
− { 12 ´ − 0 0 457 [(x2 + 1)2 − (−2, 645x)2]dx +
3 1,12 (2 + 2 cosθ ) senθdθ
[(x + 1) M x = 31 (8, 2) M x = 2, 37
0
My = 0
´ 1,93
1 3 2
+ x +
−
A = 1, 36
M x =
x3
− {( 3
2
√ (−0 943) +(−1)
V = 2 π (1, 36)(1, 17) V = 9 , 98
,
2
2
8a2 y2 = a 2 x2
y =
±
a2
x a
2
−x 8
2
2
−2x y = a√ a8√ a −x 2
Sx = 2π
´ a 0
− x4
y
2
1 + ( f (x) )2
´ a √ x ∗ 1 +√ a√ a8√ −a2x−x Sx = 2π 0 x aa√ − 8 ´ a √ 2 −12 π √ a x +4x Sx = a2√ x a − x2 ∗ 9aa√ 0 8 8 a −x ´ a π 3 2 2
Sx =
Sx = Sx = a 2 π
(2 x
a2 0 π x4 ( a2 2
x = acos3 t
y = asen3 t
−
2
− 3a x)dx
3a2 x2 a 2 )0
2
2
2
4
2
2
2
2
2
4
A =
´ 0 π
2
asen3 t( 3acos2 tsent )dt
−
π
´ A = 3a2 0 sen4 t ∗ cos2 tdt ´ 1−cos2t sen 2t 2 2
π
A = 3a
2
2
0
(
A = 3a8 ( 2t A = 0, 294a2 M x = M x = M x =
−
2 sen4t 8
2
)(
)dt 4 sen3t 2 6 )0 π
−
1 0 3 2 2 2 2 (asen t) ( 3acos tsent )dt 3a3 2 cos2 t)3 sentcos 2 tdt 2 0 (1 3a3 2 2 3sentcos 4 t + 3sentcos6 t 2 0 (sentcos t 3 3 5 3a −cos t 3cos t 3cos7 t cos9 t 2 ( + + 2 3 5 7 9 )0 2
´
−
π
´ ´
π
π
−
M x = M x = 0, 076a x = y x = y¯ =
0,076a2 0,29a2
− −
π
⇒ M y = M x = 0, 076a2
= 0 , 2587
(¯x; y¯) = (0, 258 2587; 7; 0, 2587)
− sentcos8t)dt
x2 + y 2 z
=4 = y
x = rcosθ = rcosθ = = 2cosθ y = rsenθ = rsenθ = = 2senθ x = rsenθ = rsenθ = = 2senθ r(θ) = 2cosθ, 2senθ, 2senθ
r (θ ( θ ) = −2senθ, 2cosθ, 2cosθ √ 2 2 r (θ ( θ ) = (−2senθ) senθ)2 + (2cosθ (2cosθ)) + (2cosθ (2cosθ)) = 4sen2 θ + 4cos 4 cos2 θ + 4cos 4 cos2 θ √ 4cos2θ = 2√ 1 + cos r (θ ( θ ) = 4 + 4cos + cos2 θ
L =
√ θ2 θ1
r (θ ( θ ) dθ
2π 0
L =2
1 + cos + cos2 θdθ = θdθ = 2(7. 2(7.64)
n = 6
√ 1 + cos + cos2 θ
θ
2π 6
√
( 2) 2( 5) 5 4( 2) 5 2( 5) ( 2)
π
3 2π 3
√ √
π 4π 3 5π 3
2π
I = (21.902) = 3 (21. L =2
√ 2π 0
π
(21.902) 9 (21.
= 7. 7.64
1 + cos + cos2 θdθ = θdθ = 2(7. 2(7.64) = 15. 15.28 28u u
√ − − − − − − √ − √ r (t) =
√ 4t−t , 1, 1
r (t) = d dt
t2 , t , t
4
2
√ 4t−t = ( t) 4 2
t2
−
1 2
4 (4 t2 )3
−
r (t) =
4 , 0, 0 (4 t2 )3
−
i
r (t)xr (t) =
t
− √ 4− − √ (4−4
t2
t2 )3
j 1 0
−√ 41−
=
k 1 0
t2
− √ (4−−2
+( t) .
t t2 )3
.
1 2
−
=
4− )− −(√ (4− ) t2
t2
t2
3
=
r (t)xr (t) =0 = 0i
− √ (4−4
− r (t ( t) =
k =
t
( √ 4−t
j + j +
t2 )3
2
)2
√ (4−4
t2 )3
2
2
+ (1) + (1) =
r (t)xr (t) 3
( ) 0 − √ − ( r
t
4
k =
3 t2
)
j+
√ √ (84 −2t )
2 3
√ ( − 4
t2
4
−t2 4−t2 8
f (x, y ) = ln l n
−
Dominio
−
2=
−
3
)
k
3
0 +
3 t2
( − ) 4
8 t2 4 t2
− −
+
−t2 4−t2 8
x−y
−
√ −
y x2 x y
−
> 0
4 t2
3
( − ) 4
3
y −x2
(x, y)R2 /
4
2
x2 > 0 x y > 0 y x > 0 x y > 0 y > x2 x > y
−
t2 (4 t2 ) +
√ √ =
√
y
2
2
4
i
k
2
√ ( √ − = 4 √ √ (( −− 4
2
4
t2
)3
8
t2
)3
4
t2
)3
=
Rg : Rg :
R
(2x−4y) f ( f (x) = 1−cos 3x−6y
z = x = x (x, y)
− 2y
→ (0, (0 , 0)
z
→ 0
g (z ) =
1 cos(2z ) 3
−
z =0 z = 0
0
(2z ) lim g (z ) = lim 1−cos = lim 2sen3(2z) = 3z
z
→0
z
→0
lim 2sen3(2z ) =
z
→0
lim
(x,y )
z
0 3
→0
=0
f ( f (x, y) = 0 = f = f (0 (0,, 0)
→(0,0)
0 0
V = πr 2 (2z ) x2 + y 2 + z 2 = R 2
∗
r 2 + z 2 = R 2
4πrz ; 2πr 2 = λ g(r,z) 4πrz ; 2πr 2 = λ 2r; 2z r,z) =
∇V (
4πrz = 2rλ 2πr 2 = 2zλ r2 + z 2 = R 2 4πrz 2πr 2 2
rλ = 22zλ 2z = r 2
2z 2 + z 2 = R 2
∇
z =
R
± √ 3 R2 = R 2 √ 3 6 3 R
r2 + r =
T (x, y) = x 2 + 2y2 + R : x 2 + y 2
2x + 1
• f x = 2x + 2 f y = 4y f xx xx = 2 f yy yy = 4 f xy xy = 0 f yx yx = 0 H (2,4) = [
2 0 ]=8 0 4 H (2,4) > 0
f xx xx > 0
• ∇−→T = λ∇g 2x + 22;; 4y = λ 2x; 2y 2x + 2 = 2λx 4y = 2λy x2 + y 2 = 4 λ = 2 x + 1 = 2x x = 1 2
12 + y = 4 y = 3
±√
P (1;
±√ 3)
≤4
r(t) = 2sen(3t); t; 2cos(3t)
r(t) = 2sen(3t); t; 2cos(3t) r( t) = 6cos(3t);1; 6sen(3t) 18sen(3t);0; 18cos(3t) r(t) =
−
= K =
|
r(t)
−
−
⊗r( ) |
(r
t
)3 (t)
i j k − 6 (3 ) 1 6 (3 ) cos t sen t r( t) ⊗ r(t) = −18sen(3t) 0 −18cos(3t) r ⊗ r = (−18cos(3t))i − 108 j + (18sen(3t))k (t)
(t)
P (0, π,
−2)
t = π
√ ⊗ r r (t) (t) = −18i − 108 j = 18 37 r = (6cos(3t))2 + 12 + (6sen(3t))2 (t) √ r = 37 (t)
√
√
(r( t) )3 = ( 37)3 = 37 37
= = K
√ √ =
18 37 37 37
18 37
u(x,t) = δu dt
δu dx 2
=
δ u dx2 δu dt δu dt δu dt δu dt
2
= a 2 δδxu2
1 √ ∗ e− 2a πt
(x−b)2 4a2 t
1 √ ∗ e− 2a πt
a2 t
x−b)2 a2 t
x−b)2 a2 t
−(4 1 √ ∗ e 2a πt −(4 1 ∗ = 2a√ e πt
x−b)2 a2 t
2
2
a3 t πt
a2 t
x b 2 a2 t2
∗ (− (4 − ) ) + e− 2
( (x−b4)a2−t22atπ )
= δδxu2 ( (x−b4)a2−t22atπ ) 2 = a 2 δδxu2
∗
(x−b)2 4a2 t
∗ (− 4 1 ) ∗ (2x) = − 4 1√ ∗ e− ∗ ∗(− 4 1 ) ∗ (2x) = 8 1√ ∗ e−
( = − 4a3 t1√ πt ∗ e− 4
=
(x−b)2 4a2 t
a3 t2
(x−b)2 4a2 t
πt
∗ 21 ( −2√ a
πt ) πt
(x−b)2 4a2 t
= E =
w = xy x = 1 y = 2 z = 1
1, 041,99 + ln(1, 02)
+ ln (z )
→ x = 0, 04 → y = −0, 01 → z = 0, 02 x + y + z δw = ∗ ( ) y + √ 1 δw = √ x + √ 2 + ( ) 2 + ( ) 2 + δw dx
δw dy
δw dz y ln x
yx y−1 xy
xy
ln z 1
ln z
z
xy
ln(z )
z
√ 2∗(1) (0, 04) + 2√ 21∗ +(1)(1) (−0, 01) + 2(1)√ 11 + 2 1 + (1) δw = 0, 04 − 0 + 0 , 01 = 0, 05 z = f ( + ; + ) = 1, 0493 − 1 = 0, 0493 δw =
2
x
ln
2
ln
xy
y
´ = πr Area π r2 h + 2 πr 2
ln
2
ln(1)
(0, 02)
x2 + y 2 + z 2 = 5 2 x,y,z) =
∇g( ∇A ∇A
x,y,z ) =
2x; 2y; 2z
4πxz + 4 πx; 0 ; 2x2 π
x,y,z ) = λ
∇g(
x,y,z )
4πxz + 4 πx; 0 ; 2x2 π = λ 2x; 2y; 2z
4πxz + 4 πx = 2xλ 0 = 2yλ 2x2 π = 2zλ x2 + y 2 + z 2 = 52 2π (z + 1) = λ x2 π = zλ 2π (z+1)
λ = zλ 2z 2 + 2z = x 2 x2 π
(2z 2 + 2z ) + 0 2 + z 2 = 25 3z 2 + 2z 25 = 0 z1 = 2, 572 z2 = 3, 23
−
−
2
z1
x = 2(2, 572)2 + 2(2, 572) x = 4, 286
˜
(x
1
0
− 1)√ 1 + e2 dxdy
≤ ≤ x
y
e
y
≤ y ≤ ln(x)⇒x = e 0 ≤ y ≤ 1 e ≤´ x ≤´ e √ √ ´ 1 ´ 1 1 I = 0 ( x − 1) 1 + e2 dx dy = 0 ( 2 −x) | · 1 + e2 dy = 0 ( √ e + e ) 1 + e2 dy y
y
e ey
y
y
x
2
e ey
y
2
e
−e2y 2
−
I = 2 .029 x2 + y 2 + z 2 = 5
z 2 + 4z 5 = 0 (z + 5)(z 1) = 0 z = 1 x2 + y 2 = 4 x = r.cos(θ√ ) y = r.sin(θ)
x2 + y 2 = 4 z
− −
⇒
V =
´ 2 0
r
´ 2 ´ 2π ´ 0
0
r2
5−r2
rdzdθdr =
4
´ 2 ´ 2π 0
0
3
4 dr ) V = 2 π (3.39
√ − r2−
r( 5
r
2
4 ) dθdr =
− 1) = 15.04 dxdydz D (2+x2 +y2 +z 2 )1/2
´ x = p.sin(φ).cos(θ) y = p.sin(φ).sin(θ) z = p.cos(φ)
2π (
´ 2 0
√ − r2dr−
r 5
J = p 2 .sin(φ) p2
√ 2+1 )(J )(dp.dp.dφ.dθ ) = ´ 0 1 ´ 0 2 ´ 0 √ 2+ ´ 1 ´ 2 ´ 1 I = 0 0 √ . − cos(φ) |0 dθdp = 4π 0 √ 2+ 2+ I =
π
´
( D
π
p2
π
p2
2
p
π
2
p
I = 4 π (0.207) = 2.608
2
p dp
p2
.sen(φ)dφdθdp
−→ F =< −y ; x ; −z > z = 1 − x − y C : 3
3
3
x2 + y 2 = 1 x = cos (t) y = sen (t) z = 1 cos(t) sen(t)
−
−
r (t) = < cos(t); sen s en(t); 1 r (t) = <
− cos(t) − sen(t) >
−sen(t); cos (t); sen (t) − cos(t) > π
2 I =
≤ t ≤ 0
ˆ
F.r (t).dt
c
I
4
π
=
ˆ
2
<
0
3
3
3
−sen(t) ; cos (t) ; −(1−cos(t)−sen(t))
>.<
se n(t)−cos(t) > .dt −sen(t); cos (t); sen
π
I = =
ˆ 0
+3(1
I
4
2
−sen(t)
4
+ cos(t)4
3
− (sen(t) − cos(t)).(1 − cos(t) − sen(t)
3
− cos(t) − sen(t)).(−cos(t) − sen(t) + cos(t)sen(t)) + 3cos(t)sen(t).dt π
=
ˆ
2
((cos(t)2 sen(t)2 )(cos(t)2 +sen(t)2 ) (sen(t) cos(t))(1 cos(t)3 sen(t)3
−
0
−
−
−
−
+3( cos(t)+ cos(t)2 + cos(t)sen(t) sen(t)+ cos(t)sen(t)+ sen(t)2 + cos(t)sen(t)
−
−
2
2
−cos(t) sen(t) − cos(t)sen(t) ) + 3cos(t)sen(t)))dt I
4
π
=
ˆ 0
2
2
2
4
3
3
4
−2cos(t) +2sen(t) +cos(t)−sen(t)−cos(t) −2cos(t) sen(t)+2cos(t)sen(t) −sen(t)
3
2
2
3
−6cos(t)sen(t) + 3cos(t) − 6cos(t) sen(t) + 3sen(t) cos(t) − 3sen(t) )dt I
4
=(
π
+
2
3 π 1 1 3 π − 1+1+ + − + + 3 − 2 + 2 − 1 + 2) 2 16 2 2 16
π
I
4
=
I = =
3π 8
I = =
ˆ ˆ
3π 2
rotF.N.dA
s
rotF =
−
i
j
k
d dx
d dy
d dz
y3
x3
−z
3
=< 0; 0; 0; 3x2 + 3y 2 >
f (x , y , z ) = z + x + y
−1
N =< 1 , 1, 1 > I
4
ˆ ˆ √ − 1
=
0
1
x2
< 0 , 0, 3x2 + 3y 2 > . < 1 , 1, 1 > dydx
0
I
4
ˆ ˆ √ − 1
=3
I
4
x2 + y 2 dydx
0
=3
ˆ ˆ
π
1
0
4 4
x2
0
I
I
1
=
2
r3 dθdr
0
1
=
ˆ 3π 2
r3 dr
0
3π r 4 1 3π ( ) = 2 4 0 8
|
I =
3π 2
x2 + y 2 + z 2 = a2
S : : x 2 + y 2 + z 2 = a 2 z = a2 x2 y 2 x N =< ,
− √ || a2
−
2
−x −y x2 x2
||N = − ||N || = √ − a2 a a r2
y
√ −
2
a2
−y + 2
a2
x = rcos(θ)
x2
y2 x2
, 1 >
2
− −y + 1 0 0
y = rsen(θ )
−y
||N || =
2
≤r≤a ≤ θ ≤ˆ ˆ
A = a
ˆ ˆ 0
A =
2
a
aπ
2
√ ardr− r 2
0
2
2
0
0
√ rdθdr a −r 2
2
r = asen(u) dr = acos(u)
π
ˆ
2
2
||N ||.dA π
a
A =
2
−x −y = − −y
2
s
ˆ
x2 +y 2 +a2 a2 x2
π
A =
aπ
2
asen(u).du =
a2 π
2
0
≤ u ≤ π2
π
( cos(u)) 0 =
−
|
2
a2 π
2
a
√ − a2
x2
−y
2
ˆ ˆ
M x =
|| ||
x. N .dA
s
π
a
ˆ ˆ
M x = a
o
a
M x = a a
M x = a
2
0
2
0
2
2
2
2
2
√ a r − r
2
r cos(θ ) √ dθdr a −r
√ a r − r
ˆ 0
ˆ
2
π
2
d r (sen(θ )) 0 dr
|
r = asen(u) dr = acos(u)
dr
π
ˆ
M x = a
π
2
2
2
(a sen(u) )du = a
3
0
sen(2u) − )| 2 4
u
ˆ ˆ π
a
M y = a
ˆ ˆ 0
M y = a
ˆ
a
− cos(2u) )du 2
2
ˆ
2
a3 π
π
2
=
4
2
r sen(θ ) √ dθdr a −r 2
0
2
2
√ a r − r (−cos(θ))| dr π
2
2
√ a r − r
2
2
0
2
r = asen(u) dr = acos(u)
dr
π
M y = a
1
|| ||
2
0
0
(
≤ u ≤ π2
y. N .dA
s
a
2
0
M y =
M y = a
ˆ
0
0
M x = a 3 (
ˆ
2
π
2
2
a sen(u) du = a
0
3
ˆ
2
(
1
2
3
sen se n(2u) a π − )= 2 4 4
u
Mx x ¯ = = a
a3 π 4
a2 π
=
4
a a
a
2
x, ¯ y) = ( , ) (¯x, 2 2
≤ u ≤ π2
− cos(2u) )du
0
M y = a 3 (
0
= y¯
−→ F =< xz , x y − z , 2xy + 2
y2z > z = a2
− − − − − − −− √ √ x2
S : : z = a2 x = x
2
3
y 2 , z = 0
x2
y2
y = y
a2 i
z =
x2 j
y2
1 0
T x xT y =
0 1
I
4
ˆ ˆ √ − a2
a
=
0
4
a2
a
=
−x −y
2
a2
−x −y
2
2
0
2
y
x
√ −
=<
a2
x2
−y
2
y
√ −
,
a2
x2
−y
2
, 1 >
3 2
< x(a2 x2 y 2 ), x2 y (a2 x2 y2 ) , 2xy +y 2
− −
0
ˆ ˆ √ −
x
a2
x2
<
I
k
− − −
x
− a2
x2
−y
2
,
y
− a2
x2
(x(a2 x2 y2 )+
− −
0
x2
−y
2
2
(a
a2
x2
−y
2
2
2
− x − y ) −y(a −x −y )+2xy+y 2
2
>.
, 1 > dydx
x2 y 2
−
2
2
( a2
2
2
− x − y ))dydx
x = rcos(θ)
y = rsen(θ )
I
4
π
a
=
ˆ ˆ 0
2
2
(r cos(θ)
2
0
− a
2
2
r +
r 4 cos(θ)2 sen(θ )2
√ a − r 2
+2r2 cos(θ )sen(θ) + r 2 sen(θ)2
I
4 I
4
π
a
=
ˆ ˆ 0
0
a
=
ˆ 0
2
[r3
r3
− a2
r2 +
r2 θ +
I
a
a2
4
ˆ 0
√ a − r 2
5
− =
r5 cos(θ )2 sen(θ )2
[r3
r √ 8 a −r 2
− a2
r2
2
π
2
(θ +
2
− a2
2
r2 )rdrdθ
2 2
4
3
−
r π √ −a r 16 a − r
r = asen(u) dr = acos(u)du
3
)+a2 r2 cos(θ) r4 cos(θ ) r4 cos(θ)
5
+
2
− rsen(θ)(a − r )
−a r sen(θ)+r sen(θ)+r sen(2θ))dθdr
sen(4θ)
4
2
2 2
2
2
0
≤ u ≤ π2
+ r 4 + r 3 ]dr
−
− r cos2(2θ) ]
π
2
0
I
4
π
=
ˆ
2
π
5
3
a 5 π
2
sen(u) )du + ( ( a sen(u) cos(u) + 2 16
0
I
4
=
a5 π
15 I
4
+
a5 π
30
= a 5 (
x = x z = 0 I = =
I
ˆ ˆ
4
=
ˆ ˆ s
+
4
2 − ) + a 10 15
π
4
< 0 , x2 y,
ˆ ˆ 0
−2xy > . < 0, 0, 1 > dA π
a
−2xydA =
ˆ ˆ o
0
2
2
2
−2r cos(θ)sen(θ)rdrdθ
0
3
−r sen(2θ)dθdr = I =
σ = 4a5 (
I =
divF = (
ˆ
−a
σ = σS σ S 1 + σS 2 = 4a2 (
a
0
4
−r
3
x = ρsen(φ)cos(θ ) y = ρsen(φ)sen(θ) z = ρcos(φ)
=(
−r )| = −a 4 4 a
0
4
2 − ) + a − a 10 15
π
4
4
2 − ) 10 15
π
ˆ ˆ ˆ
divF.dv
d d d )(xz 2 , x2 y , , dx dy dz
3
2
− z , 2xy + y z)
divF = (z 2 + x2 + y 2 )
5
− 13 + 15 ) + a4
π
a
=
+
a 2 − )+ 16 15 4
s
4
−3
a 5
1 0 0 T x xT y =< 0 , 0, 1 > 0 1 0
y = y
I
a5
4
+ a5 (
π
I = 4 a2 (
S 2 ;
5
4
a 4
4
)
2π
I = =
π
a
ˆ ˆ ˆ 0
0
π
ˆ ˆ 0
2
ρ .ρ sen(φ)dφdθ =
0
2π
I = =
2π
2
0
0
ρ5
a5
a
( ) 0 sen(φ)dφdθ = 5 5
|
I = =
a5
5
ˆ 0
2π
π
a
ˆ ˆ ˆ 0
ˆ
ρ4 sen(φ)dφdθ
0
2π
0
4πa 5 2dθ = 5
( cos(φ)) 0π dθ
−
|
Q :
x =
16 0
x
− y2 − 4z2
0 = 16 y 2 4z 2 y 2 + 4z 2 = 16 y 4z 16 + 16 = 1 y z 16 + 4 = 1
− −
2
2
2
V 4 V 4
2
4
=
´ ´ 0
4
√
16−y 2 2
0√
´ 16−y 0
2
−4z2
dxdzdy
16−y 2 2
− y2 −√ 4z2dzdy ´ 4 V = 16z − y 2 z − 4 z3 |0 dzdy 4 0 √ ´ 4 √ 16−y V 2 16−y − y 2 − 4 (16−y3 ) dy 4 = 0 16 2 V 16 − y 2 − y2 16 − y 2 − 61 (16 − y 2 )3/2 dy 4 =8 ´ V = 16 − y 2 (4 − y 2 ) − 34 (16 − y 2 )3/2 dy 4 0 =
´ ´ 0
0
16
3
2
4
16−y 2 2
2
2
2
3/2
V 4
=
´ π/
2
8 16cos2 θ
0
´ π/2 π/ 2
V 4
√
− 16sin2θ
√
16cos2 θ
= 0 32Cosθ 32Sin 2 θCosθ = 4[32π 8π 8π] = 64π V =
− −
−
− 61
√
− Cosθdθ
16cos2 θ 4Cosθdθ
F (x,y,z ) = < x 2 y ; (x + y 2 ); (xy2 z ) >
y
z y
= 2x = 9 y 2 =3
−
‚
· ·
·
rotF n dA i j ∂ ∂ rotF = ∂x ∂y x2 y x + y 2 0 , 2y, 1 > N :<‚ T = R rotF n dA R
· · ·
k ∂ ∂z 2
xy z
= (2xyz
− 0)i + (0 − y 2z) j + (1 − x2)k
2
‚ T = ‚ R < 2 xyz, −y z, 1 − x2 >< 0 , 2y, 1 > dA T = R −2y3 z + 1 − x2 dA ´ 3 ´ y/2 y/ 2 T = 0 0 −2y 3 (9 − y 2 ) + 1 − x2 dxdy ´ 3 ´ y/ T = 0 0 −18y3 + 2y5 + 1 − x2 dxdy ´ 3 y/ T = 0 − 18y 3 x + 2 xy 5 + x − x3 |0 dy ´ 3 T = 0 − 18y 3 y2 + 2 y2 y5 − y24 dy ´ y y 4 6 2
3
3
T =
2
3
3
−9y + y + 2 − 24 dy 0 T = − 94 y 5 + y7 + y4 − y96 | 30 T = −123, 56 7
2
4
F (x , y , z) = < z, z, x,y >
z x2 + y 2 + z
2
= x2 + y 2 =1
2
x2 + y + x2 + y 2 = 1 2(x2 + y 2 ) = 1 x2 +‚ y 2 = 21 I = R rotF n dA i j k ∂ ∂ ∂ rotF = ∂x ∂y ∂z z x y
· · · −
r (x.y) = < x,y, 1 y N :< , 2
x2
= i,j,k
− y 2 >
x , 1 > 1−x2 −y 2
√ 1−x −y √ 2
I =
‚
I 4
=
´ √ 2/2 ´ π/2 π/ 2
=
´ √ 2/2
I 4 I 4
R < 1 , 1, 1 ><
0 0
´ √ 2/2
= 0 I = π2
0
√ 1−xy −y , √ 1−xx −y
rCosθ √ 2 1−r
2
2
2
2
rSinθ + √ + 1 rdθdr 1−r 2
π/2 π/2 − √ rCosθ + θ |0 rdr 1−r √ r − π r dr 2 1−r
rSinθ √ 2 1−r 2
2
, 1 > dxdy
2
