Section 16.7 Surface Integrals

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Section 16.7 Surface Integrals

684 ¤ CHAPTER 16 VECTOR CALCULUS

16.7 Surface Integrals

1. The box is a cube where each face has surface area 4. The centers of the faces are (±1 0 0), (0 ±1 0), (0 0 ±1). For each face we take the point _^∗to be the center of the face and (  ) = cos( + 2 + 3), so by Deﬁnition 1,



 (  )  ≈ [(1 0 0)](4) + [(−1 0 0)](4) + [(0 1 0)](4)

+ [ (0 −1 0)](4) + [(0 0 1)](4) + [(0 0 −1)](4)

= 4 [cos 1 + cos(−1) + cos 2 + cos(−2) + cos 3 + cos(−3)] ≈ −693

2. Each quarter-cylinder has surface area¹₄[2(1)(2)] = and the top and bottom disks have surface area (1)²= . We can take (0 0 1) as a sample point in the top disk, (0 0 −1) in the bottom disk, and (±1 0 0), (0 ±1 0) in the four

quarter-cylinders. Then

 (  ) can be approximated by the Riemann sum

 (1 0 0)() +  (−1 0 0)() + (0 1 0) () + (0 −1 0)() + (0 0 1)() + (0 0 −1)()

= (2 + 2 + 3 + 3 + 4 + 4) = 18 ≈ 565.

3. We can use the - and -planes to divide  into four patches of equal size, each with surface area equal to¹₈the surface area of a sphere with radius√50, so ∆ =¹₈(4)√

502

= 25. Then (±3 ±4 5) are sample points in the four patches, and using a Riemann sum as in Deﬁnition 1, we have



 (  )  ≈ (3 4 5) ∆ + (3 −4 5) ∆ + (−3 4 5) ∆ + (−3 −4 5) ∆

= (7 + 8 + 9 + 12)(25) = 900 ≈ 2827

4. On the surface, (  ) = 

²+ ²+ ²

= (2) = −5. So since the area of a sphere is 4²,



 (  )  =

(2)  = −5

  = −5[4(2)²] = −80.

5. r( ) = ( + ) i + ( − ) j + (1 + 2 + ) k, 0 ≤  ≤ 2, 0 ≤  ≤ 1 and r× r^= (i + j + 2 k) × (i − j + k) = 3 i + j − 2 k ⇒ |r^× r^| =

3²+ 1²+ (−2)²=√

14. Then by Formula 2,



( +  + )  =

( +  +  −  + 1 + 2 + ) |r^× r^|  =1 0

2

0(4 +  + 1) ·√ 14  

=√ 141

0

2²+  + =2

=0 =√ 141

0 (2 + 10)  =√ 14

²+ 101 0= 11√

14

6. r( ) =  cos  i +  sin  j +  k, 0 ≤  ≤ 1, 0 ≤  ≤ 2 and

r× r^= (cos  i + sin  j + k) × (− sin  i +  cos  j) = − cos  i −  sin  j +  k ⇒

|r^× r^| =

²cos² + ²sin² + ²=√

2²=√

2 [since  ≥ 0]. Then by Formula 2,



  =

( cos )( sin )() |r^× r^|  =1 0

2

0 (³sin  cos ) ·√

2   

=√ 21

0 ⁴2

0 sin  cos   =√ 2₁

5⁵1 0

₁

2sin²2

0 =√

2 ·¹5·¹2 =₁₀¹√ 2 7. r( ) = h cos   sin  i, 0 ≤  ≤ 1, 0 ≤  ≤  and

r× r^= hcos  sin  0i × h− sin   cos  1i = hsin  − cos  i ⇒

SECTION 16.7 SURFACE INTEGRALS ¤ 685

|r^× r^| =

sin² + cos² + ²=√

²+ 1. Then



  =

( sin ) |r^× r^|  =1 0



0( sin ) ·√

²+ 1   =1 0 √

²+ 1  0 sin  

=

1

3(²+ 1)^321

0 [− cos ]^0 = ¹₃(2^32− 1) · 2 =²3(2√ 2 − 1)

8. r( ) =

2 ²− ² ²+ ²

, ²+ ² ≤ 1 and r× r^= h2 2 2i × h2 −2 2i =

8 4²− 4² −4²− 4², so

|r^× r^| =

(8)²+ (4²− 4²)²+ (−4²− 4²)²=√

64²²+ 32⁴+ 32⁴

=

32(²+ ²)²= 4√

2 (²+ ²) Then



(²+ ²)  =



(2)²+ (²− ²)²

|r^× r^|  =

(4²²+ ⁴− 2²²+ ⁴) · 4√

2 (²+ ²) 

= 4√ 2

(⁴+ 2²²+ ⁴) (²+ ²)  = 4√ 2

(²+ ²)³ = 4√ 22

0

1

0(²)³  

= 4√ 22

0 1

0 ⁷ = 4√ 2

2

0

₁

8⁸1 0= 4√

2 · 2 ·¹8 =√ 2 

9.  = 1 + 2 + 3 so 

= 2 and 

 = 3. Then by Formula 4,



²  =





²





2

+





2

+ 1  =3 0

2

0 ²(1 + 2 + 3)√

4 + 9 + 1  

=√ 143

0

2

0(² + 2³ + 3²²)   =√ 143

0

1

2²²+ ³²+ ²³=2

=0 

=√ 143

0(10²+ 4³)  =√ 1410

3³+ ⁴3

0= 171√ 14

10. is the part of the plane  = 4 − 2 − 2 over the region  = {( ) | 0 ≤  ≤ 2 0 ≤  ≤ 2 − }. Thus



  =

 (4 − 2 − 2)

(−2)²+ (−2)²+ 1  = 32 0

2−

0

4 − 2²− 2

 

= 32 0

4 − 2² − ²=2−

=0  = 32 0

4(2 − ) − 2²(2 − ) − (2 − )²



= 32 0

³− 4²+ 4

 = 3₁

4⁴−⁴3³+ 2²2 0= 3

4 −³²3 + 8

= 4

11. An equation of the plane through the points (1 0 0), (0 −2 0), and (0 0 4) is 4 − 2 +  = 4, so  is the region in the plane  = 4 − 4 + 2 over  = {( ) | 0 ≤  ≤ 1 2 − 2 ≤  ≤ 0}. Thus by Formula 4,



   =

 

(−4)²+ (2)²+ 1  =√ 211

0

0

2−2    =√ 211

0 []^=0_=2₋₂

=√ 211

0(−2²+ 2)  =√ 21

−²3³+ ²1 0=√

21

−²3+ 1

= ^√₃²¹ 12.  =²₃(^32+ ^32)and



  =





(√ )²+√2

+ 1  =1 0

1

0 √ +  + 1  

=1 0 

2

3( +  + 1)^32=1

=0 =1 0

2 3

( + 2)^32− ( + 1)^32



[continued]

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+ ²j



×

 

√²+ ²j+ k



= 

√²+ ²i− j + 

√²+ ²k 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,  = √¹

2, 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





17.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). Here  is the portion of the sphere corresponding to 0 ≤  ≤ 2, so



(² + ²)  =

(²+ ²)  =2

0

2

0 (4 sin²)(2 cos )(4 sin )  

= 322

0 2

0 sin³ cos   = 32 (2)₁

4sin⁴2

0 = 16(1 − 0) = 16

18.is given by r( ) = cos  i +  j + sin  k, 0 ≤  ≤ 2, 0 ≤  ≤ . Then r× r^= j × (− sin  i + cos  k) = cos  i + sin  k and |r^× r^| =

cos² + sin² = 1, so



( +  + )  = 0

2

0(cos  +  + sin )(1)   = 0

(cos  + sin ) +¹₂²=2

=0

=

0(2 cos  + 2 sin  + 2)  = [2 sin  − 2 cos  + 2]^0 = 2 + 2 + 2 = 4 + 2

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

4²+ 1 + 4²=

1 + 4(²+ ²). Thus



  = 

²+²≤4

(²+ ²)

1 + 4(²+ ²)  =2

0

2 0 ²√

1 + 4²   =2

0 2 0 ²√

1 + 4² 

= 22 0 ²√

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

= 217 1

1

4( − 1)√

 ·¹8 =₁₆¹17

1 (^32− ^12) 

=₁₆¹

2

5^52−²3^3217 1 = ₁₆¹

2

5(17)^52−²3(17)^32−²5 +²₃

=  60

391√ 17 + 1

20.Here  consists of three surfaces: 1, the lateral surface of the cylinder; 2, the front formed by the plane  +  = 5;

and the back, 3, in the plane  = 0.

On 1: the surface is given by r( ) =  i + 3 cos  j + 3 sin  k, 0 ≤  ≤ 2, and 0 ≤  ≤ 5 −  ⇒ 0 ≤  ≤ 5 − 3 cos . Then r^× r^= −3 cos  j − 3 sin  k and |r^× r^| =

9 cos² + 9 sin² = 3, so



₁  =2

0

5− 3 cos 

0 (3 sin )(3)   = 92

0

₁

2²=5−3 cos 

=0 sin  

= ⁹₂2

0 (5 − 3 cos )²sin   =⁹₂₁

9(5 − 3 cos )³2

0 = 0

On 2: r( ) = (5 − ) i +  j +  k and |r^× r^| = |i + j| =√

2, where ²+ ² ≤ 9 and



2  = 

²+ ²≤ 9

(5 − )√

2  =√ 22

0

3

0 (5 −  cos )( sin )   

=√ 22

0

3

0 (5²− ³cos )(sin )   =√ 22

0

₅

3³−¹4⁴cos =3

=0sin  

=√ 22

0

45 −⁸¹4 cos 

sin   =√ 2₄

81

·¹2

45 −⁸¹4 cos 22

0 = 0

On 3:  = 0 so

₃  = 0. Hence

  = 0 + 0 + 0 = 0.

21.From Exercise 5, r( ) = ( + ) i + ( − ) j + (1 + 2 + ) k, 0 ≤  ≤ 2, 0 ≤  ≤ 1, and r^× r^= 3 i + j − 2 k.

Then

F(r( )) = (1 + 2 + )^{(+)(}^−)i− 3(1 + 2 + )^{(+)(}^−)j+ ( + )( − ) k

= (1 + 2 + )^²^−²i− 3(1 + 2 + )^²^−²j+ (²− ²) k [continued]

21. From Exercise 5, r( ) = ( + ) i + ( − ) j + (1 + 2 + ) k, 0 ≤  ≤ 2, 0 ≤  ≤ 1, and r^× r^= 3 i + j − 2 k.

Then

F(r( )) = (1 + 2 + )^{(+)(}^−)i− 3(1 + 2 + )^{(+)(}^−)j+ ( + )( − ) k

= (1 + 2 + )^²^−²i− 3(1 + 2 + )^²^−²j+ (²− ²) k

Because the -component of r× r^is negative we use −(r^× r^)in Formula 9 for the upward orientation:



F· S =

 F· (−(r^× r^))  =1 0

2 0



−3(1 + 2 + )^²^−²+ 3(1 + 2 + )^²^−²+ 2(²− ²)

 

=1 0

2

0 2(²− ²)   = 21 0

1

3³− ²=2

=0  = 21 0

8 3 − 2²



= 28

3 −²3³1 0= 28

3−²3

= 4

22. r( ) = h cos   sin  i, 0 ≤  ≤ 1, 0 ≤  ≤  and

r× r^= hcos  sin  0i × h− sin   cos  1i = hsin  − cos  i. Here F(r( )) =  i +  sin  j +  cos  k and, by Formula 9,



F· S =

F· (r^× r^)  =1 0



0( sin  −  sin  cos  + ²cos )  

=1 0

sin  −  cos  −¹2 sin² + ²sin =

=0  =1

0   = ]¹₀=  23. F(  ) =  i +  j +  k,  = ( ) = 4 − ²− ², and  is the square [0 1] × [0 1], so by Equation 10



F· S =

[−(−2) − (−2) + ]  =1 0

1

0[2² + 2²(4 − ²− ²) + (4 − ²− ²)]  

=1 0

²²+⁸₃³−²3²³−²5⁵+ 4 − ³ −¹3³=1

=0

=1 0

₁

3²+¹¹₃ − ³+³⁴₁₅

 =₁

9³+¹¹₆²−¹4⁴+³⁴₁₅1 0 =⁷¹³₁₈₀ 24. F(  ) = − i −  j + ³k,  = ( ) =

²+ ², and  is the annular region

( ) | 1 ≤ ²+ ²≤ 9. Since  has downward orientation, we have





F· S = −







−(−)

 

²+ ²



− (−)

 

²+ ²

 + ³





= −





 ²+ ²

²+ ² +

²+ ²3

 = −

 2

0

 3 1

²

 + ³



  

= −2

0 3

1(²+ ⁴)  = −

2

0

1

3³+¹₅⁵3 1

= −2

9 +²⁴³₅ −¹3−¹5

= −¹⁷¹²15 

25. F(  ) =  i +  j + ²k, and using spherical coordinates,  is given by  = sin  cos ,  = sin  sin ,  = cos , 0 ≤  ≤ 2, 0 ≤  ≤ . F(r( )) = (sin  cos ) i + (sin  sin ) j + (cos²) kand, from Example 4,

r× r^ = sin² cos  i + sin² sin  j + sin  cos  k. Thus

F(r( )) · (r^× r^) = sin³ cos² + sin³ sin² + sin  cos³ = sin³ + sin  cos³

1

(2)



F· S = −





− ·¹2(4 − ²− ²)^−12(−2) − (−) ·¹2(4 − ²− ²)^−12(−2) + 



= −





 ²

4 − ²− ² −

4 − ²− ²· 

4 − ²− ² + 





= −

²(4 − (²+ ²))^−12 = −2 0

2

0( cos )²(4 − ²)^−12  

= −2

0 cos² 2

0 ³(4 − ²)^−12 

let  = 4 − ² ⇒ ²= 4 −  and −¹2 =  

= −2 0

1

2 +¹₂cos 2

0

4 −¹2(4 − )()^−12

= −₁

2 +¹₄sin 22 0

−¹2

8√

 −²3^320 4 = −^4

−¹2

−16 +¹⁶3

= −⁴3

30. F(  ) =  i + 4²j+  k,  = ( ) = ^, and  is the square [0 1] × [0 1], so by Equation 10



F· S =

[−(^) − 4²(^) + ] =1 0

1

0(−^− 4³^+ ^)  

=1 0

−4³^=1

=0  = ( − 1)1

0(−4³)  = 1 − 

31. 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 half-disk in the plane  = 2, and 4, the back half-disk in the plane  = 0.

On 1: The surface is  =

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



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 with downward orientation, so



₂F· 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. Regarding  and  as parameters, we have r× r^= iand



₃F· 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. Regarding  and  as parameters, we use − (r^× r^) = −i and



₄F· S =1

−1

 √1−²

0 ²  =1

−1

 √1−²

0 (0)   = 0 Thus

F· S =⁸3 + 0 + 2 + 0 = 2 +⁸₃.

32. 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, so



₁F· S =1 0

1−

0 [− (−1) − ( − ) (−1) + ]   =1 0

1−

0 ( + )   =1 0

1−

0 (1 − )  

=1 0

 −¹2²=1−

=0  =¹₂1 0

1 − ²

 =¹₂

 −¹3³1 0= ¹₃

(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.