Solution Manual - Calculus 8th Edition Varberg, Purcell, Rigdon Ch14 PDF

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CHAPTER

14

14.1 Concepts Review

Vector Calculus 6.

1. vector-valued function of three real variables or a vector field 2. gradient field 3. gravitational fields; electric fields 4. ∇ ⋅ F, ∇ × F

Problem Set 14.1 1.

7.

2 x − 3 y , −3x, 2

8. (cos x yz) yz, xz, xy 9.

f ( x, y, z) = ln x + ln y + ln z ;

∇f ( x, y, z) = x –1, y –1, z –1

2.

10.

x, y , z

11. e y cos z, x cos z, – x sin z 12. ∇f ( x, y, z) = 0, 2 ye–2z , – 2 y2 e–2z = 2ye –2 z 0, 1, – y

3. 13. div F = 2x – 2x + 2yz = 2yz curl F = z 2 , 0, – 2 y

14. div F = 2x + 2y + 2z curl F = 0, 0, 0 = 0 4.

15. div F =∇ ⋅ F = 0 + 0 + 0 = 0 curl F =∇ × F = x − x, y − y, z − z = 0 16. div F = –sin x + cos y + 0 curl F = 0, 0, 0 = 0 17. div F=e x cos y + e x cos y + 1 = 2e x cos y + 1

5.

curl F = 0, 0, 2e x sin y

18. div F = ∇ ⋅ F = 0 + 0 + 0 = 0 curl F = ∇ × F = 1-1,1-1,1-1 = 0

858 Section 14.1 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

19. a.

meaningless

g.

b. vector field

vector field

h. meaningless

vector field

i.

meaningless

d. scalar field

j.

scalar field

c.

e.

vector field

f.

vector field

20. a.

k. meaningless

div(curl F) = div ⋅ Py – N z , M z – Px , N x – M y

= ( Pyx – N zx ) + ( M zy – Pxy ) + ( N xz – M yz ) = 0

b.

curl(grad f ) = curl f x, f y , f z

= fzy – f yz , fxz – fzx , fyx – fxy = 0

c.

div( fF)=div fM , fN , fP = ( fM x + f x M ) + ( fN y + f y N ) + ( fPz + f z P) = ( f )( M x + N y + Pz ) + ( fx M + f y N + fz P) = ( f )(div F) + (grad f ) ⋅ F

d.

curl( fF)=curl fM , fN , fP = ( f Py + f y P) – ( fN z + f z N), ( fM z + fz M ) – ( fPx + f x P), ( fN x + f x N ) – ( fM y + f y M ) = ( f ) Py – N z , M z – Px , N x – M y + fx , f y , f z × M , N , P = (f)(curl F) + (grad f) × F

21. Let f ( x, y , z ) = – c r grad( f ) = 3c r

–4

–3

23. grad f = f ′(r ) xr –1/ 2 , f ′(r ) yr –1/ 2 , f ′( r ) zr –1/ 2

, so

(if r ≠ 0) = f ′(r )r –1/ 2 x, y, z = f ′( r ) r–1/ 2 r

r –5 = 3c r r . r

(

Then curl F = curl ⎡ –c r ⎣⎢

–3

) r⎤⎦⎥ r ) × r (by 20d)

curl F = [ f (r )][curl r] + [ f ′(r )r –1/ 2 r] × r = [ f (r )][curl r] + [ f ′(r )r –1/ 2r] × r =0+0=0

( ) ( = ( – c r ) (0) + (3c r ) (r × r) = 0 + 0 = 0 div F =div ⎡ ( –c r )r⎤ ⎢⎣ ⎥⎦ = ( – c r ) (div r) + (3c r r ) ⋅ r (by 20c) = ( – c r ) (1 + 1 + 1) + ( 3c r ) r = ( –3c r ) + 3c r = 0 = –c r

–3

–5

(curl r) + 3c r

–3

–5

24. div F = div[f(r)r] = [f(r)](div r) + grad[f(r)] ⋅ r = [ f (r )](div r) + [ f ′(r ) r –1r] ⋅ r

–3

–3

= [ f (r )](3) +[ f ′(r )r –1]( r ⋅ r)

–5

–3

–5

–3

3

22. curl ⎡−c r ⎣⎢ =0

−m

div ⎡– c r ⎣⎢

–m

= ( m – 3) c r

2

(

−m

(

–m

r ⎤⎥ = −c r ⎦

r ⎤⎥ = – c r ⎦

) (0) + mc r

) (3) + mc r

− m −2

– m–2

( 0)

= 3 f ( r) +[ f ′( r) r –1]( r 2) = 3 f ( r) + rf ′( r) Now if div F = 0, and we let y = f(r), we have the dy differential equation 3 y + r = 0, which can be dr solved as follows: dy dr = –3 ; ln y = –3ln r + ln C = ln Cr – 3 , y r

for each C ≠ 0. Then y = Cr –3 , or r

2

f (r ) = Cr –3 , is a solution (even for C = 0).

–m

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Let P = ( x 0 , y 0 ). div F = div H = 0 since there is no tendency toward P except along the line x = x0 , and along that line the tendencies toward and away from P are balanced; div G < 0 since there is no tendency toward P except along the line x = x 0 , and along that line there is more tendency toward than away from P; div L > 0 since the tendency away from P is greater than the tendency toward P.

25. a.

No rotation for F, G, L; clockwise rotation for H since the magnitudes of the forces to the right of P are less than those to the left.

b.

b. If a paddle wheel is placed at the point (with its axis perpendicular to the plane), the velocities over the top half of the wheel will exceed those over the bottom, resulting in a net clockwise motion. Using the right-hand rule, we would expect curl F to point into the plane (negative z). By calculating curl F = (0 − 0) i + (0 − 0) j+ (0 − 1) k = − k 27. F ( x, y , z ) = M i + N j+ P k , where y x , , N =− M =− 2 2 2 32 (1 + x + y 2) 3 2 (1 + x + y ) P =0 1

div F = 0; curl F = 0

c.

2

div G = –2ye – y < 0 since y > 0 at P; curl G=0

0.5

div L = ( x2 + y2 )–1/ 2 ; curl L = 0

div H = 0; curl H = 0, 0, – 2 xe –x

2

which

-1

-0.5

0.5

1

points downward at P, so the rotation is clockwise in a right-hand system. -0.5

26. F (x , y , z ) = M i + N j+ P k , where M (x , y , z ) = y , N ( x, y , z) = 0, P( x, y, z) = 0

-1

2

0.5

a. Since all the vectors are directed toward the origin, we would expect accumulation at that point; thus div F (0, 0, 0) .should be negative. Calculating, 3( x2 + y2 ) divF ( x, y, z) = (1 + x 2 + y 2) 5 2 2 − +0 2+ 2 32 + y ) (1 x so that div F (0,0,0) = −2

a. Since the velocity into (1, 1, 0) equals the velocity out, there is no tendency to diverge from or accumulate to the point. Geometrically, it appears that div F (1,1, 0) = 0 . Calculating,

b. If a paddle wheel is placed at the origin (with its axis perpendicular to the plane), the force vectors all act radially along the wheel and so will have no component acting tangentially along the wheel. Thus the wheel will not turn at all, and we would expect curl F = 0 . By calculating curl F = (0 − 0) i + (0 − 0) j

1.5

1

1.2

1.4

div F( x, y, z) =

1.6

1.8

2

∂M ∂N ∂P + + = 0 +0 +0 = 0 ∂x ∂y ∂z

⎛ ⎞ 3yx 3xy ⎟k +⎜ − ⎜ (1+ x2 + y2 )3 2 (1+ x2 + y2 )3 2 ⎟ ⎝ ⎠ =0

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28. div v = 0 + 0 + 0 = 0; curl v = 0, 0, w + w = 2ω k 29. ∇f ( x, y, z ) =

30. ∇ 2 f = div(grad f ) = div f x , f y , f z = f xx + f yy + f zz

1 mω 2 2 x, 2 y, 2 z = mω 2 x, y, z 2

a.

∇2 f = 4 – 2 – 2 = 0

b.

∇ 2 f = 0 +0 +0 = 0

c.

∇ 2 f = 6 x – 6x + 0 = 0

d.

∇ 2 f = div(grad f ) = div grad r

= F(x, y, z)

(

)

(

–1

)

= div – r –3 r = 0 (by problem 21)

Hence, each is harmonic. F × G = ( f yg z − f z g y )i − ( f x g z − f z g x )j + ( f x g z − f z g x )k . Therefore,

31. a.

∂ ∂ ∂ ( f yg z − f z g y ) − ( f x g z − f z gx ) + ( fx g y − f y g x ) . ∂x ∂y ∂z Using the product rule for partials and some algebra gives ∂f y ⎤ ⎡∂ f ⎡ ∂f y ∂fx ⎤ ⎡ ∂fx ∂ fz ⎤ − + gz ⎢ − div ( F × G) = g x ⎢ z − ⎥+gy ⎢ ⎥ ⎥ ⎣ ∂z ∂ x ⎦ ⎣ ∂y ∂ z ⎦ ⎣ ∂ x ∂y ⎦ ∂gy ⎤ ⎡ ∂g ⎡ ∂ g y ∂g x ⎤ ⎡ ∂g x ∂g z ⎤ +fx ⎢ z − − + fz ⎢ − ⎥ + fy ⎢ ⎥ ⎥ ∂z ⎦ ∂x ⎦ ∂y ⎦ ⎣ ∂z ⎣ ∂y ⎣ ∂x = G ⋅ curl( F) − F ⋅ curl( G) div ( F × G) =

⎛ ∂f ∂g ∂f ∂g ⎞ ⎛ ∂f ∂g ∂f ∂g ⎞ ⎛ ∂f ∂g ∂f ∂g ⎞ ∇f × ∇g = ⎜ − − − ⎟i− ⎜ ⎟ j+⎜ ⎟k ⎝ ∂y ∂z ∂z ∂y ⎠ ⎝ ∂x ∂z ∂z ∂x ⎠ ⎝ ∂x ∂y ∂y ∂x ⎠ ∂ ⎛ ∂f ∂ g ∂ f ∂g ⎞ ∂ ⎛ ∂f ∂g ∂f ∂g ⎞ ∂ ⎛ ∂f ∂g ∂f ∂g ⎞ − − − + − . Therefore, div (∇f × ∇g ) = ⎜ ∂x ⎝ ∂ y ∂ z ∂ z ∂ y ⎟⎠ ∂ y⎜⎝ ∂ x ∂ z ∂ z ∂ x ⎟⎠ ∂ z⎜⎝ ∂ x ∂ y ∂ y ∂ x⎟⎠ Using the product rule for partials and some algebra will yield the result div (∇f × ∇g ) = 0

b.

32.

lim

( x, y, z) →( a, b, c)

F (x , y , z ) = L if for each ε > 0 there is a δ > 0 such that 0 < x, y, z – a, b, c < δ implies that

F ( x, y, z ) – L < ε.

F is continuous at (a, b, c) if and only if

lim

( x , y , z )→ (a , b , c )

= F (a, b, c ).

14.2 Concepts Review 1. Increasing values of t n

2.

∑ f ( x i, y i ) Δs i i =1

3.

f (x (t ), y (t )) [ x′ (t )]2 + [ y′ (t )]2

4. F ⋅

dr dt

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Problem Set 14.2 1.

1

∫0 (27t

3

(

)

+ t3 )(9 + 9t 4 )1/ 2 dt = 14 2 2 –1 ≈ 25.5980 1/ 2

⎞ ⎛ 1 25t 3 ⎞ 2. ∫ ⎜ ⎟ (t ) ⎜ + ⎟ ⎜4 0⎝ 2 ⎠ 4 ⎟⎠ ⎝ 1⎛ t

⎛ 1 ⎞ 3/ 2 –1) ≈ 0.2924 dt = ⎜ ⎟ (26 ⎝ 450 ⎠

3. Let x = t, y = 2t, t in [0, π ] . π

∫C (sin x + cos y )ds = ∫0 (sin t + cos 2t )

Then

1 + 4dt = 2 5 ≈ 4.4721

4. Vector equation of the segment is x , y = −1, 2 + t 2, −1 , t in [0, 1]. 1

∫0 (–1+ 2t )e 1

5.

∫0 (2 t + 9 t

6.

∫0

2π

3

2–t

(4 + 1)1/ 2 dt = 5e2 (1 – 3e –1 ) ≈ −1.7124

⎛ 1⎞ )(1 + 4 t 2 + 9 t 4 )1/ 2 dt = ⎜ ⎟ (143 / 2 –1) ≈ 8.5639 ⎝ 6⎠

(16cos2 t +16sin2 t + 9t 2 )(16sin t 2 +16cos2 t + 9)1/ 2 dt = ∫

2π 0

(16 + 9t 2 )(5)dt

2π

= ⎡80t +15t 3 ⎤ = 160 π +120 π3 ≈ 4878.11 ⎣ ⎦0 2

2

–1)(2) + (4t 2 )(2t )]dt =

7.

∫0 [(t

8.

∫0 (–1)dx + ∫–1(4)

9.

∫C y

10.

∫–2[(t

4

3

3

2

dy = 60

dx + x3 dy = ∫

C1

1

2

100 3

y3 dx + x3 dy + ∫

C2

– 3) 3(2) + (2 t) 3(2 t)] dt =

y3 dx + x3 dy = ∫

–2

1

2

(–4)3 dy + ∫ (–2)3 dx = 192 + (–48) = 144 –4

828 ≈ 23.6571 35

11. y = –x + 2 3

∫1 ([x + 2(–x + 2)](1) + [x – 2(– x + 2)](–1))dx = 0 12.

1

∫0 [x

2

1

+ (x )2x ]dx = ∫ 3x 2dx = 1 0

(letting x be the parameter; i.e., x = x, y = x 2 ) 13.

x, y , z = 1, 2,1 + t 1, -1, 0 1

∫0 [(4 – t)(1) + (1 + t)(–1) – (2 – 3 t + t 14

1

2 )(–1)]

∫0 [(e t )(et ) + (e 3

–t

dt =

17 ≈ 2.8333 6

5 ⎛1 ⎞ ⎛2 ⎞ ⎛1 ⎞ ≈ 23.9726 + e2t )(–3 –t ) + ( et )(2 e2t )] dt = ⎜ ⎟ e 4 + ⎜ ⎟ e3 – e + ⎜ ⎟ e –2 – 4 3 2 12 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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15. On C1 : y = z = dy= dz = 0 On C2 : x = 2, z = dx = dz = 0 On C3 : x = 2, y = 3, dx = dy = 0

2 4 ⎡ x2 ⎤ ⎡ 3 4 z2 ⎤ 2 3 y y z x [2 – ] 7 – (2 – 2 ) (4 3 – ) y dy + ∫ + z dz = ⎢ ⎥ + ⎥ = 2 + (–3) + 20 = 19 0 +⎢ ∫0 dx + ∫0 0 2 ⎦⎥ ⎢⎣ ⎣⎢ 2 ⎦⎥0 0 2

16.

x, y, z = t 2,3, 4 , t in [0, 1]. 1

∫0 [(9t )(2) + (8t )(3) + (3t )(4)]dt = 27 17. m = ∫ k x ds = ∫ C

⎛k ⎞ 2 1/ 2 k x (1 + 4 x ) dx = ⎜ ⎟ (173 / 2 –1) ≈ 11.6821k ⎝6 ⎠

2 –2

18. Let δ(x, y, z) = k (a constant). m = k ∫ 1 ds = k ∫ C

3π 0

1( a2 sin 2 t + a 2 cos 2 t + b2 )1/ 2 dt = 3πk (a2 + b2 )1/ 2

M xy = k ∫ z ds = k ( a 2 + b 2) 1/ 2 ∫ C

3π 0

M xz = k ∫ y ds = k( a2 + b2 )1/ 2 ∫ C

M yz = k ∫ x ds = C

Therefore, x =

3π

2 2 2 1/ 2 9π bk (a + b ) 2

bt dt =

a sin t dt = ak (a 2 + b 2 )1/ 2 (2) = 2ak (a 2 + b 2 )1/ 2

0 2 2 1/ 2 3π k (a + b ) a cos t dt 0

∫

M yz

= 0; y =

m

= ak (a 2 + b 2) 1/ 2(0) = 0

Mxy 3 π b M xz 2a . ; z = = = m m 3π 2

0

19

∫C (x

20.

∫C ex dx – e –y dy = ∫1 ⎣⎢ (t 3 )⎜⎝ t ⎟⎠ – ⎜⎝ 2t ⎟⎜⎠⎝ t ⎟⎠⎥⎦ dt = 123.6

3

3

2

– y )dx + xy dy = ∫ [(t 6 – t 9 )(2t ) + (t 2 )(t 6 )(3t 2 )]dt = − –1

5⎡

⎛3 ⎞ ⎛ 1 ⎞⎛1 ⎞⎤

21. W = ∫ F ⋅ d r = ∫ (x + y )dx + (x – y )dy = ∫ C

=∫

C

[( a cos t + b sin t )(– a sin t ) + (a cos t – b sin t )(b cos t )]dt

[–( a 2 + b 2) sin t cos t + ab(cos 2 t – sin 2 t )]dt = π/ 2

⎡( a 2 + b 2) cos 2t ab sin 2t ⎤ =⎢ + ⎥ 4 2 ⎣⎢ ⎦⎥ 0

22.

π/2

0

π/ 2

0

7 ≈ −0.1591 44

=

π / 2 –(a 2

∫0

+ b 2 )sin 2t + ab cos 2t dt 2

a 2 +b 2 –2

x, y, z = t 1,1,1 , t in [0, 1]. 1

∫C (2x – y) dx + 2 z dy + ( y – z) dz = ∫0 ( t + 2 t + 0) dt = 1.5

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23.

π ⎡⎛π

⎞

⎛πt⎞

⎛πt⎞

⎛ π t⎞

2 ⎛ πt⎞ ⎤ = ⎟ – t⎥ dt 2 – π ≈ 1.3634 2⎠ ⎦

∫0 ⎢⎣ ⎜⎝ 2 ⎟⎠ sin⎜⎝ 2 ⎟⎠ cos⎜⎝ 2 ⎟⎠ + π t cos⎜⎝ 2 ⎟⎠ + sin⎜⎝

24. W = ∫ F ⋅ d r = ∫ y dx + z dy + x dz = C

=

64 5

+ 12 +

C

8 3

=

25. The line integral

412 15

2

∫0 [(t

2

3

2

2

)(1) + (t )(2t ) + (t )(3t )]dt = ∫ (2t 4 + 3t 3 + t 2 ) dt 0

≈ 27.4667

∫ F • dr represents the work done in moving a particle through the force field F along the curve Ci

Ci , i = 1, 2, 3,

a. In the first quadrant, the tangential component to C1 of each force vector is in the positive y-direction , the same direction as the object moves along C1 . Thus the line integral (work) should be positive. b. The force vector at each point on C2 appears to be tangential to the curve, but in the opposite direction as the object moves along C2 . Thus the line integral (work) should be negative. c. The force vector at each point on C3 appears to be perpendicular to the curve, and hence has no component in the direction the object is moving. Thus the line integral (work) should be zero 26. The line integral

∫ F • dr represents the work done in moving a particle through the force field F along the curve Ci

Ci , i = 1, 2, 3,

a. In the first quadrant, the tangential component to C1 of each force vector is in the positive y-direction , the same direction as the object moves along C1 . Thus the line integral (work) should be positive. b. The force vector at each point on C2 appears to be perpendicular to the curve, and hence has no component in the direction the object is moving. Thus the line integral (work) should be zero c. The force vector at each point on C3 is along the curve, and in the same direction as the movement of the object. Thus the line integral (work) should be positive. 27.

⎛

y⎞

Christy needs

28.

2

∫ C ⎜⎝1 + 3 ⎟⎠ ds = ∫0 (1 +10sin

∫C

3

t)[( −90cos 2 tsin t) 2 +(90sin 2 tcos t) 2] 1/ 2dt = 225

450 = 2.25 gal of paint. 200

0, 0, 1.2 ⋅ dx , dy , dz =

8π

∫C1.2 dz = ∫0

1.2(4)dt = 38.4π ≈ 120.64 ft-lb

Trivial way: The squirrel ends up 32π ft immediately above where it started. ( 32π ft)(1.2 lb) ≈ 120.64 ft-lb 29. C: x + y = a Let x = t, y = a – t, t in [0, a]. Cylinder: x + y = a; ( x + y) 2 = a2 ; x 2 + 2 xy + y 2 = a 2 Sphere: x2 + y2 + z2 = a2 The curve of intersection satisfies: z 2 = 2 xy; z = 2 xy .

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a

Area = 8 ∫

2xyds = 8∫

⎡ a ⎢t – = 16 ⎢ 2 2 ⎢ ⎣

a 2 (2 ) at – t +

C

2t (a – t ) (1) 2 + (–1) 2dt = 16∫

0

a

0

at – t 2 dt

a

⎤ ⎛ ⎡ ⎛ a2 ⎞ ⎛ t – a ⎞⎥ ⎤ ⎡ ⎛ a2 ⎞ ⎤⎞ –1 ⎜ 2 ⎟ ⎜ ⎢ 0 + ⎜ ⎟ ⎛⎜ π ⎞⎟ ⎥ − ⎢ 0 + ⎜ ⎟ ⎛⎜ −π ⎞⎟ ⎥ ⎟ = sin 16 ⎜ a ⎟⎥ ⎜ ⎢⎣ ⎜⎝ 8 ⎟⎠ ⎝ 2 ⎠ ⎦⎥ ⎣⎢ ⎜⎝ 8 ⎟⎠ ⎝ 2 ⎠ ⎦⎥ ⎟ 2 ⎝ 2 ⎠⎥ ⎝ ⎠ ⎦0 = 2a2 π Trivial way: Each side of the cylinder is part of a plane that intersects the sphere in a circle. The radius of each a ⎛ a ⎞⎛ a ⎞ a 2 . Therefore, the circle is the value of z in z = 2 xy when x = y = . That is, the radius is 2 ⎜ ⎟⎜ ⎟ = 2 2 ⎝ 2 ⎠⎝ 2 ⎠ 2

2⎤ ⎡ ⎛ a 2⎞ ⎥ total area of the part cut out is r ⎢ π ⎜ = 2a2 π. ⎟ ⎢ ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎦

ka 3 ∫c 0 3 (using same parametric equations as in Problem 29) I x = I y (symmetry)

30. I y =

a

kx 2ds = 4k ∫ t 2 2dt = 4 2

Iz = Ix + Iy = 8 2

ka 3 3

31. C: x 2 + y2 = a2 ⎡ π⎤ Let x = a cos θ, y = a sin θ, θ in ⎢0, ⎥ . ⎣ 2⎦

Area = 8 ∫

C

= 8∫

π/2

0 π/2

= 8∫

0 2

a2 – x2 ds

(a sin θ ) (– a sinθ )2 + (a cosθ )2 dθ (a sin θ ) a2 dθ = 8a2 [– cos θ ]π0 / 2

= 8a

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32. Note that r = a cos θ along C. Then (a2 – x2 – y2 )1/ 2 = ( a2 – r2 )1/ 2 = a cos θ .

⎧ x = r cosθ = (a sin θ) cos θ⎫ ⎡ π⎤ Let ⎨ ⎬ , θ in ⎢0, ⎥. ⎣ 2⎦ ⎩ y = r sinθ = (a sin θ ) sin θ ⎭ Therefore, x ′(θ ) = a cos 2θ ; y ′(θ ) = a sin 2θ . Then Area = 4∫ (a 2 – x 2 – y 2 )1/ 2 ds = 4∫ C

33. a.

∫C

b.

∫C

x 2 y ds = ∫

π/ 2

0

4

π/ 2

0

( a cosθ )[( a sin 2θ )2 + ( a cos 2θ )2 ]1/ 2 dθ = 4a2 .

(3sin t) 2 (3cos t)[(3cos t) 2 + (–3sin t) 2 ]1/ 2 dt = 81∫

π/ 2

0

π/ 2

⎡⎛ 1 ⎞ ⎤ sin 2 t cos t dt = 81⎢⎜ ⎟ sin3 t ⎥ ⎣⎝ 3 ⎠ ⎦0

= 27

3 3 3 xy 2 dx + xy 2 dy = ∫ (3 – t)(5 – t) 2 (–1) dt + ∫ (3 – t)(5 – t) 2(–1) dt = 2∫ (t 3 –13t 2 + 55t – 75)dt = –148.5 0

14.3 Concepts Review 1. f(b) – f(a) 2. gradient; ∇f (r ) 3. 0; 0 4. F is conservative.

0