ਚੌਕੀ
12 i − j 12 i − j
ਘੁੰਮਣਹਾਰ
6565 ਮੀ/ਸੈ
ਨਹੀਂ।
–1.49063 × 10 −18 , 4.96876 × 10 −19 , –9.93752 × 10 −19 N
ਨਹੀਂ
∇ f = v
P y = x ≠ Q x = −2 x y
ਨਹੀਂ
2 2
2 10 π + 2 10 π 2
ਦੋਵੇਂ ਰੇਖਾ ਸਮਾਕਲ −1000303.−1000303. ਦੇ ਬਰਾਬਰ ਹਨ।
4 17
∫ C F · T d s
−26
0
182π2 ਕਿਲੋਗ੍ਰਾਮ
3/2
2 π
0
ਹਾਂ
ਚਿੱਤਰ ਵਿੱਚ ਖੇਤਰ ਜੁੜਿਆ ਹੋਇਆ ਹੈ। ਚਿੱਤਰ ਵਿੱਚ ਖੇਤਰ ਸਧਾਰਨ ਤੌਰ 'ਤੇ ਜੁੜਿਆ ਹੋਇਆ ਨਹੀਂ ਹੈ।
2
If C1C1 and C2C2 represent the two curves, then ∫C1F·dr≠∫C2F·dr.∫C1F·dr≠∫C2F·dr.
f ( x , y ) = e x y 3 + x y f ( x , y ) = e x y 3 + x y
f ( x , y , z ) = 4 x 3 + sin y cos z + z f ( x , y , z ) = 4 x 3 + sin y cos z + z
f ( x , y , z ) = G x 2 + y 2 + z 2 f ( x , y , z ) = G x 2 + y 2 + z 2
It is conservative.
−10 π −10 π
Negative
45 2 45 2
2 3 2 3
3 π 2 3 π 2
g ( x , y ) = − x cos y g ( x , y ) = − x cos y
No
105 π 105 π
y − z 2 y − z 2
Yes
All points on line y=1.y=1.
− i − i
curl v = 0 curl v = 0
No
Yes
Cylinder x2+y2=4x2+y2=4
Cone x2+y2=z2x2+y2=z2
r(u,v)=〈ucosv,usinv,u〉,r(u,v)=〈ucosv,usinv,u〉, 0<u<∞,0≤v<π20<u<∞,0≤v<π2
Yes
≈ 43.02 ≈ 43.02
With the standard parameterization of a cylinder, Equation 6.18 shows that the surface area is 2πrh.2πrh.
2 π ( 2 + sinh −1 ( 1 ) ) 2 π ( 2 + sinh −1 ( 1 ) )
24
0
38.401 π ≈ 120.640 38.401 π ≈ 120.640
N ( x , y ) = 〈 − y 1 + x 2 + y 2 , − x 1 + x 2 + y 2 , 1 1 + x 2 + y 2 〉 N ( x , y ) = 〈 − y 1 + x 2 + y 2 , − x 1 + x 2 + y 2 , 1 1 + x 2 + y 2 〉
0
400 kg/sec/m
− 440 π 3 − 440 π 3
Both integrals give 00
− π − π
3 2 3 2
curl E = 〈 x , y , −2 z 〉 curl E = 〈 x , y , −2 z 〉
Both integrals equal 6π.6π.
30
9 ln ( 16 ) 9 ln ( 16 )
≈ 6.777 × 10 9 ≈ 6.777 × 10 9
Section 6.1 Exercises
Vectors
False
F ( x , y ) = sin ( y ) i + ( x cos y − sin y ) j F ( x , y ) = sin ( y ) i + ( x cos y − sin y ) j
F ( x , y , z ) = ( 2 x y + y ) i + ( x 2 + x + 2 y z ) j + y 2 k F ( x , y , z ) = ( 2 x y + y ) i + ( x 2 + x + 2 y z ) j + y 2 k
F ( x , y ) = ( 2 x 1 + x 2 + 2 y 2 ) i + ( 4 y 1 + x 2 + 2 y 2 ) j F ( x , y ) = ( 2 x 1 + x 2 + 2 y 2 ) i + ( 4 y 1 + x 2 + 2 y 2 ) j
F ( x , y ) = ( 1 − x ) i − y j ( 1 − x ) 2 + y 2 F ( x , y ) = ( 1 − x ) i − y j ( 1 − x ) 2 + y 2
F ( x , y ) = - x i − y j x 2 + y 2 F ( x , y ) = - x i − y j x 2 + y 2
F ( x , y ) = y i − x j F ( x , y ) = y i − x j
F ( x , y ) = −10 ( x 2 + y 2 ) 3 / 2 ( x i + y j ) F ( x , y ) = −10 ( x 2 + y 2 ) 3 / 2 ( x i + y j )
E = c x 2 + y 2 x 2 + y 2 = c x 2 + y 2 = c r E = c x 2 + y 2 x 2 + y 2 = c x 2 + y 2 = c r
c ′ ( t ) = ( cos t , − sin t , e − t ) = F ( c ( t ) ) c ′ ( t ) = ( cos t , − sin t , e − t ) = F ( c ( t ) )
H
d. −F+G−F+G
a. F+GF+G
Section 6.2 Exercises
True
False
False
∫ C ( x − y ) d s = 10 ∫ C ( x − y ) d s = 10
∫ C x y 4 d s = 8192 5 ∫ C x y 4 d s = 8192 5
W = 8 W = 8
W = 3 π 4 W = 3 π 4
W = π W = π
∫ C F · d r = 4 ∫ C F · d r = 4
∫ C y z d x + x z d y + x y d z = −1 ∫ C y z d x + x z d y + x y d z = −1
∫ C ( y 2 ) d x + ( x ) d y = 245 6 ∫ C ( y 2 ) d x + ( x ) d y = 245 6
∫ C x y d x + y d y = 190 3 ∫ C x y d x + y d y = 190 3
∫ C y 2 x 2 − y 2 d s = 2 ln 5 ∫ C y 2 x 2 − y 2 d s = 2 ln 5
W = −66 W = −66
W = −10 π 2 W = −10 π 2
W = 2 W = 2
a. W=11;W=11; b. W=394;W=394; c. No
W = 2 π W = 2 π
∫ C x y d s = 25 5 + 1 120 ∫ C x y d s = 25 5 + 1 120
∫ C y 2 d x + ( x y − x 2 ) d y = 6.15 ∫ C y 2 d x + ( x y − x 2 ) d y = 6.15
∫ γ x e y d s ≈ 7.157 ∫ γ x e y d s ≈ 7.157
∫ γ ( y 2 − x y ) d x ≈ −1.379 ∫ γ ( y 2 − x y ) d x ≈ −1.379
∫ C F · d r ≈ −1.133 ∫ C F · d r ≈ −1.133
∫ C F · d r ≈ 2 2 . 8 5 7 ∫ C F · d r ≈ 2 2 . 8 5 7
flux = − 1 3 flux = − 1 3
flux = −20 flux = −20
flux = 0 flux = 0
m = 4 π ρ 5 m = 4 π ρ 5
W = 0 W = 0
W = k 2 W = k 2
Section 6.3 Exercises
True
True
∫ C F · d r = 24 ∫ C F · d r = 24
∫ C F · d r = e − 3 π 2 ∫ C F · d r = e − 3 π 2
Not conservative
Conservative, f(x,y)=3x2+5xy+2y2f(x,y)=3x2+5xy+2y2
Conservative, f(x,y)=yex+xsin(y)f(x,y)=yex+xsin(y)
∫ C ( 2 y d x + 2 x d y ) = 32 ∫ C ( 2 y d x + 2 x d y ) = 32
F ( x , y ) = ( 10 x + 3 y ) i + ( 3 x + 20 y ) j F ( x , y ) = ( 10 x + 3 y ) i + ( 3 x + 20 y ) j
F is not conservative.
F is conservative and a potential function is f(x,y,z)=xyez.f(x,y,z)=xyez.
F is conservative and a potential function is f(x,y,z)=z2–z–xy.f(x,y,z)=z2–z–xy.
F is conservative and a potential function is f(x,y,z)=x2y+y2z.f(x,y,z)=x2y+y2z.
F is conservative and a potential function is f(x,y)=ex2yf(x,y)=ex2y
∫ C F · d r = e 2 + 1 ∫ C F · d r = e 2 + 1
∫ C F · d r = -2 ∫ C F · d r = -2
∫ C 1 G · d r = −8 π ∫ C 1 G · d r = −8 π
∫ C 2 G · d r = 7 ∫ C 2 G · d r = 7
∫ C F · d r = 159 ∫ C F · d r = 159
∫ C F · d r = −1 ∫ C F · d r = −1
4 × 10 29 erg 4 × 10 29 erg
∫ C F · d r ≈ 2 . 9923 ∫ C F · d r ≈ 2 . 9923
circulation = π a 2 and flux = 0 circulation = π a 2 and flux = 0
Section 6.4 Exercises
∫ C 2 x y d x + ( x + y ) d y = 32 3 ∫ C 2 x y d x + ( x + y ) d y = 32 3
∫ C sin x cos y d x + ( x y + cos x sin y ) d y = 1 12 ∫ C sin x cos y d x + ( x y + cos x sin y ) d y = 1 12
∫ C ( − y d x + x d y ) = π ∫ C ( − y d x + x d y ) = π
∫ C x e −2 x d x + ( x 4 + 2 x 2 y 2 ) d y = 0 ∫ C x e −2 x d x + ( x 4 + 2 x 2 y 2 ) d y = 0
∫Cy3dx−x3ydy=−20π∫Cy3dx−x3ydy=−20π
∫ C − x 2 y d x + x y 2 d y = 8 π ∫ C − x 2 y d x + x y 2 d y = 8 π
∫ C ( x 2 + y 2 ) d x + 2 x y d y = 0 ∫ C ( x 2 + y 2 ) d x + 2 x y d y = 0
A = 19 π A = 19 π
A = 3 π 8 A = 3 π 8
∫ C + ( y 2 + x 3 ) d x + x 4 d y = 0 ∫ C + ( y 2 + x 3 ) d x + x 4 d y = 0
A = 9 π 8 A = 9 π 8
A = 8 3 5 A = 8 3 5
∫ C ( x 2 y − 2 x y + y 2 ) d s = 1 2 ∫ C ( x 2 y − 2 x y + y 2 ) d s = 1 2
∫ C x d x + y d y x 2 + y 2 = 0 ∫ C x d x + y d y x 2 + y 2 = 0
W = 225 2 W = 225 2
W = 12 π W = 12 π
W = 2 π W = 2 π
∫ C y 2 d x + x 2 d y = 1 3 ∫ C y 2 d x + x 2 d y = 1 3
∫ C 1 + x 3 d x + 2 x y d y = -3 ∫ C 1 + x 3 d x + 2 x y d y = -3
∫ C ( 3 y − e sin x ) d x + ( 7 x + y 4 + 1 ) d y = 36 π ∫ C ( 3 y − e sin x ) d x + ( 7 x + y 4 + 1 ) d y = 36 π
∫ C F · d r = 2 ∫ C F · d r = 2
∫ C ( y + x ) d x + ( x + sin y ) d y = 0 ∫ C ( y + x ) d x + ( x + sin y ) d y = 0
∫ C x y d x + x 3 y 3 d y = 22 21 ∫ C x y d x + x 3 y 3 d y = 22 21
∫ C F · d r = 15 π 4 ∫ C F · d r = 15 π 4
∫ C sin ( x + y ) d x + cos ( x + y ) d y = 4 ∫ C sin ( x + y ) d x + cos ( x + y ) d y = 4
∫ C F · d r = π ∫ C F · d r = π
∫ C F · N ^ d s = 4 ∫ C F · N ^ d s = 4
∫ C F · N d s = 0 ∫ C F · N d s = 0
∫ C [ − y 3 + sin ( x y ) + x y cos ( x y ) ] d x + [ x 3 + x 2 cos ( x y ) ] d y = 4.7124 ∫ C [ − y 3 + sin ( x y ) + x y cos ( x y ) ] d x + [ x 3 + x 2 cos ( x y ) ] d y = 4.7124
∫ C ( y + e x ) d x + ( 2 x + cos ( y 2 ) ) d y = 1 3 ∫ C ( y + e x ) d x + ( 2 x + cos ( y 2 ) ) d y = 1 3
Section 6.5 Exercises
False
True
True
curl F = i + x 2 j + y 2 k curl F = i + x 2 j + y 2 k
curl F = ( x z 2 − x y 2 ) i + ( x 2 y − y z 2 ) j + ( y 2 z − x 2 z ) k curl F = ( x z 2 − x y 2 ) i + ( x 2 y − y z 2 ) j + ( y 2 z − x 2 z ) k
curl F = i + j + k curl F = i + j + k
curl F = − y i − z j − x k curl F = − y i − z j − x k
curl F = 0 curl F = 0
div F = 3 y z 2 + 2 y sin z + 2 x e 2 z div F = 3 y z 2 + 2 y sin z + 2 x e 2 z
div F = 2 ( x + y + z ) div F = 2 ( x + y + z )
div F = 1 x 2 + y 2 div F = 1 x 2 + y 2
div F = a + b div F = a + b
div F = x + y + z div F = x + y + z
Harmonic
div ( F × G ) = 2 z + 3 x div ( F × G ) = 2 z + 3 x
div F = 2 ( x 2 + y 2 + z 2 ) div F = 2 ( x 2 + y 2 + z 2 )
curl r = 0 curl r = 0
curl r r 3 = 0 curl r r 3 = 0
curl F = 2 x x 2 + y 2 k curl F = 2 x x 2 + y 2 k
div F = 0 div F = 0
div F = 2 − 2 e −6 div F = 2 − 2 e −6
div F = 0 div F = 0
curl F = j − 3 k curl F = j − 3 k
curl F = 2 j − k curl F = 2 j − k
a = 3 a = 3
F is conservative.
div F = cosh x + sinh y − x y div F = cosh x + sinh y − x y
( b z − c y ) i + ( c x − a z ) j + ( a y − b x ) k ( b z − c y ) i + ( c x − a z ) j + ( a y − b x ) k
curl F = 2 ω curl F = 2 ω
F×GF×G does not have zero divergence.
∇ · F = −200 k [ 1 + 2 ( x 2 + y 2 + z 2 ) ] e − x 2 + y 2 + z 2 ∇ · F = −200 k [ 1 + 2 ( x 2 + y 2 + z 2 ) ] e − x 2 + y 2 + z 2
Section 6.6 Exercises
True
True
r(u,v)=〈u,v,2−3u+2v〉r(u,v)=〈u,v,2−3u+2v〉 for −∞≤u<∞−∞≤u<∞ and −∞≤v<∞.−∞≤v<∞.
r(u,v)=〈u,v,13(16−2u+4v)〉r(u,v)=〈u,v,13(16−2u+4v)〉 for |u|<∞|u|<∞ and |v|<∞.|v|<∞.
r(u,v)=〈3cosu,3sinu,v〉r(u,v)=〈3cosu,3sinu,v〉 for 0≤u≤π2,0≤v≤30≤u≤π2,0≤v≤3
A = 28 π = 87.9646 A = 28 π = 87.9646
∬ S z d S = 8 π ∬ S z d S = 8 π
∬ S ( x 2 + y 2 ) z d S = 16 π ∬ S ( x 2 + y 2 ) z d S = 16 π
∬ S F · N d S = 4 π 3 ∬ S F · N d S = 4 π 3
m ≈ 13.0639 m ≈ 13.0639
m ≈ 228.5313 m ≈ 228.5313
∬ S g d S = 3 14 ∬ S g d S = 3 14
∬ S ( x - y 2 + z ) d S ≈ 0.9617 ∬ S ( x - y 2 + z ) d S ≈ 0.9617
∬ S ( x 2 + y 2 ) d S = 4 π 3 ∬ S ( x 2 + y 2 ) d S = 4 π 3
∬ S x 2 z d S = 1023 π 2 5 ∬ S x 2 z d S = 1023 π 2 5
∬ S ( z + y ) d S ≈ 10.1 ∬ S ( z + y ) d S ≈ 10.1
m = π a 3 m = π a 3
∬ S F · N d S = 13 24 ∬ S F · N d S = 13 24
∬ S F · N d S = 3 4 ∬ S F · N d S = 3 4
∫ 0 8 ∫ 0 6 ( 4 − 3 y + 1 16 y 2 + z ) ( 1 4 17 ) d z d y ∫ 0 8 ∫ 0 6 ( 4 − 3 y + 1 16 y 2 + z ) ( 1 4 17 ) d z d y
∫ 0 2 ∫ 0 6 [ x 2 − 2 ( 8 − 4 x ) + z ] 17 d z d x ∫ 0 2 ∫ 0 6 [ x 2 − 2 ( 8 − 4 x ) + z ] 17 d z d x
∬ S ( x 2 z + y 2 z ) d S = π a 5 2 ∬ S ( x 2 z + y 2 z ) d S = π a 5 2
∬ S x 2 y z d S = 171 14 ∬ S x 2 y z d S = 171 14
∬ S y z d S = 2 π 4 ∬ S y z d S = 2 π 4
∬ S ( x i + y j ) · d S = 16 π ∬ S ( x i + y j ) · d S = 16 π
m = π a 7 192 m = π a 7 192
F ≈ 4.57 lb . F ≈ 4.57 lb .
8 π a 8 π a
The net flux is zero.
Section 6.7 Exercises
∬ S ( curl F · N ) d S = π a 2 ∬ S ( curl F · N ) d S = π a 2
∬ S ( curl F · N ) d S = 18 π ∬ S ( curl F · N ) d S = 18 π
∬ S ( curl F · N ) d S = −8 π ∬ S ( curl F · N ) d S = −8 π
∬ S ( curl F · N ) d S = 0 ∬ S ( curl F · N ) d S = 0
∫ C F · d r = 0 ∫ C F · d r = 0
∫ s F · d r = - 3 π ≈ - 9 . 4248 ∫ s F · d r = - 3 π ≈ - 9 . 4248
∬ S curl F · d S = 0 ∬ S curl F · d S = 0
∬ S curl F · d S = 2.6667 ∬ S curl F · d S = 2.6667
∬ S ( curl F · N ) d S = − 1 6 ∬ S ( curl F · N ) d S = − 1 6
∫ C ( 1 2 y 2 d x + z d y + x d z ) = − π 4 ∫ C ( 1 2 y 2 d x + z d y + x d z ) = − π 4
∬ S ( curl F · N ) d S = 3 π ∬ S ( curl F · N ) d S = 3 π
∫ C ( c k × R ) · d r = 2 π c ∫ C ( c k × R ) · d r = 2 π c
∬ S curl F · d S = 0 ∬ S curl F · d S = 0
∫ C F · d r = −4 ∫ C F · d r = −4
∬ S curl F · d S = 0 ∬ S curl F · d S = 0
∬ S curl F · d S = −36 π ∬ S curl F · d S = −36 π
∬ S curl F · d S = 0 ∬ S curl F · d S = 0
∫ C F · d r = 0 ∫ C F · d r = 0
∬ S curl ( F ) · d S = 84.8230 ∬ S curl ( F ) · d S = 84.8230
A = ∬ S curl F · d S = 0 A = ∬ S curl F · d S = 0
∬ S curl F · d S = 2 π ∬ S curl F · d S = 2 π
C = π ( cos φ − sin φ ) C = π ( cos φ − sin φ )
∫ C F · d r = 48 π ∫ C F · d r = 48 π
∬ S curl F · d S = 0 ∬ S curl F · d S = 0
0
Section 6.8 Exercises
∫ S F · N d s = 24 π ≈ 75.3982 ∫ S F · N d s = 24 π ≈ 75.3982
∫ S F · N d s = 243 π 2 ≈ 381 . 704 ∫ S F · N d s = 243 π 2 ≈ 381 . 704
∫ S F · N d s = 12 π ≈ 37.6991 ∫ S F · N d s = 12 π ≈ 37.6991
∫ S F · N d s = 9 π a 4 2 ∫ S F · N d s = 9 π a 4 2
∬ S F · d S = 4 π 3 ∬ S F · d S = 4 π 3
∬ S F · d S = 0 ∬ S F · d S = 0
∬ S F · d S = 384 π 5 ≈ 241.2743 ∬ S F · d S = 384 π 5 ≈ 241.2743
∬DF·dS=∬DF·dS= Net flux =0=0; flux through the paraboloid =–π=–π
∬ S F · d S = 2 π 3 ∬ S F · d S = 2 π 3
16 6 π 16 6 π
− 128 π 3 − 128 π 3
– 224 π ≈ −703.7168 – 224 π ≈ −703.7168
20
∬ S F · d S = 8 ∬ S F · d S = 8
∬ S F · N d S = 1 8 ∬ S F · N d S = 1 8
∬ S ‖ R ‖ R · n d S = 4 π a 4 ∬ S ‖ R ‖ R · n d S = 4 π a 4
∭ R z 2 d V = 4 π 15 ∭ R z 2 d V = 4 π 15
∬ S F · d S = 3 e - 2 + 1 2 sin 1 ≈ 6.5759 ∬ S F · d S = 3 e - 2 + 1 2 sin 1 ≈ 6.5759
∬ S F · d S = 21 ∬ S F · d S = 21
∬ S F · d S = 72 ∬ S F · d S = 72
∬ S F · d S = - 32 π 3 ≈ −33.5103 ∬ S F · d S = - 32 π 3 ≈ −33.5103
∬ S F · d S = π a 4 b 2 ∬ S F · d S = π a 4 b 2
∬ S F · d S = 5 π 2 ∬ S F · d S = 5 π 2
∬ S F · d S = 21 π 2 ∬ S F · d S = 21 π 2
e - 1 - 1 e - 1 - 1
Review Exercises
False
False
Conservative, f(x,y)=xy−2eyf(x,y)=xy−2ey
Conservative, f(x,y,z)=x2y+y2z+z2xf(x,y,z)=x2y+y2z+z2x
− 16 3 − 16 3
32 2 9 ( 3 3 − 1 ) 32 2 9 ( 3 3 − 1 )
Divergence: ex+xexy+xyexyz,ex+xexy+xyexyz, curl: xzexyzi−yzexyzj+yexykxzexyzi−yzexyzj+yexyk
−18 π −18 π
− π − π
24 π 24 π
2 ( 2 2 + π ) = 4 + π 2 2 ( 2 2 + π ) = 4 + π 2
8 π / 3 8 π / 3