Section 16.9 The Divergence Theorem

(1)

Section 16.9 The Divergence Theorem

700 ¤ CHAPTER 16 VECTOR CALCULUS

On 1: The surface is  = ²+ ², ²+ ²≤ 9, with downward orientation, and F(  ) = ²³i+ 2 j + 4²k.

Then



₁F· S = −

[−(²³)(2) − (2)(2) + (4²)] 

=



2²(²+ ²)³+ 4²(²+ ²) − 4(²+ ²)²



=2

0

3

0 (2³cos  sin² · ⁶+ 4²sin² · ²− 4⁴)   

=2

0

3

0 (2¹⁰sin² cos  + 4⁵sin² − 4⁵)  

=2

0

2

11¹¹sin² cos  +²₃⁶sin² −²3⁶=3

=0

=2

0

_354,294

11 sin² cos  + 486 sin² − 486



=_354,294

11 ·¹3sin³ + 486₁

2 −¹4sin 2

− 4862

0

= 0 + 486( − 0) − 486(2) = −486

On 2: The surface is  = 9, ²+ ²≤ 9, with upward orientation, so F(  ) = ²³i+ 2 j + 4²kand



₂F· S =

[−(²³)(0) − (2)(0) + (4²)]  =

4(9)²

= 324 () = 324 · (3)²= 2916

Thus

F· S =

₁F· S +

₂F· S = −486 + 2916 = 2430.

3. div F = 0 + 1 + 0 = 1, so

div F  =

1  =  () = ⁴₃ · 4³=²⁵⁶₃ .

is a sphere of radius 4 centered at the origin which can be parametrized by r( ) = h4 sin  cos  4 sin  sin  4 cos i, 0 ≤  ≤ , 0 ≤  ≤ 2 (similar to Example 16.6.10). Then

r× r^= h4 cos  cos  4 cos  sin  −4 sin i × h−4 sin  sin  4 sin  cos  0i

=

16 sin² cos  16 sin² sin  16 cos  sin  and F(r( )) = h4 cos  4 sin  sin  4 sin  cos i. Thus

F· (r^× r^) = 64 cos  sin² cos  + 64 sin³ sin² + 64 cos  sin² cos  = 128 cos  sin² cos  + 64 sin³ sin²

and 

F· S =

F· (r^× r^)  =2

0



0 (128 cos  sin² cos  + 64 sin³ sin²)  

=2

0

₁₂₈

3 sin³ cos  + 64₁

3cos³ − cos 

sin²=

=0 

=2

0 256

3 sin²  =²⁵⁶₃ 1

2 −¹4sin 22

0 =²⁵⁶₃  4. div F = 2 − 1 + 1 = 2, so





div F  =



²+²≤9

 2 0

2 



 =



²+²≤9

4  = 4(area of circle) = 4( · 3²) = 36

Let 1be the front of the cylinder (in the plane  = 2), 2the back (in the -plane), and 3the lateral surface of the cylinder.

1is the disk  = 2, ²+ ² ≤ 9. A unit normal vector is n = h1 0 0i and F = h4 − i on ¹, so



₁F· S =

₁F· n  =

₁4  = 4(surface area of 1) = 4( · 3²) = 36. 2is the disk  = 0, ²+ ²≤ 9.

SECTION 16.9 THE DIVERGENCE THEOREM ¤ 701 Here n = h−1 0 0i and F = h0 − i, so

₂F· S =

₂F· n  =

₂0  = 0.

3can be parametrized by r( ) = h 3 cos  3 sin i, 0 ≤  ≤ 2, 0 ≤  ≤ 2. Then

r× r^= h1 0 0i × h0 −3 sin  3 cos i = h0 −3 cos  −3 sin i. For the outward (positive) orientation we use

−(r^× r^)and F(r( )) =

² −3 cos  3 sin , so



₃F· S =

F· (−(r^× r^))  =2 0

2

0 (0 − 9 cos² + 9 sin²)  

= −92 0 2

0 cos 2  = −9 (2)1

2sin 22

0 = 0

Thus

F· S = 36 + 0 + 0 = 36.

5. div F =_^ (^) +_^(²³) +_^(−^) = ^+ 2³− ^ = 2³, so by the Divergence Theorem,



F· S =

div F  =3 0

2 0

1

0 2³   = 23 0  2

0  1 0 ³

= 2₁

2²3 0

₁

2²2 0

₁

4⁴1 0= 2₉

2

(2)₁

4

=⁹₂

6. div F =_^ (²) +_^(²) +_^ (²) = 2 + 2 + 2 = 6, so by the Divergence Theorem,



F· S =

div F  = 0

 0



06    = 6 0  

0   0  

= 6₁

2² 0

₁

2² 0

₁

2² 0= 6₁

2² ₁

2² ₁

2²

=³₄²²² 7. div F = 3²+ 0 + 3², so using cylindrical coordinates with  =  cos ,  =  sin ,  =  we have



F· S =

(3²+ 3²)  =2

0

1 0

2

−1(3²cos² + 3²sin²)    

= 32

0 1

0 ³ 2

−1 = 3

2

0

₁

4⁴1 0

2

−1= 3(2)₁

4

(3) = ^9₂

8. div F = 3²+ 3²+ 3², so by the Divergence Theorem,



F· S =

3(²+ ²+ ²)  = 0

2

0

2

0 3²· ²sin     = 3

0 sin  2

0 2 0 ⁴

= 3 [− cos ]^0

2

0

1 5⁵2

0= 3 (2) (2)32 5

=³⁸⁴₅ 

9. div F = ^+ (−^) + 0 = 0, so by the Divergence Theorem,

F· S =

0  = 0.

10. The tetrahedron has vertices (0 0 0), ( 0 0), (0  0), (0 0 ) and is described by

 =

(  ) | 0 ≤  ≤ , 0 ≤  ≤  1 −^

, 0 ≤  ≤ 

1 −^−^

. Here we have div F = 0 + 1 +  =  + 1, so



F· S =

( + 1)  = 0

^(¹−^)

0

^(¹−^−^)

0 ( + 1)   

= 0

^(¹−^_)

0 ( + 1)



1 −^−^

  = 

0 ( + 1)

1 −^

 −2¹²=(¹−^_)

=0 

= 

0 ( + 1)

1 −^

·  1 −^

−2¹ · ² 1 −^

2

 =¹₂

0 ( + 1) 1 −^

2



=¹₂ 0

₁

²³+_¹2²−²²+  −² + 1



=¹₂ 1

4²⁴+_3¹2³−3²³+¹₂²−¹²+  0

=¹₂1

4²+¹₃ −²3²+¹₂²−  + 

= ¹₂1

12²+¹₃

= ₂₄¹( + 4)

SECTION 16.9 THE DIVERGENCE THEOREM ¤ 701 Here n = h−1 0 0i and F = h0 − i, so

₂F· S =

₂F· n  =

₂0  = 0.

3can be parametrized by r( ) = h 3 cos  3 sin i, 0 ≤  ≤ 2, 0 ≤  ≤ 2. Then

r× r^= h1 0 0i × h0 −3 sin  3 cos i = h0 −3 cos  −3 sin i. For the outward (positive) orientation we use

−(r^× r^)and F(r( )) =

² −3 cos  3 sin , so



3F· S =

F· (−(r^× r^))  =2 0

2

0 (0 − 9 cos² + 9 sin²)  

= −92 0 2

0 cos 2  = −9 (2)₁

2sin 22

0 = 0

Thus

F· S = 36 + 0 + 0 = 36.

5. div F =_^ (^) +_^(²³) +_^(−^) = ^+ 2³− ^ = 2³, so by the Divergence Theorem,



F· S =

div F  =3 0

2 0

1

0 2³   = 23 0  2

0  1 0 ³

= 2₁

2²3 0

₁

2²2 0

₁

4⁴1 0= 2₉

2

(2)₁

4

=⁹₂

6. div F =_^ (²) +_^(²) +_^ (²) = 2 + 2 + 2 = 6, so by the Divergence Theorem,



F· S =

div F  = 0

 0



06    = 6 0  

0   0  

= 61 2²

0

1 2²

0

1 2²

0= 61 2² 1

2² 1 2²

=³₄²²² 7. div F = 3²+ 0 + 3², so using cylindrical coordinates with  =  cos ,  =  sin ,  =  we have



F· S =

(3²+ 3²)  =2

0

1 0

2

−1(3²cos² + 3²sin²)    

= 32

0 1

0 ³ 2

−1 = 3

2

0

1 4⁴1

0

2

−1= 3(2)1 4

(3) = ^9₂

8. div F = 3²+ 3²+ 3², so by the Divergence Theorem,



F· S =

3(²+ ²+ ²)  = 0

2

0

2

0 3²· ²sin     = 3

0 sin  2

0 2 0 ⁴

= 3 [− cos ]^0

2

0

₁

5⁵2

0= 3 (2) (2)₃₂

5

=³⁸⁴₅ 

9. div F = ^+ (−^) + 0 = 0, so by the Divergence Theorem,

F· S =

0  = 0.

10. The tetrahedron has vertices (0 0 0), ( 0 0), (0  0), (0 0 ) and is described by

 =

(  ) | 0 ≤  ≤ , 0 ≤  ≤  1 −^

, 0 ≤  ≤ 

1 −^−^

. Here we have div F = 0 + 1 +  =  + 1, so



F· S =

( + 1)  = 0

^(¹−^_)

0

^(¹−^_−^_)

0 ( + 1)   

= 0

^(¹−^_)

0 ( + 1)



1 −^−^

  = 

0 ( + 1)

1 −^

 −2¹²=(¹−^_)

=0 

= 

0 ( + 1)

1 −^

·  1 −^

−2¹ · ² 1 −^

2

 =¹₂

0 ( + 1) 1 −^

2



=¹₂ 0

1

²³+_¹2²−²²+  −² + 1



=¹₂ ₁

4²⁴+_3¹2³−3²³+¹₂²−¹²+  0

=¹₂₁

4²+¹₃ −²3²+¹₂²−  + 

= ¹₂₁

12²+¹₃

= ₂₄¹( + 4)

702 ¤ CHAPTER 16 VECTOR CALCULUS 11. div F = 6²+ 3²+ 3²= 6²+ 6²so



F· S =

6(²+ ²)  =2

0

1 0

1−²

0 6²·     =2

0

1

0 6³(1 − ²)  

=2

0 1

0(6³− 6⁵)  =

2

0

3

2⁴− ⁶1 0= 23

2 − 1

=  12. For ²+ ²≤ 4 the plane  =  − 2 is below the -plane, so the solid  bounded by  is

 =

(  ) | ²+ ²≤ 4  − 2 ≤  ≤ 0

. Here div F =  + 2 + 2 − 2 = 3 so



F· S =

3  =2

0

2 0

0

 sin −2(3 sin )    

=2

0

2

0(3²sin )(0 −  sin  + 2)   =2

0

2 0

−3³sin² + 6²sin 

 

=2

0

−³4⁴sin² + 2³sin =2

=0 =2

0

−12 sin² + 16 sin 



=

−12₁

2 −¹4sin 2

− 16 cos 2

0 = −12 − 16 + 16 = −12

13. F(  ) = 

²+ ²+ ²i+ 

²+ ²+ ²j+ 

²+ ²+ ²k, so

div F =  ·¹2(²+ ²+ ²)^−12(2) + (²+ ²+ ²)^12+  ·¹2(²+ ²+ ²)^−12(2) + (²+ ²+ ²)^12 +  ·¹2(²+ ²+ ²)^−12(2) + (²+ ²+ ²)^12

= (²+ ²+ ²)^−12

²+ (²+ ²+ ²) + ²+ (²+ ²+ ²) + ²+ (²+ ²+ ²)

=4(²+ ²+ ²)

²+ ²+ ² = 4

²+ ²+ ².

Then 



F· S =





4

²+ ²+ ² =

 2 0

2

0

1 0

4

²· ²sin    

=2

0 sin  2

0 1

0 4³ = [− cos ]^20 []^2₀ 

⁴1

0= (1) (2) (1) = 2

14. F(  ) = (²+ ²+ ²) i + (²+ ²+ ²) j + (²+ ²+ ²) k, so

div F =  · 2 + (²+ ²+ ²) +  · 2 + (²+ ²+ ²) +  · 2 + (²+ ²+ ²) = 5(²+ ²+ ²). Then





F· S =





5(²+ ²+ ²)  =

 0

 2

0

  0

5²· ²sin    

= 5

0 sin  2

0 

0 ⁴ = 5 [− cos ]^0[]^2₀ 1 5⁵

0 = 5 (2) (2)1 5⁵

= 4⁵ 15. 

F· S =



√3 − ² =1

−1

1

−1

2− ⁴− ⁴ 0

√3 − ²   =³⁴¹₆₀ √

2 +⁸¹₂₀sin⁻¹√ 3 3



16.

By the Divergence Theorem, the ﬂux of F across the surface of the cube is



F· S =2 0

2 0

cos  cos² + 3 sin² cos  cos⁴ + 5 sin⁴ cos  cos⁶

   = ¹⁹₆₄².

1

(2)

SECTION 16.9 THE DIVERGENCE THEOREM ¤ 703 17. For 1we have n = −k, so F · n = F · (−k) = −² − ²= −²(since  = 0 on 1). So if  is the unit disk, we get



₁F· S =

₁F· n  =

(−²)  = −2

0

1

0 ²(sin²)    = −¹4. Now since 2is closed, we can use the Divergence Theorem. Since div F =_^ (²) +_^ ₁

3³+ tan 

+_^ (² + ²) = ²+ ²+ ², we use spherical coordinates to get

₂F· S =

div F  =^2

0

^2

0

¹

0 ²· ²sin     = ²₅. Finally



F· S =

₂F· S −

₁F· S =²5 −

−¹4

= ¹³₂₀.

18. As in the hint to Exercise 17, we create a closed surface 2 =  ∪ ¹, where  is the part of the paraboloid ²+ ²+  = 2 that lies above the plane  = 1, and 1is the disk ²+ ²= 1on the plane  = 1 oriented downward, and we then apply the Divergence Theorem. Since the disk 1is oriented downward, its unit normal vector is n = −k and F · (−k) = − = −1 on

1. So

₁F· S =

₁F· n  =

₁(−1)  = −(¹) = −. Let  be the region bounded by ². Then



₂F· S =

div F  =

1  =1 0

2

0

2−²

1     =1 0

2

0 ( − ³)   = (2)¹₄ = ^₂. Thus the ﬂux of F across  is

F· S =

₂F· S −

₁F· S =^2 − (−) = ^32 .

19. The vectors that end near 1are longer than the vectors that start near 1, so the net ﬂow is inward near 1and div F(1)is negative. The vectors that end near 2are shorter than the vectors that start near 2, so the net ﬂow is outward near 2and div F(2)is positive.

20. (a) The vectors that end near 1are shorter than the vectors that start near 1, so the net ﬂow is outward and 1is a source.

The vectors that end near 2are longer than the vectors that start near 2, so the net ﬂow is inward and 2is a sink.

(b) F( ) =

 ²

⇒ div F = ∇ · F = 1 + 2. The -value at ¹is positive, so div F = 1 + 2 is positive, thus 1

is a source. At 2,   −1, so div F = 1 + 2 is negative, and ²is a sink.

21. From the graph it appears that for points above the -axis, vectors starting near a particular point are longer than vectors ending there, so divergence is positive.

The opposite is true at points below the -axis, where divergence is negative.

F( ) =

  + ²

⇒ div F = ^ () +_^ 

 + ²=  + 2 = 3.

Thus div F  0 for   0, and div F  0 for   0.

22. From the graph it appears that for points above the line  = −, vectors starting near a particular point are longer than vectors ending there, so divergence is positive. The opposite is true at points below the line  = −, where divergence is negative.

F( ) =

² ²

⇒ div F = ^ (²) +_^ (²) = 2 + 2. Then div F  0 for 2 + 2  0 ⇒   −, and div F  0 for   −.

704 ¤ CHAPTER 16 VECTOR CALCULUS 23. Since x

|x|³ =  i +  j +  k

(²+ ²+ ²)^32and 



 

(²+ ²+ ²)^32



=(²+ ²+ ²) − 3²

(²+ ²+ ²)^52 with similar expressions for 



 

(²+ ²+ ²)^32

 and 



 

(²+ ²+ ²)^32



, we have

div

 x

|x|³



=3(²+ ²+ ²) − 3(²+ ²+ ²)

(²+ ²+ ²)^52 = 0, except at (0 0 0) where it is undeﬁned.

24. We ﬁrst need to ﬁnd F so that

F· n  =

(2 + 2 + ²) , so F · n = 2 + 2 + ². But for , n=  i +  j +  k

²+ ²+ ² =  i +  j +  k. Thus F = 2 i + 2 j +  k and div F = 1.

If  =

(  ) | ²+ ²+ ²≤ 1, then

(2 + 2 + ²)  =

 =  () = ⁴₃(1)³= ⁴₃.

25. 

a· n  =

div a  = 0since div a = 0.

26. ¹₃

F· S =¹3



div F  =¹₃

3  =  () 27. 

curl F · S =

div(curl F)  = 0by Theorem 16.5.11.

28. 

 n  =

(∇ · n)  =

div(∇)  =

∇² 

29. 

( ∇) · n  =

div( ∇)  =

( ∇² + ∇ · ∇)  by Exercise 16.5.25.

30. 

( ∇ − ∇) · n  =



( ∇² + ∇ · ∇) − (∇² + ∇ · ∇)

 [by Exercise 29].

But ∇ · ∇ = ∇ · ∇, so that

( ∇ − ∇) · n  =

( ∇² − ∇² ) .

31. If c = 1i+ 2j+ 3kis an arbitrary constant vector, we deﬁne F = c = 1i+  2j+  3k. Then

div F = div  c =

1+

2+

3= ∇ · c and the Divergence Theorem says

F· S =

div F  ⇒



F· n  =

∇ · c  . In particular, if c = i then

 i · n  =

∇ · i  ⇒





 1 =







 (where n = 1i+ 2j+ 3k). Similarly, if c = j we have





 2 =







, and c = k gives



 3 =







. Then



 n  =

 1 i+

 2 j+

 3 k

=









 i+









 j+









 k=







i+

j+

k





=

∇  as desired.

704 ¤ CHAPTER 16 VECTOR CALCULUS 23. Since x

|x|³ =  i +  j +  k

(²+ ²+ ²)^32and 



 

(²+ ²+ ²)^32



=(²+ ²+ ²) − 3²

(²+ ²+ ²)^52 with similar expressions for 



 

(²+ ²+ ²)^32

 and 



 

(²+ ²+ ²)^32



, we have

div

 x

|x|³



=3(²+ ²+ ²) − 3(²+ ²+ ²)

(²+ ²+ ²)^52 = 0, except at (0 0 0) where it is undeﬁned.

24. We ﬁrst need to ﬁnd F so that

F· n  =

(2 + 2 + ²) , so F · n = 2 + 2 + ². But for , n=  i +  j +  k

²+ ²+ ² =  i +  j +  k. Thus F = 2 i + 2 j +  k and div F = 1.

If  =

(  ) | ²+ ²+ ²≤ 1, then

(2 + 2 + ²)  =

 =  () = ⁴₃(1)³= ⁴₃.

25. 

a· n  =

div a  = 0since div a = 0.

26. ¹₃

F· S =¹3



div F  =¹₃

3  =  () 27. 

curl F · S =

div(curl F)  = 0by Theorem 16.5.11.

28. 

 n  =

(∇ · n)  =

div(∇)  =

∇² 

29. 

( ∇) · n  =

div( ∇)  =

( ∇² + ∇ · ∇)  by Exercise 16.5.25.

30. 

( ∇ − ∇) · n  =



( ∇² + ∇ · ∇) − (∇² + ∇ · ∇)

 [by Exercise 29].

But ∇ · ∇ = ∇ · ∇, so that

( ∇ − ∇) · n  =

( ∇² − ∇² ) .

31. If c = 1i+ 2j+ 3kis an arbitrary constant vector, we deﬁne F = c = 1i+  2j+  3k. Then

div F = div  c =

1+

2+

3= ∇ · c and the Divergence Theorem says

F· S =

div F  ⇒



F· n  =

∇ · c  . In particular, if c = i then

 i · n  =

∇ · i  ⇒





 1 =







 (where n = 1i+ 2j+ 3k). Similarly, if c = j we have





 2 =







, and c = k gives





 3 =







. Then



 n  =

 1 i+

 2 j+

 3 k

=









 i+









 j+









 k=







i+

j+

k





=

∇  as desired.