Section 16.8 Stokes’ Theorem

(1)

Section 16.8 Stokes’ Theorem

696 ¤ CHAPTER 16 VECTOR CALCULUS

6. The boundary curve  is the circle ²+ ² = 1,  = 0 which should be oriented in the counterclockwise direction when viewed from the right, so a vector equation of  is r() = cos(−) i + sin(−) k = cos  i − sin  k, 0 ≤  ≤ 2. Then F(r()) = i + − cos  sin j− cos² sin  k, r⁰() = − sin  i − cos  k, and F(r()) · r⁰() = − sin  + cos³ sin . Thus



curl F · S =

F· r =2

0 F(r()) · r⁰()  =2

0 (− sin  + cos³ sin ) 

=

cos  −¹4cos⁴2

0 = 0

7. curl F = −2 i − 2 j − 2 k and we take the surface  to be the planar region enclosed by , so  is the portion of the plane

 +  +  = 1over  = {( ) | 0 ≤  ≤ 1, 0 ≤  ≤ 1 − }. Since  is oriented counterclockwise, we orient  upward.

Using Equation 16.7.10, we have  = ( ) = 1 −  − ,  = −2,  = −2,  = −2, and



F· r =

curl F · S =

[−(−2)(−1) − (−2)(−1) + (−2)] 

=1 0

1−

0 (−2)   = −21

0(1 − )  = −1 8. curl F = ( − ) i −  j + k and  is the portion of the plane 3 + 2 +  = 1 over

 =

( ) | 0 ≤  ≤ ¹3 0 ≤  ≤ ¹2(1 − 3). We orient  upward and use Equation 16.7.10 with

 = ( ) = 1 − 3 − 2:



F· r =

curl F · S =

[−( − )(−3) − (−)(−2) + 1]  =13 0

(1−3)2

0 (1 + 3 − 5)  

=13 0

(1 + 3) −⁵2²=(1−3)2

=0  =13

0

₁

2(1 + 3)(1 − 3) −⁵2·¹4(1 − 3)²



=13 0

−⁸¹8²+¹⁵₄ −¹8

 =

−²⁷8³+¹⁵₈²−¹813

0 = −¹8 +₂₄⁵ −24¹ = ₂₄¹

9. curl F = − i −  j −  k and we take  to be the part of the paraboloid  = 1 − ²− ²in the ﬁrst octant. Since  is oriented counterclockwise (from above), we orient  upward. Then using Equation 16.7.10 with  = ( ) = 1 − ²− ² we have



F· r =

curl F · S =

[−(−)(−2) − (−)(−2) + (−)]  =



−2 − 2(1 − ²− ²) − 



=2 0

1 0

−2( cos )( sin ) − 2( sin )(1 − ²) −  cos 

  

=2 0

1 0

−2³sin  cos  − 2(²− ⁴) sin  − ²cos 

 

=2 0

−¹2⁴sin  cos  − 21

3³−¹5⁵

sin  −¹3³cos =1

=0 

=2 0

−¹2sin  cos  −15⁴ sin  −¹3cos 

 =

−¹4sin² +₁₅⁴ cos  −¹3sin 2 0

= −¹4−15⁴ −¹3 = −¹⁷20

10. The curve of intersection is an ellipse in the plane  =  + 2. curl F = (1 − ) i − j + ( − 2) k and we take the surface  to be the planar region enclosed by  with upward orientation. From Equation 16.7.10 with  = ( ) =  + 2 we have



F· r =

curl F · S = 

²+²≤1

[−(1 − ) (0) − (−1)(1) + ( + 2 − 2)] 

= 

²+²≤1

( + 1)  =2

0

1

0 ( sin  + 1)    =2

0

1

3³sin  +¹₂²=1

=0 

=2

0

1

3sin  +¹₂

 =

−¹3cos  +¹₂2

0 = 

6. The boundary curve  is the circle ²+ ² = 1,  = 0 which should be oriented in the counterclockwise direction when viewed from the right, so a vector equation of  is r() = cos(−) i + sin(−) k = cos  i − sin  k, 0 ≤  ≤ 2. Then F(r()) = i + − cos  sin j− cos² sin  k, r⁰() = − sin  i − cos  k, and F(r()) · r⁰() = − sin  + cos³ sin . Thus



curl F · S =

F· r =2

0 F(r()) · r⁰()  =2

0 (− sin  + cos³ sin ) 

=

cos  −¹4cos⁴2

0 = 0

7. curl F = −2 i − 2 j − 2 k and we take the surface  to be the planar region enclosed by , so  is the portion of the plane

 +  +  = 1over  = {( ) | 0 ≤  ≤ 1, 0 ≤  ≤ 1 − }. Since  is oriented counterclockwise, we orient  upward.

Using Equation 16.7.10, we have  = ( ) = 1 −  − ,  = −2,  = −2,  = −2, and



F· r =

curl F · S =

[−(−2)(−1) − (−2)(−1) + (−2)] 

=1 0

1−

0 (−2)   = −21

0(1 − )  = −1 8. curl F = ( − ) i −  j + k and  is the portion of the plane 3 + 2 +  = 1 over

 =

( ) | 0 ≤  ≤ ¹3 0 ≤  ≤ ¹2(1 − 3). We orient  upward and use Equation 16.7.10 with

 = ( ) = 1 − 3 − 2:



F· r =

curl F · S =

[−( − )(−3) − (−)(−2) + 1]  =13 0

(1−3)2

0 (1 + 3 − 5)  

=13 0

(1 + 3) −⁵2²=(1−3)2

=0  =13

0

1

2(1 + 3)(1 − 3) −⁵2·¹4(1 − 3)²



=13 0

−⁸¹8²+¹⁵₄ −¹8

 =

−²⁷8³+¹⁵₈²−¹813

0 = −¹8 +₂₄⁵ −24¹ = ₂₄¹

9. curl F = − i −  j −  k and we take  to be the part of the paraboloid  = 1 − ²− ²in the ﬁrst octant. Since  is oriented counterclockwise (from above), we orient  upward. Then using Equation 16.7.10 with  = ( ) = 1 − ²− ² we have



F· r =

curl F · S =

[−(−)(−2) − (−)(−2) + (−)]  =



−2 − 2(1 − ²− ²) − 



=2 0

1 0

−2( cos )( sin ) − 2( sin )(1 − ²) −  cos 

  

=2 0

1 0

−2³sin  cos  − 2(²− ⁴) sin  − ²cos 

 

=2 0

−¹2⁴sin  cos  − 2₁

3³−¹5⁵

sin  −¹3³cos =1

=0 

=2 0

−¹2sin  cos  −15⁴ sin  −¹3cos 

 =

−¹4sin² +₁₅⁴ cos  −¹3sin 2 0

= −¹4−15⁴ −¹3 = −¹⁷20

10. The curve of intersection is an ellipse in the plane  =  + 2. curl F = (1 − ) i − j + ( − 2) k and we take the surface  to be the planar region enclosed by  with upward orientation. From Equation 16.7.10 with  = ( ) =  + 2 we have



F· r =

curl F · S = 

²+²≤1

[−(1 − ) (0) − (−1)(1) + ( + 2 − 2)] 

= 

²+²≤1

( + 1)  =2

0

1

0 ( sin  + 1)    =2

0

1

3³sin  +¹₂²=1

=0 

=2

0

₁

3sin  +¹₂

 =

−¹3cos  +¹₂2

0 = 

Now curl F = (−3 − 2) j + 2 k, and the projection  of  on the -plane is the disk ²+ ²≤ 4, so by Equation 16.7.10 with  = ( ) = 5 − ²− ²we have



curl F · S =

[−0 − (−3 − 2)(−2) + 2]  =

[−6 − 4²+ 2(5 − ²− ²)] 

=2

0

2 0

−6 sin  − 4²sin² + 2(5 − ²)

   =2

0

−2³sin  − ⁴sin² + 5²−¹2⁴=2

=0 

=2

0

−16 sin  − 16 sin² + 20 − 8

 = 16 cos  − 16₁

2 −¹4sin 2

+ 122

0 = 8

15. The boundary curve  is the circle ²+ ²= 1,  = 0 oriented in the counterclockwise direction as viewed from the positive

-axis. Then  can be described by r() = cos  i − sin  k, 0 ≤  ≤ 2, and r⁰() = − sin  i − cos  k. Thus F(r()) = − sin  j + cos  k, F(r()) · r⁰() = − cos², and

F· r =2

0 (− cos²)  = −¹2 −¹4sin 22

0 = −.

Now curl F = −i − j − k, and  can be parametrized (see Example 16.6.10) by r( ) = sin  cos  i + sin  sin  j + cos  k, 0 ≤  ≤ , 0 ≤  ≤ . Then r× r^ = sin² cos  i + sin² sin  j + sin  cos  kand



curl F · S = 

²+²≤1

curl F · (r^× r^)  = 0



0(− sin² cos  − sin² sin  − sin  cos )  

=

0(−2 sin² −  sin  cos )  =1

2sin 2 −  −^2 sin² 0 = −

16. Let  be the surface in the plane  +  +  = 1 with upward orientation enclosed by . Then an upward unit normal vector for  is n =√¹

3(i + j + k). Orient  in the counterclockwise direction, as viewed from above.

  − 2  + 3  is equivalent to

F· r for F(  ) =  i − 2 j + 3 k, and the components of F are polynomials, which have continuous partial derivatives throughout R³. We have curl F = 3 i + j − 2 k, so by Stokes’ Theorem,



  − 2  + 3  =

F· r =

curl F · n  =

(3 i + j − 2 k) ·^√¹₃(i + j + k) 

=^√²₃

 =^√²₃(surface area of ) Thus the value of

  − 2  + 3  is always^√²₃ times the area of the region enclosed by , regardless of its shape or location. [Notice that because n is normal to a plane, it is constant. But curl F is also constant, so the dot product curl F · n is constant and we could have simply argued that

curl F · n  is a constant multple of

, the surface area of .]

17. It is easier to use Stokes’ Theorem than to compute the work directly. Let  be the planar region enclosed by the path of the particle, so  is the portion of the plane  =¹₂for 0 ≤  ≤ 1, 0 ≤  ≤ 2, with upward orientation.

curl F = 8 i + 2 j + 2 k and



F· r =

curl F · S =



−8 (0) − 2₁

2

+ 2

 =1 0

2 0

2 −¹2

 

=1 0

2 0

3

2   =1 0

3 4²=2

=0 =1

0 3  = 3 18. 

( + sin )  + (²+ cos )  + ³ =

F· r, where F(  ) = ( + sin ) i + (²+ cos ) j + ³k ⇒ curl F = −2 i − 3²j− k. Since sin 2 = 2 sin  cos ,  lies on the surface  = 2. Let  be the part of this surface that is bounded by . Then the projection of  onto the -plane is the unit disk  [²+ ² ≤ 1].  is traversed clockwise (when viewed from above) so  is oriented downward. Using Equation 16.7.10 with ( ) = 2,

Now curl F = (−3 − 2) j + 2 k, and the projection  of  on the -plane is the disk ²+ ²≤ 4, so by Equation 16.7.10 with  = ( ) = 5 − ²− ²we have



curl F · S =

[−0 − (−3 − 2)(−2) + 2]  =

[−6 − 4²+ 2(5 − ²− ²)] 

=2

0

2 0

−6 sin  − 4²sin² + 2(5 − ²)

   =2

0

−2³sin  − ⁴sin² + 5²−¹2⁴=2

=0 

=2

0

−16 sin  − 16 sin² + 20 − 8

 = 16 cos  − 161

2 −¹4sin 2

+ 122

0 = 8

15. The boundary curve  is the circle ²+ ²= 1,  = 0 oriented in the counterclockwise direction as viewed from the positive

-axis. Then  can be described by r() = cos  i − sin  k, 0 ≤  ≤ 2, and r⁰() = − sin  i − cos  k. Thus F(r()) = − sin  j + cos  k, F(r()) · r⁰() = − cos², and

F· r =2

0 (− cos²)  = −¹2 −¹4sin 22

0 = −.

Now curl F = −i − j − k, and  can be parametrized (see Example 16.6.10) by r( ) = sin  cos  i + sin  sin  j + cos  k, 0 ≤  ≤ , 0 ≤  ≤ . Then r× r^ = sin² cos  i + sin² sin  j + sin  cos  kand



curl F · S = 

²+²≤1

curl F · (r^× r^)  = 0



0(− sin² cos  − sin² sin  − sin  cos )  

=

0(−2 sin² −  sin  cos )  =₁

2sin 2 −  −^2 sin² 0 = −

16. Let  be the surface in the plane  +  +  = 1 with upward orientation enclosed by . Then an upward unit normal vector for  is n =√¹

3(i + j + k). Orient  in the counterclockwise direction, as viewed from above.

  − 2  + 3  is equivalent to

F· r for F(  ) =  i − 2 j + 3 k, and the components of F are polynomials, which have continuous partial derivatives throughout R³. We have curl F = 3 i + j − 2 k, so by Stokes’ Theorem,



  − 2  + 3  =

F· r =

curl F · n  =

(3 i + j − 2 k) ·^√¹₃(i + j + k) 

=√² 3



 =√²

3(surface area of ) Thus the value of

  − 2  + 3  is always^√²₃ times the area of the region enclosed by , regardless of its shape or location. [Notice that because n is normal to a plane, it is constant. But curl F is also constant, so the dot product curl F · n is constant and we could have simply argued that

curl F · n  is a constant multple of

, the surface area of .]

17. It is easier to use Stokes’ Theorem than to compute the work directly. Let  be the planar region enclosed by the path of the particle, so  is the portion of the plane  =¹₂for 0 ≤  ≤ 1, 0 ≤  ≤ 2, with upward orientation.

curl F = 8 i + 2 j + 2 k and



F· r =

curl F · S =



−8 (0) − 21 2

+ 2

 =1 0

2 0

2 −¹2

 

=1 0

2 0

3

2   =1 0

₃

4²=2

=0 =1

0 3  = 3 18. 

( + sin )  + (²+ cos )  + ³ =

F· r, where F(  ) = ( + sin ) i + (²+ cos ) j + ³k ⇒ curl F = −2 i − 3²j− k. Since sin 2 = 2 sin  cos ,  lies on the surface  = 2. Let  be the part of this surface that is bounded by . Then the projection of  onto the -plane is the unit disk  [²+ ² ≤ 1].  is traversed clockwise (when viewed from above) so  is oriented downward. Using Equation 16.7.10 with ( ) = 2,

SECTION 16.9 THE DIVERGENCE THEOREM ¤ 699

 = −2 = −2(2) = −4,  = −3²,  = −1 and multiplying by −1 for the downward orientation, we have



F· r = −

curl F · S = −



−(−4)(2) − (−3²)(2) − 1



= −

(8²+ 6³− 1)  = −2

0

1

0(8³cos  sin² + 6³cos³ − 1)   

= −2

0

8

5cos  sin² +⁶₅cos³ −¹2

 = −8

15sin³ +⁶₅

sin  −¹3sin³

−¹22

0 = 

19. Assume  is centered at the origin with radius  and let 1and 2be the upper and lower hemispheres, respectively, of .

Then

curl F · S =

₁curl F · S +

₂curl F · S =

₁F· r +

₂F· r by Stokes’ Theorem. But ¹is the circle ²+ ²= ²oriented in the counterclockwise direction while 2is the same circle oriented in the clockwise direction.

Hence

₂F· r = −

₁F· r so

curl F · S = 0 as desired.

20. (a) By Exercise 16.5.26, curl(∇) =  curl(∇) + ∇ × ∇ = ∇ × ∇ since curl(∇) = 0. Hence by Stokes’

Theorem

( ∇) · r =

(∇ × ∇) · S.

(b) As in (a), curl(∇) = ∇ × ∇ = 0, so by Stokes’ Theorem,

( ∇) · r =

[curl( ∇)] · S = 0.

(c) As in part (a),

curl( ∇ + ∇) = curl(∇) + curl(∇) [by Exercise 16.5.24]

= (∇ × ∇) + (∇ × ∇) = 0 [since u × v = −(v × u)]

Hence by Stokes’ Theorem,

( ∇ + ∇) · r =

curl( ∇ + ∇) · S = 0.

16.9 The Divergence Theorem

1. div F = 3 +  + 2 = 3 + 3, so



div F  =1 0

1 0

1

0(3 + 3)    =⁹₂ (notice the triple integral is three times the volume of the cube plus three times ).

To compute

F· S, on

1: n = i, F = 3 i +  j + 2 k, and

₁F· S =

₁3  = 3;

2: F = 3 i +  j + 2 k, n = j and

₂F· S =

₂  =¹₂;

3: F = 3 i +  j + 2 k, n = k and

₃F· S =

₃2  = 1;

4: F = 0,

4F· S = 0; ⁵: F = 3 i + 2 k, n = −j and

5F· S =

50  = 0;

6: F = 3 i +  j, n = −k and

₆F· S =

₆0  = 0. Thus

F· S =⁹2. 2. div F = 0 + 2 + 8 = 10so, using cylindrical coordinates,



div F  =

10  =2

0

3 0

9

²(10)    

=2

0

3 0

5²=9

=²   =2

0

3

0(405 − 5⁵)  

=2

0 3

0(405 − 5⁵)  =

2

0

405

2 ²−⁵6⁶3 0

= 2₃₆₄₅

2 −¹²¹⁵2

= 2430

[continued]

1

(2)

SECTION 16.9 THE DIVERGENCE THEOREM ¤ 699

 = −2 = −2(2) = −4,  = −3²,  = −1 and multiplying by −1 for the downward orientation, we have



F· r = −

curl F · S = −



−(−4)(2) − (−3²)(2) − 1



= −

(8²+ 6³− 1)  = −2

0

1

0(8³cos  sin² + 6³cos³ − 1)   

= −2

0

8

5cos  sin² +⁶₅cos³ −¹2

 = −8

15sin³ +⁶₅

sin  −¹3sin³

−¹22

0 = 

19. Assume  is centered at the origin with radius  and let 1and 2be the upper and lower hemispheres, respectively, of .

Then

curl F · S =

₁curl F · S +

₂curl F · S =

₁F· r +

₂F· r by Stokes’ Theorem. But ¹is the circle ²+ ²= ²oriented in the counterclockwise direction while 2is the same circle oriented in the clockwise direction.

Hence

2F· r = −

1F· r so

curl F · S = 0 as desired.

20. (a) By Exercise 16.5.26, curl(∇) =  curl(∇) + ∇ × ∇ = ∇ × ∇ since curl(∇) = 0. Hence by Stokes’

Theorem

( ∇) · r =

(∇ × ∇) · S.

(b) As in (a), curl(∇) = ∇ × ∇ = 0, so by Stokes’ Theorem,

( ∇) · r =

[curl( ∇)] · S = 0.

(c) As in part (a),

curl( ∇ + ∇) = curl(∇) + curl(∇) [by Exercise 16.5.24]

= (∇ × ∇) + (∇ × ∇) = 0 [since u × v = −(v × u)]

Hence by Stokes’ Theorem,

( ∇ + ∇) · r =

curl( ∇ + ∇) · S = 0.

16.9 The Divergence Theorem

1. div F = 3 +  + 2 = 3 + 3, so



div F  =1 0

1 0

1

0(3 + 3)    =⁹₂ (notice the triple integral is three times the volume of the cube plus three times ).

To compute

F· S, on

1: n = i, F = 3 i +  j + 2 k, and

1F· S =

13  = 3;

2: F = 3 i +  j + 2 k, n = j and

2F· S =

2  =¹₂;

3: F = 3 i +  j + 2 k, n = k and

₃F· S =

₃2  = 1;

4: F = 0,

₄F· S = 0; ⁵: F = 3 i + 2 k, n = −j and

₅F· S =

₅0  = 0;

6: F = 3 i +  j, n = −k and

₆F· S =

₆0  = 0. Thus

F· S =⁹2. 2. div F = 0 + 2 + 8 = 10so, using cylindrical coordinates,



div F  =

10  =2

0

3 0

9

²(10)    

=2

0

3 0

5²=9

=²   =2

0

3

0(405 − 5⁵)  

=2

0 3

0(405 − 5⁵)  =

2

0

₄₀₅

2 ²−⁵6⁶3 0

= 23645 2 −¹²¹⁵2

= 2430

[continued]