Section 16.7 Surface Integrals

Section 16.7 Surface Integrals

14. Evaluate the surface integral. RR

Sy²z²dS, S is the part of the cone y =√

x²+ z² given by 0 ≤ y ≤ 5.

Solution:

686 ¤ CHAPTER 16 VECTOR CALCULUS

12. The sphere intersects the cylinder in the circle ²+ ²= 1,  =√

3, so  is the portion of the sphere where  ≥√ 3.

Using spherical coordinates to parametrize the sphere we have r( ) = 2 sin  cos  i + 2 sin  sin  j + 2 cos  k, and

|r^× r^| = 4 sin  (see Example 16.6.10). The portion where  ≥√

3corresponds to 0 ≤  ≤^6, 0 ≤  ≤ 2 so



² =2

0

6

0 (2 sin  sin )²(4 sin )   = 162

0 sin² 6

0 sin³ 

= 16₁

2 −¹4sin 22

0

₁

3cos³ − cos 6

0 = 16()√ 3

8 − ^√2³ −¹3+ 1

=₃₂

3 − 6√ 3



13. Using  and  as parameters, we have r( ) = (²+ ²) i +  j +  k, ²+ ²≤ 1. Then r× r^= (2 i + j) × (2 i + k) = i − 2 j − 2 k and |r^× r^| =

1 + 4²+ 4²=

1 + 4(²+ ²). Thus



 ² = 

²+²≤1

²

1 + 4(²+ ²)  =2

0

1

0( sin )²√

1 + 4²  

=2

0 sin² 1 0 ³√

1 + 4² let  = 1 + 4² ⇒ ²= ¹₄( − 1) and   = ¹8

=1

2 −¹4sin 22

0

5 1

1

4( − 1)√

 ·¹8 =  · 32¹

5

1(^32− ^12)  = ₃₂¹

2

5^52−²3^325 1

= ₃₂¹

2

5(5)^52−²3(5)^32−²5 +²₃

= ₃₂¹₂₀

3

√5 +₁₅⁴

= ₁₂₀¹  25√

5 + 1

14. Using  and  as parameters, we have r( ) =  i +√

²+ ²j+  k, ²+ ²≤ 25. Then r× r^ =



i+ 

√²+ ² j



×

 

√²+ ²j+ k



= 

√²+ ² i− j + 

√²+ ² k and

|r^× r^| =

 ²

²+ ² + 1 + ²

²+ ² =

²+ ²

²+ ² + 1 =√ 2. Thus



 ²² = 

²+²≤25

(²+ ²)²√

2  =√ 22

0

5

0 ²( sin )²  

=√ 22

0 sin²  5

0 ⁵ =√ 21

2 −¹4sin 22

0

1 6⁶5

0

=√

2 () ·¹6(15,625 − 0) = 15,625√ 2

6 

15. Using  and  as parameters, we have r( ) =  i + (²+ 4) j +  k, 0 ≤  ≤ 1, 0 ≤  ≤ 1. Then r× r^ = (i + 2 j) × (4 j + k) = 2 i − j + 4 k and |r^× r^| =√

4²+ 1 + 16 =√

4²+ 17. Thus



  =1 0

1 0 √

4²+ 17   =1 0 √

4²+ 17  =

1

8 ·²3(4²+ 17)^321 0

= ₁₂¹(21^32− 17^32) = ₁₂¹  21√

21 − 17√ 17

= ⁷₄√ 21 −¹⁷12

√17

16. The sphere intersects the cone in the circle ²+ ²= ¹₂,  = ^√1₂, so  is the portion of the sphere where  ≥ ^√1₂. Using spherical coordinates to parametrize the sphere we have r( ) = sin  cos  i + sin  sin  j + cos  k, and

|r^× r^| = sin  (as in Example 1). The portion where  ≥ ^√1₂ corresponds to 0 ≤  ≤^4, 0 ≤  ≤ 2 so



 ² =2

0

4

0 (sin  sin )²(sin )   =2

0 sin²  4

0 sin³  =2

0 sin² 4

0 (1 − cos²) sin  

=₁

2 −¹4sin 22

0

₁

3cos³ − cos 4

0 = √ 2

12 −^√2²−¹3 + 1

=

2 3−⁵12^√2



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28. Evaluate the surface integralRR

SF·dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = yz i + zx j + xy k. S is the surface z = x sin y, 0 ≤ x ≤ 2, 0 ≤ y ≤ π, with upward orientation.

Solution:

SECTION 16.7 SURFACE INTEGRALS ¤ 689

and 

F· S =

[F(r( )) · (r^× r^)]  =2

0



0(sin³ + sin  cos³)  

=2

0  

0(1 − cos² + cos³) sin   = (2)

− cos  +¹3cos³ −¹4cos⁴ 0

= 2

1 −¹3−¹4 + 1 −¹3 +¹₄

= ⁸₃

26. F(  ) =  i −  j + 2 k,  = ( ) =

4 − ²− ²and  is the disk

( ) ²+ ²≤ 4.  has downward orientation, so by Equation 10,



F· S = −

[− ·¹2(4 − ²− ²)^−12(−2) − (−) · ¹2(4 − ²− ²)^−12(−2) + 2] 

= −





 

4 − ²− ² − 

4 − ²− ² + 2

4 − ²− ²





= −

2

4 − ²− ² = −22

0

2 0

√4 − ²  

= −22

0  2 0 √

4 − ² = −2(2)

−¹2 ·²3(4 − ²)^322 0

= −4

0 +¹₃(4)^32

= −4 · ⁸3 = −³²3

27. Let 1be the paraboloid  = ²+ ², 0 ≤  ≤ 1 and ²the disk ²+ ²≤ 1,  = 1. Since  is a closed surface, we use the outward orientation.

On 1: F(r( )) = (²+ ²) j −  k and r^× r^= 2 i − j + 2 k (since the j-component must be negative on ¹). Then



₁F· S = 

²+ ²≤ 1

[−(²+ ²) − 2²]  = −2

0

1

0(²+ 2²sin²)   

= −2

0

1

0 ³(1 + 2 sin²)   = −2

0 (1 + 1 − cos 2) 1 0 ³

= −

2 −¹2sin 2^2

0

1 4⁴¹

0= −4 ·¹4 = −

On 2: F(r( )) = j −  k and r^× r^= j. Then

₂F· S = 

²+ ²≤ 1

(1)  = .

Hence

F· S = − +  = 0.

28. F(  ) =  i +  j +  k,  = ( ) =  sin , and  is the rectangle [0 2] × [0 ], so by Equation 10



F· S =

[−(sin ) − ( cos ) + ] 

= 0

2

0(− sin² − ³sin  cos  + )  

= 0

−¹2² sin² − ¹4⁴sin  cos  +¹₂²=2

=0 

= 0

−2 sin² − 4 sin  cos  + 2

 [integrate by parts in the first term]

=

−¹2²+ ¹₂ sin 2 +¹₄cos 2

− 2 sin² + ²

0 = −¹2²+¹₄ + ²−¹4 = ¹₂² 29. F(  ) =  i −  j +  k,  = ( ) =

4 − ²− ²and  is the quarter disk

( )

 0 ≤  ≤ 2 0 ≤  ≤√

4 − ²

.  has downward orientation, so by Formula 10,

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30. Evaluate the surface integralRR

SF·dS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = x i + y j + 5 k, S is the boundary of the region enclosed by the cylinder x²+ z²= 1 and the planes y = 0 and x + y = 2.

Solution:

1

1674 ¤ CHAPTER 16 VECTOR CALCULUS

30. F(  ) =  i +  j + 5 k. Here  consists of three surfaces: 1, the lateral surface of the cylinder ²+ ²= 1;

2, the front formed by the plane  +  = 2; and the back, 3, in the plane  = 0.

On 1: r( ) = sin  i +  j + cos  k. F(r( )) = sin  i +  j + 5 k and r× r^= sin  i + cos  k ⇒



₁F· S =2

0

2− sin 

0 (sin² + 5 cos )  

=2

0 (2 sin² + 10 cos  − sin³ − 5 sin  cos )  = 2

On 2: r( ) =  i + (2 − ) j +  k. F(r( )) =  i + (2 − ) j + 5 k and r^× r^= i + j.



₂F· S = 

²+ ²≤ 1

[ + (2 − )]  = 2

On 3: F(r( )) =  i + 5 k. The surface is oriented in the negative ­direction so that n = −j and by (8),



₃F· S =

₃F· n S = 0. Hence,

F· S = 4.

31. F(  ) = ²i+ ²j+ ²k. Here  consists of four surfaces: 1, the top surface (a portion of the circular cylinder

²+ ²= 1); 2, the bottom surface (a portion of the ­plane); 3, the front half­disk in the plane  = 2, and 4, the back half­disk in the plane  = 0.

On 1: The surface is  =

1 − ²for 0 ≤  ≤ 2, −1 ≤  ≤ 1 with upward orientation, so by Equation 10,



₁

F· S =

 2 0

 1

−1



−²(0) − ²



− 

1 − ²

 + ²



  =

 2 0

 1

−1

 ³

1 − ² + 1 − ²



 

=2 0



−

1 − ²+¹₃(1 − ²)^32+  −¹3³=1

=−1 =2 0

4 3 =⁸₃

On 2: The surface is  = 0 for 0 ≤  ≤ 2, −1 ≤  ≤ 1 with downward orientation, so that n = −k and by (8),



₂F· S =

₂F· n S =2 0

1

−1

−²

  =2 0

1

−1(0)   = 0 On 3: The surface is  = 2 for −1 ≤  ≤ 1, 0 ≤  ≤

1 − ², oriented in the positive ­direction, so that n = i and by (8),



₃F· S =

₃F· n S =1

−1

 √1−²

0 ²  =1

−1

 √1−²

0 4   = 4 (3) = 2

On 4: The surface is  = 0 for −1 ≤  ≤ 1, 0 ≤  ≤

1 − ², oriented in the negative ­direction, so that n = −i and by (8),



₄F· S =

₄F· n S =1

−1

 √1−²

0 (−²)   =1

−1

 √1−²

0 (0)   = 0 Thus,

F· S = ⁸3 + 0 + 2 + 0 = 2 +⁸₃.

32. F(  ) =  i + ( − ) j +  k. Here  consists of four surfaces: ¹, the triangular face with vertices (1 0 0), (0 1 0), and (0 0 1); 2, the face of the tetrahedron in the ­plane; 3, the face in the ­plane; and 4, the face in the ­plane.

On 1: The face is the portion of the plane  = 1 −  −  for 0 ≤  ≤ 1, 0 ≤  ≤ 1 −  with upward orientation,

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44. Seawater has density 1025 kg/m³ and flows in a velocity field v = y i + x j, where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the hemisphere x²+ y²+ z²= 9, z ≥ 0.

Solution:

SECTION 16.7 SURFACE INTEGRALS ¤ 693

(a)  =

(  ) =2

0

tan⁻¹(34)

0 (25 sin )   = 252

0 tan⁻¹(34)

0 sin  

= 25(2)

− cos

tan^{−1 3}₄ + 1

= 50

−⁴5 + 1= 10.

Because  has constant density,  =  = 0 by symmetry, and

 = _¹ 

(  ) = _10¹ 2

0

tan⁻¹(34)

0 (5 cos )(25 sin )  

= _10¹ (125)2

0  tan⁻¹(34)

0 sin  cos   = _10¹ (125) (2)₁

2sin²tan⁻¹(34)

0 = 25 ·¹2

₃

5

2

= ⁹₂, so the center of mass is (  ) =

0 0⁹₂

 (b) =

(²+ ²)(  ) =2

0

tan⁻¹(34)

0 (25 sin²)(25 sin )  

= 6252

0  tan⁻¹(34)

0 sin³  = 625(2)1

3cos³ − cos ^{tan−1(34)}

0

= 1250

1 3

4 5

3

−⁴5−¹3 + 1

= 125014 375

= ¹⁴⁰₃ 

43. The rate of flow through the cylinder is the flux

v · n  =

v · S. We use the parametric representation r( ) = 2 cos  i + 2 sin  j +  kfor , where 0 ≤  ≤ 2, 0 ≤  ≤ 1, so r^= −2 sin  i + 2 cos  j, r^ = k, and the outward orientation is given by r× r^= 2 cos  i + 2 sin  j. Then



v · S = 2

0

1 0

 i + 4 sin² j + 4 cos² k

· (2 cos  i + 2 sin  j)  

= 2

0

1 0

2 cos  + 8 sin³

  = 2

0

cos  + 8 sin³



= 

sin  + 8

−¹3

(2 + sin²) cos 2

0 = 0kgs

44. A parametric representation for the hemisphere  is r( ) = 3 sin  cos  i + 3 sin  sin  j + 3 cos  k, 0 ≤  ≤ 2, 0 ≤  ≤ 2. Then r^= 3 cos  cos  i + 3 cos  sin  j − 3 sin  k, r^ = −3 sin  sin  i + 3 sin  cos  j, and the outward orientation is given by r× r^= 9 sin² cos  i + 9 sin² sin  j + 9 sin  cos  k. The rate of flow through  is



v · S = 2 0

2

0 (3 sin  sin  i + 3 sin  cos  j) ·

9 sin² cos  i + 9 sin² sin  j + 9 sin  cos  k

 

= 272 0

2

0

sin³ sin  cos  + sin³ sin  cos 

  = 542

0 sin³ 2

0 sin  cos  

= 54

−¹3(2 + sin²) cos 2 0

1

2sin²2

0 = 0kgs 45. consists of the hemisphere 1given by  =

²− ²− ²and the disk 2given by 0 ≤ ²+ ²≤ ²,  = 0.

On 1: E =  sin  cos  i +  sin  sin  j + 2 cos  k,

T× T^= ²sin² cos  i + ²sin² sin  j + ²sin  cos  k. Thus



₁E· S =2

0

2

0 (³sin³ + 2³sin  cos²)  

=2

0

2

0 (³sin  + ³sin  cos²)   = (2)³ 1 +¹₃

= ⁸₃³ On 2: E =  i +  j, and r× r^= −k so

₂E· S = 0. Hence the total charge is  = ⁰

E· S = ⁸3³0.

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