Section 16.5 Curl and Divergence

(1)

Section 16.5 Curl and Divergence

33. Let r = x i + y j + z k and r = |r|.

Verify each identity. (a) ∇r = r/r (b) ∇ × r = 0 (c) ∇(1/r) = −r/r³ (d) ∇ ln r = r/r² Solution:

668 ¤ CHAPTER 16 VECTOR CALCULUS 31. (a) ∇ = ∇

²+ ²+ ² = 

²+ ²+ ²i+ 

²+ ²+ ² j+ 

²+ ²+ ²k=  i +  j +  k

²+ ²+ ² = r



(b) ∇ × r =



i j k













  



=

 

() − 

 ()

 i+

 

 () − 

()

 j+

 

() − 

()

 k= 0

(c) ∇

1





= ∇

 1

²+ ²+ ²



=

− 1

2

²+ ²+ ²(2)

²+ ²+ ² i−

1 2

²+ ²+ ² (2)

²+ ²+ ² j−

1 2

²+ ²+ ² (2)

²+ ²+ ² k

= −  i +  j +  k

(²+ ²+ ²)^32 = − r

³

(d) ∇ ln  = ∇ ln(²+ ²+ ²)^12= ¹₂∇ ln(²+ ²+ ²)

= 

²+ ²+ ²i+ 

²+ ²+ ²j+ 

²+ ²+ ² k=  i +  j +  k

²+ ²+ ² = r

² 32. r =  i +  j +  k ⇒  = |r| =

²+ ²+ ², so F= r

^ = 

(²+ ²+ ²)^2i+ 

(²+ ²+ ²)^2j+ 

(²+ ²+ ²)^2k Then 





(²+ ²+ ²)^2 = (²+ ²+ ²) − ²

(²+ ²+ ²)^{1 + 2} = ²− ²

^{ + 2} . Similarly,







(²+ ²+ ²)^2 = ²− ²

^{ + 2} and 





(²+ ²+ ²)^2 = ²− ²

^{ +2} . Thus

div F = ∇ · F =²− ²

^{ + 2} + ²− ²

^{ + 2} +²− ²

^{ + 2} = 3²− ²− ²− ²

^{ + 2}

= 3²− (²+ ²+ ²)

^{ + 2} = 3²− ²

^{ + 2} = 3 − 

^ Consequently, if  = 3 we have div F = 0.

33. By (13),

 (∇) · n  =

div( ∇)  =

[ div(∇) + ∇ · ∇]  by Exercise 25. But div(∇) = ∇².

Hence

 ∇²  =

 (∇) · n  −

∇ · ∇ .

34. By Exercise 33,

 ∇²  =

 (∇) · n  −

∇ · ∇  and



∇²  =

(∇) · n  −

∇ · ∇ . Hence





 ∇² − ∇²

 =

[ (∇) · n − (∇) · n]  +

(∇ · ∇ − ∇ · ∇)  =

[ ∇ − ∇ ] · n .

35. Let ( ) = 1. Then ∇ = 0 and Green’s ﬁrst identity (see Exercise 33) says



∇²  =

(∇) · n  −

0· ∇  ⇒ 

∇²  =

∇ · n . But  is harmonic on , so

∇² = 0 ⇒ 

∇ · n  = 0 and

n  =

(∇ · n)  = 0.

35. Use Greens Theorem in the form of Equation 13 to prove Greens first identity:

Z Z

D

f ∇²gdA = I

C

f (∇g)·n ds − Z Z

D

∇f · ∇g dA

where D and C satisfy the hypotheses of Greens Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity ∇g·n = Dng occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of g.)

Solution:

668 ¤ CHAPTER 16 VECTOR CALCULUS 31. (a) ∇ = ∇

²+ ²+ ²= 

²+ ²+ ² i+ 

²+ ²+ ²j+ 

²+ ²+ ² k=  i +  j +  k

²+ ²+ ² = r



(b) ∇ × r =



i j k













  



=

 

 () − 

()

 i+

 

() − 

()

 j+

 

() − 

 ()

 k= 0

(c) ∇

1





= ∇

 1

²+ ²+ ²



=

− 1

2

²+ ²+ ²(2)

²+ ²+ ² i−

1 2

²+ ²+ ²(2)

²+ ²+ ² j−

1 2

²+ ²+ ²(2)

²+ ²+ ² k

= −  i +  j +  k

(²+ ²+ ²)^32 = −r

³

(d) ∇ ln  = ∇ ln(²+ ²+ ²)^12= ¹₂∇ ln(²+ ²+ ²)

= 

²+ ²+ ² i+ 

²+ ²+ ² j+ 

²+ ²+ ²k=  i +  j +  k

²+ ²+ ² = r

² 32. r =  i +  j +  k ⇒  = |r| =

²+ ²+ ², so F= r

^ = 

(²+ ²+ ²)^2i+ 

(²+ ²+ ²)^2j+ 

(²+ ²+ ²)^2k Then 





(²+ ²+ ²)^2 = (²+ ²+ ²) − ²

(²+ ²+ ²)^{1 + 2} = ²− ²

^{ + 2} . Similarly,







(²+ ²+ ²)^2 = ²− ²

^{ + 2} and 





(²+ ²+ ²)^2 = ²− ²

^{ +2} . Thus

div F = ∇ · F = ²− ²

^{ + 2} +²− ²

^{ + 2} +²− ²

^{ + 2} = 3²− ²− ²− ²

^{ + 2}

= 3²− (²+ ²+ ²)

^{ + 2} = 3²− ²

^{ + 2} = 3 − 

^ Consequently, if  = 3 we have div F = 0.

33. By (13),

 (∇) · n  =

div( ∇)  =

[ div(∇) + ∇ · ∇]  by Exercise 25. But div(∇) = ∇².

Hence

 ∇²  =

 (∇) · n  −

∇ · ∇ .

34. By Exercise 33,

 ∇²  =

 (∇) · n  −

∇ · ∇  and



∇²  =

(∇) · n  −

∇ · ∇ . Hence





 ∇² − ∇²

 =

[ (∇) · n − (∇) · n]  +

(∇ · ∇ − ∇ · ∇)  =

[ ∇ − ∇ ] · n .

35. Let ( ) = 1. Then ∇ = 0 and Green’s ﬁrst identity (see Exercise 33) says



∇²  =

(∇) · n  −

0· ∇  ⇒ 

∇²  =

∇ · n . But  is harmonic on , so

∇² = 0 ⇒ 

∇ · n  = 0 and

n  =

(∇ · n)  = 0.

38. Use Greens first identity to show that if f is harmonic on D, and if f(x, y) = 0 on the boundary curve C, then RR

D|∇f |²dA = 0. (Assume the same hypotheses as in Exercise 35.) Solution:

/HW j = i 7KHQ *UHHQ¶V ¿UVW LGHQWLW\ VHH ([HUFLVH VD\VUU

Gi Q²i gD =K

F (i )(Qi) · n gv 3UU

GQi · Qi gD

%XW i LV KDUPRQLF VR Q²i = 0 DQG Qi · Qi = |Qi|² VR ZH KDYH 0 =K

F(i ) (Qi) · n gv 3UU

G|Qi|² gD i UU

G|Qi|² gD =K

F (i ) (Qi) · n gv = 0 VLQFH i ({> |) = 0 RQ F

35

39. This exercise demonstrates a connection between the curl vector and rotations. Let B be a rigid body rotating about the z-axis. The rotation can be described by the vector w = ωk, where ω is the angular speed of B, that is, the tangential speed of any point P in B divided by the distance d from the axis of rotation. Let r =< x, y, z > be

1

(2)

the position vector of P .

(a) By considering the angle θ in the figure, show that the velocity field of B is given by v = w × r.

(b) Show that v = −ωyi + ωxj.

(c) Show that curl v = 2w.

1110 Chapter 16 Vector Calculus

23–29 Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If f is a scalar field and F, G are vector fields, then fF, F G, and F 3 G are defined by

s f Fdsx, y, zd − f sx, y, zd Fsx, y, zd sF Gdsx, y, zd − Fsx, y, zd Gsx, y, zd sF 3 Gdsx, y, zd − Fsx, y, zd 3 Gsx, y, zd 23. divsF 1 Gd − div F 1 div G

24. curlsF 1 Gd − curl F 1 curl G 25. divs f Fd − f div F 1 F =f 26. curls f Fd − f curl F 1 s=f d 3 F 27. divsF 3 Gd − G curl F 2 F curl G 28. divs=f 3 =td − 0

29. curlscurl Fd − gradsdiv Fd 2 =²F

30–32 Let r − x i 1 y j 1 z k and r −

|

^r

|

^.

30. Verify each identity.

(a) = r− 3 (b) = srrd − 4r

(c) =²r³− 12r 31. Verify each identity.

(a) =r − ryr (b) = 3r − 0

(c) =s1yrd − 2ryr³ (d) = ln r − ryr² 32. If F − ryr^p, find div F. Is there a value of p for which

div F − 0?

33. Use Green’s Theorem in the form of Equation 13 to prove Green’s first identity:

y

D

y

^{f =}²^{t dA −}

y

C fs=td n ds 2

y

D

y

^{= f =}^{t dA}

where D and C satisfy the hypotheses of Green’s Theorem and the appropriate partial derivatives of f and t exist and are continuous. (The quantity =t n − Dⁿt occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of t.) 34. Use Green’s first identity (Exercise 33) to prove Green’s

second identity:

y

D

y

^{s f =}²^{t 2 t=}²^f^{d dA −}

y

C s f =t 2 t=f d n ds where D and C satisfy the hypotheses of Green’s Theorem

and the appropriate partial derivatives of f and t exist and are continuous.

35. Recall from Section 14.3 that a function t is called harmonic on D if it satisfies Laplace’s equation, that is, =²t − 0 on D.

Use Green’s first identity (with the same hypotheses as in

Exercise 33) to show that if t is harmonic on D, then

yC Dnt ds − 0. Here Dⁿt is the normal derivative of t defined in Exercise 33.

36. Use Green’s first identity to show that if f is harmonic on D, and if fsx, yd − 0 on the boundary curve C, then yyD

|

^=f

|

² dA − 0. (Assume the same hypotheses as in Exercise 33.)

37. This exercise demonstrates a connection between the curl vector and rotations. Let B be a rigid body rotating about the z-axis. The rotation can be described by the vector w − k, where is the angular speed of B, that is, the tangential speed of any point P in B divided by the distance d from the axis of rotation. Let r −kx, y, zl be the position vector of P.

(a) By considering the angle in the figure, show that the velocity field of B is given by v − w 3 r.

(b) Show that v − 2y i 1 x j.

(c) Show that curl v − 2w.

0

¨ P d B

w

v z

y x

38. Maxwell’s equations relating the electric field E and magnetic field H as they vary with time in a region containing no charge and no current can be stated as follows:

div E − 0 div H − 0

curl E − 21 c

−H

−t curl H − 1 c

−E

−t

where c is the speed of light. Use these equations to prove the following:

(a) = 3s= 3 Ed − 2 1 c²

−²E

−t² (b) = 3s= 3 Hd − 2 1

c²

−²H

−t² (c) =²E − 1

c²

−²E

−t² [Hint: Use Exercise 29.]

(d) =²H − 1 c²

−²H

−t²

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Solution:

SECTION 16.5 CURL AND DIVERGENCE ¤ 669 36. Let  = . Then Green’s ﬁrst identity (see Exercise 33) says

 ∇²  =

 ( )(∇) · n  −

∇ · ∇ .

But  is harmonic, so ∇² = 0, and ∇ · ∇ = |∇|², so we have 0 =

 ( ) (∇) · n  −

|∇|²  ⇒



|∇|²  =

 ( ) (∇) · n  = 0 since  ( ) = 0 on .

37. (a) We know that  = , and from the diagram sin  =  ⇒  =  = (sin ) = |w × r|. But v is perpendicular to both w and r, so that v = w × r.

(b) From (a), v = w × r =



i j k 0 0 

  



= (0 ·  − ) i + ( − 0 · ) j + (0 ·  −  · 0) k = − i +  j

(c) curl v = ∇ × v =



i j k

  

−  0



=

 

 (0) − 

()

 i+

 

(−) − 

(0)

 j+

 

() − 

(−)

 k

= [ − (−)] k = 2 k = 2w 38. Let H = h¹ 2 3i and E = h¹ 2 3i.

(a) ∇ × (∇ × E) = ∇ × (curl E) = ∇ ×



−1



H





= −1





i j k

  

1 2 3



= −1



²3

 − ²2

 

 i+

²1

 − ²3

 

 j+

²2

  − ²1

 

 k



= −1







3

 −2



 i+

1

 − 3



 j+

2

 −1



 k

 [assuming that the partial derivatives are continuous so that the order of

differentiation does not matter]

= −1





curl H = −1







1



E





= −1

²

²E

²

(b) ∇ × (∇ × H) = ∇ × (curl H) = ∇ ×

1



E





= 1





i j k

  

1 2 3



= 1



²3

 −²2

 

 i+

²1

 − ²3

 

 j+

²2

 − ²1

 

 k



= 1







3

 − 2



 i+

1

 −3



 j+

2

 −1



 k

= 1





curl E = 1









−1



H





= −1

²

²H

²

40. Maxwells equations relating the electric field E and magnetic field H as they vary with time in a region containing no charge and no current can be stated as follows:

divE = 0 divH = 0

curlE = −1 c

∂H

∂t curlH = 1 c

∂E

∂t where c is the speed of light. Use these equations to prove the following:

(a) ∇ × (∇ × E) = −_c¹2

∂²E

∂t²

(b) ∇ × (∇ × H) = −_c¹2

∂²H

∂t²

(c) ∇²E = _c¹2

∂²E

∂t² [Hint: Use Exercise 31.]

(d) ∇²H = _c¹2

∂²H

∂t²

2

(3)

Solution:

SECTION 16.5 CURL AND DIVERGENCE ¤ 669 36. Let  = . Then Green’s ﬁrst identity (see Exercise 33) says

 ∇²  =

 ( )(∇) · n  −

∇ · ∇ .

But  is harmonic, so ∇² = 0, and ∇ · ∇ = |∇|², so we have 0 =

 ( ) (∇) · n  −

|∇|²  ⇒



|∇|²  =

 ( ) (∇) · n  = 0 since  ( ) = 0 on .

37. (a) We know that  = , and from the diagram sin  =  ⇒  =  = (sin ) = |w × r|. But v is perpendicular to both w and r, so that v = w × r.

(b) From (a), v = w × r =



i j k 0 0 

  



= (0 ·  − ) i + ( − 0 · ) j + (0 ·  −  · 0) k = − i +  j

(c) curl v = ∇ × v =



i j k

  

−  0



=

 

(0) − 

()

 i+



(−) − 

(0)

 j+

 

() − 

(−)

 k

= [ − (−)] k = 2 k = 2w 38. Let H = h¹ 2 3i and E = h¹ 2 3i.

(a) ∇ × (∇ × E) = ∇ × (curl E) = ∇ ×



−1



H





= −1





i j k

  

1 2 3



= −1



²3

 − ²2

 

 i+

²1

 − ²3

 

 j+

²2

 − ²1

 

 k



= −1







3

 −2



 i+

1

 −3



 j+

2

 −1



 k

= −1





curl H = −1







1



E





= −1

²

²E

²

(b) ∇ × (∇ × H) = ∇ × (curl H) = ∇ ×

1



E





= 1





i j k

  

1 2 3



= 1



²3

 −²2

 

 i+

²1

 −²3

 

 j+

²2

 −²1

 

 k



= 1







3

 −2



 i+

1

 −3



 j+

2

 − 1



 k

= 1





curl E =1









−1



H





= −1

²

²H

²

670 ¤ CHAPTER 16 VECTOR CALCULUS

(c) Using Exercise 29, we have that curl curl E = grad div E − ∇²E ⇒

∇²E= grad div E − curl curl E = grad 0 + 1

²

²E

² [from part (a)] = 1

²

²E

² . (d) As in part (c), ∇²H= grad div H − curl curl H = grad 0 + 1

²

²H

² [using part (b)] = 1

²

²H

² . 39. For any continuous function  on R³, deﬁne a vector ﬁeld G(  ) = h(  ) 0 0i where (  ) =

0  (  ) .

Then div G = 

((  )) + 

 (0) + 

(0) = 





0  (  )  =  (  )by the Fundamental Theorem of Calculus. Thus every continuous function  on R³is the divergence of some vector ﬁeld.

16.6 Parametric Surfaces and Their Areas

1.  (4 −5 1) lies on the parametric surface r( ) = h +   − 2 3 +  − i if and only if there are values for  and  where  +  = 4,  − 2 = −5, and 3 +  −  = 1. From the ﬁrst equation we have  = 4 −  and substituting into the second equation gives 4 −  − 2 = −5 ⇔  = 3. Then  = 1, and these values satisfy the third equation, so  does lie on the surface.

(0 4 6)lies on r( ) if and only if  +  = 0,  − 2 = 4, and 3 +  −  = 6, but solving the ﬁrst two equations simultaneoulsy gives  =⁴₃,  = −⁴₃and these values do not satisfy the third equation, so  does not lie on the surface.

2.  (1 2 1)lies on the parametric surface r( ) =

1 +  −   + ² ²− ²if and only if there are values for  and  where 1 +  −  = 1,  + ²= 2, and ²− ²= 1. From the ﬁrst equation we have  =  and substituting into the third equation gives 0 = 1, an impossibility, so  does not lie on the surface.

(2 3 3)lies on r( ) if and only if 1 +  −  = 2,  + ²= 3, and ²− ²= 3. From the ﬁrst equation we have

 =  + 1and substituting into the second equation gives  + 1 + ²= 3 ⇔ ²+  − 2 = 0 ⇔ ( + 2)( − 1) = 0, so  = −2 ⇒  = −1 or  = 1 ⇒  = 2. The third equation is satisﬁed by  = 2,  = 1 so  does lie on the surface.

3. r( ) = ( + ) i + (3 − ) j + (1 + 4 + 5) k = h0 3 1i +  h1 0 4i +  h1 −1 5i. From Example 3, we recognize this as a vector equation of a plane through the point (0 3 1) and containing vectors a = h1 0 4i and b = h1 −1 5i. If we

wish to ﬁnd a more conventional equation for the plane, a normal vector to the plane is a × b =





i j k 1 0 4 1 −1 5





= 4 i − j − k

and an equation of the plane is 4( − 0) − ( − 3) − ( − 1) = 0 or 4 −  −  = −4.

4. r( ) = ²i+  cos  j +  sin  k, so the corresponding parametric equations for the surface are  = ²,  =  cos ,

 =  sin . For any point (  ) on the surface, we have ²+ ²= ²cos² + ²sin² = ²= . Since no restrictions are placed on the parameters, the surface is  = ²+ ², which we recognize as a circular paraboloid whose axis is the -axis.

3