1. 17
VECTOR CALCULUS

2. 17.7 Surface Integrals In this section, we will learn about:
VECTOR CALCULUS
17.7
Surface Integrals
In this section, we will learn about:
Integration of different types of surfaces.

3. SURFACE INTEGRALS
The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length.

4. SURFACE INTEGRALS
Suppose f is a function of three variables whose domain includes a surface S.
We will define the surface integral of f over S such that, in the case where f(x, y, z) = 1, the value of the surface integral is equal to the surface area of S.

5. We start with parametric surfaces.
SURFACE INTEGRALS
We start with parametric surfaces.
Then, we deal with the special case where S is the graph of a function of two variables.

6. r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
PARAMETRIC SURFACES
Suppose a surface S has a vector equation
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
(u, v) D

7. PARAMETRIC SURFACES
We first assume that the parameter domain D is a rectangle and we divide it into subrectangles Rij with dimensions ∆u and ∆v.

8. Then, the surface S is divided into corresponding patches Sij.
PARAMETRIC SURFACES
Then, the surface S is divided into corresponding patches Sij.

9. PARAMETRIC SURFACES
We evaluate f at a point Pij* in each patch, multiply by the area ∆Sij of the patch, and form the Riemann sum

10. SURFACE INTEGRAL
Equation 1
Then, we take the limit as the number of patches increases and define the surface integral of f over the surface S as:

11. Notice the analogy with:
SURFACE INTEGRALS
Notice the analogy with:
The definition of a line integral (Definition 2 in Section 16.2)
The definition of a double integral (Definition 5 in Section 15.1)

12. SURFACE INTEGRALS
To evaluate the surface integral in Equation 1, we approximate the patch area ∆Sij by the area of an approximating parallelogram in the tangent plane.

13. where: are the tangent vectors at a corner of Sij.
SURFACE INTEGRALS
In our discussion of surface area in Section 16.6, we made the approximation ∆Sij ≈ |ru x rv| ∆u ∆v
where: are the tangent vectors at a corner of Sij.

14. SURFACE INTEGRALS
Formula 2
If the components are continuous and ru and rv are nonzero and nonparallel in the interior of D, it can be shown from Definition 1—even when D is not a rectangle—that:

15. This should be compared with the formula for a line integral:
SURFACE INTEGRALS
This should be compared with the formula for a line integral:
Observe also that:

16. SURFACE INTEGRALS
Formula 2 allows us to compute a surface integral by converting it into a double integral over the parameter domain D.
When using this formula, remember that f(r(u, v) is evaluated by writing x = x(u, v), y = y(u, v), z = z(u, v) in the formula for f(x, y, z)

17. SURFACE INTEGRALS
Example 1
Compute the surface integral , where S is the unit sphere x2 + y2 + z2 = 1.

18. SURFACE INTEGRALS
Example 1
As in Example 4 in Section 16.6, we use the parametric representation x = sin Φ cos θ, y = sin Φ sin θ, z = cos Φ ≤ Φ ≤ π, 0 ≤ θ ≤ 2π
That is, r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k

19. As in Example 10 in Section 16.6, we can compute that:
SURFACE INTEGRALS
Example 1
As in Example 10 in Section 16.6, we can compute that:
|rΦ x rθ| = sin Φ

20. SURFACE INTEGRALS
Example 1
Therefore, by Formula 2,

21. SURFACE INTEGRALS
Example 1

22. APPLICATIONS
Surface integrals have applications similar to those for the integrals we have previously considered.

23. For example, suppose a thin sheet (say, of aluminum foil) has:
APPLICATIONS
For example, suppose a thin sheet (say, of aluminum foil) has:
The shape of a surface S.
The density (mass per unit area) at the point (x, y, z) as ρ(x, y, z).

24. Then, the total mass of the sheet is:

25. The center of mass is: where

26. Moments of inertia can also be defined as before.
See Exercise 39.

27. GRAPHS
Any surface S with equation z = g(x, y) can be regarded as a parametric surface with parametric equations
x = x y = y z = g(x, y)
So, we have:

28. GRAPHS
Equation 3
Thus,
and

29. Therefore, in this case, Formula 2 becomes:
GRAPHS
Formula 4
Therefore, in this case, Formula 2 becomes:

30. GRAPHS
Similar formulas apply when it is more convenient to project S onto the yz-plane or xy-plane.

31. GRAPHS
For instance, if S is a surface with equation y = h(x, z) and D is its projection on the xz-plane, then

32. Evaluate where S is the surface z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2
GRAPHS
Example 2
Evaluate where S is the surface z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

33. GRAPHS
Example 2
So, Formula 4 gives:

34. GRAPHS
If S is a piecewise-smooth surface—a finite union of smooth surfaces S1, S2, , Sn that intersect only along their boundaries—then the surface integral of f over S is defined by:

35. Evaluate , where S is the surface whose:
GRAPHS
Example 3
Evaluate , where S is the surface whose:
Sides S1 are given by the cylinder x2 + y2 = 1.
Bottom S2 is the disk x2 + y2 ≤ 1 in the plane z = 0.
Top S3 is the part of the plane z = 1 + x that lies above S2.

36. GRAPHS
Example 3
The surface S is shown.
We have changed the usual position of the axes to get a better look at S.

37. x = cos θ y = sin θ z = z where:
GRAPHS
Example 3
For S1, we use θ and z as parameters (Example 5 in Section 16.6) and write its parametric equations as:
x = cos θ y = sin θ z = z where:
0 ≤ θ ≤ 2π
0 ≤ z ≤ 1 + x = 1 + cos θ

38. GRAPHS
Example 3
Therefore,
and

39. Thus, the surface integral over S1 is:
GRAPHS
Example 3
Thus, the surface integral over S1 is:

40. Since S2 lies in the plane z = 0, we have:
GRAPHS
Example 3
Since S2 lies in the plane z = 0, we have:

41. S3 lies above the unit disk D and is part of the plane z = 1 + x.
GRAPHS
Example 3
S3 lies above the unit disk D and is part of the plane z = 1 + x.
So, taking g(x, y) = 1 + x in Formula 4 and converting to polar coordinates, we have the following result.

42. GRAPHS
Example 3

43. GRAPHS
Example 3
Therefore,

44. ORIENTED SURFACES
To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown.
It is named after the German geometer August Möbius (1790– 1868).

45. You can construct one for yourself by:
MOBIUS STRIP
You can construct one for yourself by:
Taking a long rectangular strip of paper.
Giving it a half-twist.
Taping the short edges together.

46. MOBIUS STRIP
If an ant were to crawl along the Möbius strip starting at a point P, it would end up on the “other side” of the strip—that is, with its upper side pointing in the opposite direction.

47. MOBIUS STRIP
Then, if it continued to crawl in the same direction, it would end up back at the same point P without ever having crossed an edge.
If you have constructed a Möbius strip, try drawing a pencil line down the middle.

48. Therefore, a Möbius strip really has only one side.
MOBIUS STRIP
Therefore, a Möbius strip really has only one side.
You can graph the Möbius strip using the parametric equations in Exercise 32 in Section 16.6.

49. From now on, we consider only orientable (two-sided) surfaces.
ORIENTED SURFACES
From now on, we consider only orientable (two-sided) surfaces.

50. ORIENTED SURFACES
We start with a surface S that has a tangent plane at every point (x, y, z) on S (except at any boundary point).
There are two unit normal vectors n1 and n2 = –n1 at (x, y, z).

51. ORIENTED SURFACE & ORIENTATION
If it is possible to choose a unit normal vector n at every such point (x, y, z) so that n varies continuously over S, then
S is called an oriented surface.
The given choice of n provides S with an orientation.

52. POSSIBLE ORIENTATIONS
There are two possible orientations for any orientable surface.

53. UPWARD ORIENTATION
Equation 5
For a surface z = g(x, y) given as the graph of g, we use Equation 3 to associate with the surface a natural orientation given by the unit normal vector
As the k-component is positive, this gives the upward orientation of the surface.

54. ORIENTATION
Equation 6
If S is a smooth orientable surface given in parametric form by a vector function r(u, v), then it is automatically supplied with the orientation of the unit normal vector
The opposite orientation is given by –n.

55. = a sin Φ cos θ i + a sin Φ sin θ j + a cos Φ k
ORIENTATION
For instance, in Example 4 in Section 16.6, we found the parametric representation
r(Φ, θ)
= a sin Φ cos θ i + a sin Φ sin θ j a cos Φ k
for the sphere x2 + y2 + z2 = a2

56. Then, in Example 10 in Section 16.6, we found that:
ORIENTATION
Then, in Example 10 in Section 16.6, we found that:
rΦ x rθ = a2 sin2 Φ cos θ i a2 sin2 Φ sin θ j a2 sin Φ cos Φ k
and
|rΦ x rθ| = a2 sin Φ

57. ORIENTATION
So, the orientation induced by r(Φ, θ) is defined by the unit normal vector

58. POSITIVE ORIENTATION
Observe that n points in the same direction as the position vector—that is, outward from the sphere.

59. NEGATIVE ORIENTATION
The opposite (inward) orientation would have been obtained if we had reversed the order of the parameters because rθ x rΦ = –rΦ x rθ

60. CLOSED SURFACES
For a closed surface—a surface that is the boundary of a solid region E—the convention is that:
The positive orientation is the one for which the normal vectors point outward from E.
Inward-pointing normals give the negative orientation.

61. SURFACE INTEGRALS OF VECTOR FIELDS
Suppose that S is an oriented surface with unit normal vector n.
Then, imagine a fluid with density ρ(x, y, z) and velocity field v(x, y, z) flowing through S.
Think of S as an imaginary surface that doesn’t impede the fluid flow—like a fishing net across a stream.

62. SURFACE INTEGRALS OF VECTOR FIELDS
Then, the rate of flow (mass per unit time) per unit area is ρv.

63. SURFACE INTEGRALS OF VECTOR FIELDS
If we divide S into small patches Sij , then Sij is nearly planar.

64. SURFACE INTEGRALS OF VECTOR FIELDS
So, we can approximate the mass of fluid crossing Sij in the direction of the normal n per unit time by the quantity
(ρv · n)A(Sij)
where ρ, v, and n are evaluated at some point on Sij.
Recall that the component of the vector ρv in the direction of the unit vector n is ρv · n.

65. VECTOR FIELDS
Equation 7
Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function ρv · n over S:
This is interpreted physically as the rate of flow through S.

66. If we write F = ρv, then F is also a vector field on .
VECTOR FIELDS
If we write F = ρv, then F is also a vector field on .
Then, the integral in Equation 7 becomes:

67. It is called the surface integral (or flux integral) of F over S.
A surface integral of this form occurs frequently in physics—even when F is not ρv.
It is called the surface integral (or flux integral) of F over S.

68. FLUX INTEGRAL
Definition 8
If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is:
This integral is also called the flux of F across S.

69. In words, Definition 8 says that:
FLUX INTEGRAL
In words, Definition 8 says that:
The surface integral of a vector field over S is equal to the surface integral of its normal component over S (as previously defined).

70. FLUX INTEGRAL
If S is given by a vector function r(u, v), then n is given by Equation 6.
Then, from Definition 8 and Equation 2, we have: where D is the parameter domain.

71. FLUX INTEGRAL
Formula 9
Thus, we have:

72. FLUX INTEGRALS
Example 4
Find the flux of the vector field F(x, y, z) = z i + y j + x k across the unit sphere x2 + y2 + z2 = 1

73. F(r(Φ, θ)) = cos Φ i + sin Φ sin θ j + sin Φ cos θ k
FLUX INTEGRALS
Example 4
Using the parametric representation r(Φ, θ) = sin Φ cos θ i + sin Φ sin θ j + cos Φ k
0 ≤ Φ ≤ π ≤ θ ≤ 2π
we have:
F(r(Φ, θ)) = cos Φ i + sin Φ sin θ j + sin Φ cos θ k

74. From Example 10 in Section 16.6, rΦ x rθ
FLUX INTEGRALS
Example 4
From Example 10 in Section 16.6,
rΦ x rθ
= sin2 Φ cos θ i + sin2 Φ sin θ j sin Φ cos Φ k

75. = cos Φ sin2 Φ cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ
FLUX INTEGRALS
Example 4
Therefore,
F(r(Φ, θ)) · (rΦ x rθ)
= cos Φ sin2 Φ cos θ sin3 Φ sin2 θ sin2 Φ cos Φ cos θ

76. Then, by Formula 9, the flux is:
FLUX INTEGRALS
Example 4
Then, by Formula 9, the flux is:

77. This is by the same calculation as in Example 1.
FLUX INTEGRALS
Example 4
This is by the same calculation as in Example 1.

78. FLUX INTEGRALS
The figure shows the vector field F in Example 4 at points on the unit sphere.

79. VECTOR FIELDS
If, for instance, the vector field in Example 4 is a velocity field describing the flow of a fluid with density 1, then the answer, 4π/3, represents:
The rate of flow through the unit sphere in units of mass per unit time.

80. VECTOR FIELDS
In the case of a surface S given by a graph z = g(x, y), we can think of x and y as parameters and use Equation 3 to write:

81. VECTOR FIELDS
Formula 10
Thus, Formula 9 becomes:
This formula assumes the upward orientation of S.
For a downward orientation, we multiply by –1.

82. VECTOR FIELDS
Similar formulas can be worked out if S is given by y = h(x, z) or x = k(y, z).
See Exercises 35 and 36.

83. Evaluate where: F(x, y, z) = y i + x j + z k
VECTOR FIELDS
Example 5
Evaluate
where:
F(x, y, z) = y i + x j + z k
S is the boundary of the solid region E enclosed by the paraboloid z = 1 – x2 – y2 and the plane z = 0.

84. S consists of: A parabolic top surface S1.
VECTOR FIELDS
Example 5
S consists of:
A parabolic top surface S1.
A circular bottom surface S2.

85. VECTOR FIELDS
Example 5
Since S is a closed surface, we use the convention of positive (outward) orientation.
This means that S1 is oriented upward.
So, we can use Equation 10 with D being the projection of S1 on the xy-plane, namely, the disk x2 + y2 ≤ 1.

86. P(x, y, z) = y Q(x, y, z) = x R(x, y, z) = z = 1 – x2 – y2
VECTOR FIELDS
Example 5
On S1,
P(x, y, z) = y Q(x, y, z) = x R(x, y, z) = z = 1 – x2 – y2
Also,

87. VECTOR FIELDS
Example 5
So, we have:

88. VECTOR FIELDS
Example 5

89. The disk S2 is oriented downward.
VECTOR FIELDS
Example 5
The disk S2 is oriented downward.
So, its unit normal vector is n = –k and we have: since z = 0 on S2.

90. VECTOR FIELDS
Example 5
Finally, we compute, by definition, as the sum of the surface integrals of F over the pieces S1 and S2:

91. APPLICATIONS
Although we motivated the surface integral of a vector field using the example of fluid flow, this concept also arises in other physical situations.

92. ELECTRIC FLUX
For instance, if E is an electric field (Example 5 in Section 16.1), the surface integral is called the electric flux of E through the surface S.

93. GAUSS’S LAW
Equation 11
One of the important laws of electrostatics is Gauss’s Law, which says that the net charge enclosed by a closed surface S is:
where ε0 is a constant (called the permittivity of free space) that depends on the units used.
In the SI system, ε0 ≈ x 10–12 C2/N · m2

94. GAUSS’S LAW
Thus, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is: Q = 4πε0/3

95. Another application occurs in the study of heat flow.
Suppose the temperature at a point (x, y, z) in a body is u(x, y, z).

96. Then, the heat flow is defined as the vector field F = –K ∇u
where K is an experimentally determined constant called the conductivity of the substance.

97. HEAT FLOW
Then, the rate of heat flow across the surface S in the body is given by the surface integral

98. HEAT FLOW
Example 6
The temperature u in a metal ball is proportional to the square of the distance from the center of the ball.
Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.

99. Taking the center of the ball to be at the origin, we have:
HEAT FLOW
Example 6
Taking the center of the ball to be at the origin, we have:
u(x, y, z) = C(x2 + y2 + z2)
where C is the proportionality constant.

100. HEAT FLOW
Example 6
Then, the heat flow is: F(x, y, z) = –K ∇u = –KC(2x i + 2y j + 2z k) where K is the conductivity of the metal.

101. HEAT FLOW
Example 6
Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the sphere x2 + y2 + z2 = a2 at the point (x, y, z) is:
n = 1/a (x i + y j + z k)
Thus,

102. However, on S, we have: x2 + y2 + z2 = a2
HEAT FLOW
Example 6
However, on S, we have: x2 + y2 + z2 = a2
Thus, F · n = –2aKC

103. Thus, the rate of heat flow across S is:
Example 6
Thus, the rate of heat flow across S is: