Section 16.9 The Divergence Theorem

Section 16.9 The Divergence Theorem

12. Use the Divergence Theorem to calculate the surface integral RR

SF·dS; that is, calculate the flux of F across S.

F(x, y, z) = (xy + 2xz) i + (x²+ y²) j + (xy − z²) k, S is the surface of the solid bounded by the cylinder x²+ y²= 4 and the planes z = y − 2 and z = 0.

Solution:

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. div F = 4³+ 4²so



F· S =

4(²+ ²)  =2

0

1 0

 cos +2

0 (4³cos )    

=2

0

1

0 (4⁵cos² + 8⁴cos )   =2

0

₂

3cos² +⁸₅cos 

 = ²₃

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

2 0

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

   = ¹⁹₆₄².

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

14. Use the Divergence Theorem to calculate the surface integral RR

SF·dS; that is, calculate the flux of F across S.

F(x, y, z) = (xy − z²) i + x³√

z j + (xy + z²) k, S is the surface of the solid bounded by the cylinder x = y²and the planes x + z = 1 and z = 0.

20. Let F(x, y, z) = z tan⁻¹(y²) i+z³ln(x²+1) j+z k. Find the flux of F across the part of the paraboloid x²+y²+z = 2 that lies above the plane z = 1 and is oriented upward.

Solution:

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   −.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

26. Use the Divergence Theorem to evaluate

Z Z

S

(2x + 2y + z²) dS where S is the sphere x²+ y²+ z²= 1

Solution:

704 ¤ CHAPTER 16 VECTOR CALCULUS 23. Since x

|x|³ =  i +  j +  k

(²+ ²+ ²)^32 and 



 

(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.

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

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