Section 16.8 Stokes’ Theorem

(1)

Section 16.8 Stokes’ Theorem

10. Use Stokes Theorem to evaluate R_CF·dr. In each case C is oriented counterclockwise as viewed from above, unless otherwise stated.

F(x, y, z) = 2y i + xz j + (x + y) k, C is the curve of intersection of the plane z = y + 2 and the cylinder x²+ y²= 1.

Solution:

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 = (^− 2) i − (^− ) j + (2 − ) k and we take  to be the disk ²+ ²≤ 16,  = 5. Since  is oriented counterclockwise (from above), we orient  upward. Then n = k and curl F · n = 2 −  on , where  = 5. Thus



F· r =

curl F · n =

(2 − )  =

(10 − 5)  = 5(area of ) = 5( · 4²) = 80

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 = 

14. Use Stokes Theorem to evaluate R_CF·dr. In each case C is oriented counterclockwise as viewed from above, unless otherwise stated.

F(x, y, z) =< x³− z, xy, y + z²>, C is the curve of intersection of the paraboloid z = x²+ y²and the plane z = x.

1200 CHAPTER 16 Vector Calculus

6. Fsx, y, zd − e^xy i 1 e^xzj 1 x²zk,

S is the half of the ellipsoid 4x²1y²14z²− 4 that lies to the right of the xz-plane, oriented in the direction of the positive y-axis

z

y

x 0

S

4≈+¥+4z@=4

7–14 Use Stokes’ Theorem to evaluate yC F dr. In each case C is oriented counterclockwise as viewed from above, unless otherwise stated.

7. Fsx, y, zd − sx 1 y²di 1sy 1 z²dj 1sz 1 x²dk, C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1)

8. Fsx, y, zd − i 1 sx 1 yzd j 1

(

^{xy 2}_sz

)

k,

C is the boundary of the part of the plane 3x 1 2y 1 z − 1 in the first octant

9. Fsx, y, zd − xy i 1 yz j 1 zx k,

C is the boundary of the part of the paraboloid z − 1 2 x²2y² in the first octant

z=1-≈-¥

C z

y x

10. Fsx, y, zd − 2y i 1 xz j 1 sx 1 yd k,

C is the curve of intersection of the plane z − y 1 2 and the cylinder x²1y²− 1

11. Fsx, y, zd − k2yx², xy², e^xyl, C is the circle in the xy-plane of radius 2 centered at the origin

12. Fsx, y, zd − ze^x i 1sz 2 y³d j 1 sx 2 z³d k,

C is the circle y²1z²− 4, x − 3, oriented clockwise as viewed from the origin

13. Fsx, y, zd − x²y i 1 x³ j 1 e^z tan²¹z k,

C is the curve with parametric equations x − cos t, y − sin t, z − sin t, 0 < t < 2

z

x y (_1, 0, 0)

(1, 0, 0) C

14. Fsx, y, zd − kx ³2z, xy, y 1 z²l, C is the curve of intersec- tion of the paraboloid z − x²1y² and the plane z − x

z

x y

z=≈+¥

z=x

15. (a) Use Stokes’ Theorem to evaluate yC F dr, where Fsx, y, zd − x²zi 1 xy²j 1 z²k

and C is the curve of intersection of the plane

x 1 y 1 z − 1 and the cylinder x²1y²− 9, oriented counterclockwise as viewed from above.

(b) Graph both the plane and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part (a).

(c) Find parametric equations for C and use them to graph C.

16. (a) Use Stokes’ Theorem to evaluate yC F dr, where Fsx, y, zd − x²y i 1¹₃x³j 1 xy k and C is the curve of intersection of the hyperbolic paraboloid z − y²2x² and the cylinder x²1y²− 1, oriented counterclockwise as viewed from above.

(b) Graph both the hyperbolic paraboloid and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part (a).

(c) Find parametric equations for C and use them to graph C.

17–19 Verify that Stokes’ Theorem is true for the given vector field F and surface S.

17. Fsx, y, zd − 2y i 1 x j 2 2 k,

S is the cone z²− x²1y², 0 < z < 4, oriented downward

;

Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

SECTION 16.8 STOKES’ THEOREM ¤ 1681

10. F(  ) = 2 i +  j + ( + ) k. 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 = 

11. F(  ) =

−² ² ^and curl F = ^i− ^j+ (²+ ²) k.  is the circle in the plane centered at the origin with radius 2. Choose  to be the portion of the plane enclosed by . So  =  = {( ) | ²+ ²≤ 4}.  is oriented counterclockwise, so we orient  upward and the normal vector to  is n = k. By Stokes’ Theorem (See Solution 2 of Example 2), we get



F· r =

curl F · S =

curl F · n S =

[^i− ^j+ (²+ ²) k] · k 

=

(²+ ²)  =2 0

2

0 ²   = 22

0 ³ = 2

⁴ 4

2 0

= 8

12. F(  ) = ^i+ ( − ³) j + ( − ³) kand curl F = − i − (1 − ^) j.  is the circle ²+ ²= 4,  = 3, and we choose the surface  to be the portion of the plane  = 3 enclosed by  The projection of  onto the plane is the disk

 = {( ) | ²+ ²≤ 4}.  is oriented clockwise, so we orient  to have normal vector n = i. By Stokes’ Theorem (see Solution 2 of Example 2), we get



F· r =

curl F · S =

curl F · n S =

[−i − (1 − ^) j] · i 

= −

 = −() = −(2²) = −4

13. F(  ) = ² i + ³j+ ^tan⁻¹ kand curl F = 2²k. Note that the curve  = hcos  sin  sin i is contained in the plane  =  because the j and k components of the curve are equal. We choose the surface  to be the portion of the plane

 = enclosed by . The projection of  onto the plane is the circle hcos  sin  0i, and  is the disk in the plane enclosed by . We orient  upward and use Equation 16.7.10 with  = ( ) = :



F· r =

curl F · S =

[−(0)(0) − (0)(1) + 2²]  =2

0

1

0 2( cos )²  

=

 2

0

(1 + cos 2) 

 1 0

³ =



 +sin 2

2

2

0

⁴ 4

¹

0

= (2 + 0 − 0)1 4 = 

2 14. F(  ) =

³−    + ²

and curl F = i − j +  k.  is the curve of intersection of the paraboloid  = ²+ ² and the plane  = . Let  be the portion of the plane  =  enclosed by . To find the projection of  onto the plane, note that  = ²+ ². Converting to polar coordinates, we get  cos  = ² ⇒  = cos . So  is the region in the

plane enclosed by the circle  = cos ; that is,  = {( ) | 0 ≤  ≤ cos , 0 ≤  ≤ }. We orient  upward and use Equation 16.7.10 with  = ( ) = :





F· r =



curl F · S =



[−(1)(1) − (−1)(0) + ]  =

(−1 + ) 

=

  0

 cos 

0 (−1 +  sin )    =

  0



−² 2 + ³

3 sin 

^{cos }

0



=

  0



−cos²

2 +cos³ 3 sin 



 =

  0



−1 + cos 2

4 +cos³ 3 sin 





=



−1

4 −sin 2

8 −cos⁴ 12

^

0

=



−

4 − 0 − 1 12



−



0 − 0 − 1 12



= − 4

15. (a) F(  ) = ² i + ²j+ ²k. The curve of intersection is an ellipse in the plane  +  +  = 1. The unit normal is n= √¹

3(i + j + k), curl F = ²j+ ²k, and curl F · n = ^√¹₃(²+ ²). Then, by Stokes’ Theorem,



F· r =

curl F · n S =



√1 3

²+ ²



=

²+ ²≤ 9

²+ ²

  =2

0

3

0 ³  = 281 4

= ^81₂

(b) (c) One possible parametrization is  = 3 cos ,  = 3 sin ,

 = 1 − 3 cos  − 3 sin , 0 ≤  ≤ 2.

16. (a) F(  ) = ² i +¹₃³j+  k.  is the part of the surface  = ²− ²that lies above the unit disk .

curl F =  i −  j + (²− ²) k =  i −  j. Using Equation 16.7.10 with  = ( ) = ²− ², by Stokes’ Theorem we have



F· r =

curl F · S =

[−(−2) − (−)(2)]  = 2

(²+ ²) 

= 22

0

1

0 ²   = 2(2)1 4⁴1

0= 

(b) (c) One possible set of parametric equations is  = cos ,

 = sin ,  = sin² − cos², 0 ≤  ≤ 2.

1

(2)

20. Let C be a simple closed smooth curve that lies in the plane x + y + z = 1. Show that the line integral Z

C

z dx − 2x dy + 3y dz

depends only on the area of the region enclosed by C and not on the shape of C or its location in the plane.

Solution:

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 −  −^2sin² 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,

22. Evaluate

Z

C

(y + sin x) dx + (z²+ cos y) dy + x³dz

where C is the curve r =< sin t, cos t, sin 2t >, 0 ≤ t ≤ 2π. [Hint: Observe that C lies on the surface z = 2xy.]

Solution:

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 −  −^2sin² 0 = −

16. Let  be the surface in the plane  +  +  = 1 with upward orientation enclosed by . Then an upward unit normal vector for  is n = ^√¹₃(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]

2