16.7 Surface Integrals

16.7 Surface Integrals

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Surface Integrals

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

Suppose f is a function of three variables whose domain includes a surface S.

We will define the surface integral of f over S in such a way that, in the case where f(x, y, z) = 1, the value of the

surface integral is equal to the surface area of S.

We start with parametric surfaces and then deal with the special case where S is the graph of a function of two variables.

Parametric Surfaces

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Parametric Surfaces

Suppose that 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 We first assume that the parameter

domain D is a rectangle and we divide it into subrectangles R_ij with dimensions ∆u and ∆v.

Then the surface S is divided into corresponding patches S_ij as in Figure 1.

Figure 1

Parametric Surfaces

We evaluate f at a point in each patch, multiply by the area ∆S_ij of the patch, and form the Riemann sum

Then we take the limit as the number of patches increases and define the surface integral of f over the surface S as

Notice the analogy with the definition of a line integral and

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Parametric Surfaces

To evaluate the surface integral in Equation 1 we approximate the patch area ∆S_ij by the area of an approximating parallelogram in the tangent plane.

In our discussion of surface area we made the approximation

∆S_ij ≈ |r_u × r_v| ∆u ∆v where

are the tangent vectors at a corner of S_ij.

Parametric Surfaces

If the components are continuous and r_u and r_v are nonzero and nonparallel in the interior of D, it can be shown from

Definition 1, even when D is not a rectangle, that

This should be compared with the formula for a line integral:

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Parametric Surfaces

Observe also that

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), and z = z(u, v) in the formula for f(x, y, z).

Example 1

Compute the surface integral

∫∫

Solution:

We use the parametric representation

x = sin φ cos θ y = sin φ sin θ z = cos φ 0 ≤ φ ≤ π 0 ≤ θ ≤ 2π

That is,

r(φ, θ) = sin φ cos θ i + sin φ sin θ j + cos φ k

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Example 1 – Solution

Therefore, by Formula 2,

cont’d

Parametric Surfaces

If a thin sheet (say, of aluminum foil) has the shape of a surface S and the density (mass per unit area) at the

point (x, y, z) is ρ(x, y, z), then the total mass of the sheet is

and the center of mass is , where

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Graphs of Functions

Graphs of Functions

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)

and so we have

Thus

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Graphs of Functions

Therefore, in this case, Formula 2 becomes

Similar formulas apply when it is more convenient to project S onto the yz-plane or xz-plane. For instance, if S is a

surface with equation y = h(x, z) and D is its projection onto the xz-plane, then

Example 2

Evaluate

∫∫

Figure 2

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Example 2 – Solution

Since

and Formula 4 gives

Graphs of Functions

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

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Oriented Surfaces

Oriented Surfaces

To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Möbius strip shown in Figure 4. [It is named after the German geometer August Möbius (1790–1868).]

A Möbius strip

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Oriented Surfaces

You can construct one for yourself by taking a long

rectangular strip of paper, giving it a half-twist, and taping the short edges together as in Figure 5.

Figure 5

Constructing a Möbius strip

Oriented Surfaces

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).

Then, if the ant 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.)

Therefore a Möbius strip really has only one side.

From now on we consider only orientable (two-sided) surfaces.

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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). (See Figure 6.)

Figure 6

Oriented Surfaces

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 and the given choice of n provides S with an orientation.

There are two possible orientations for any orientable surface (see Figure 7).

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Oriented Surfaces

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

Since the k-component is positive, this gives the upward orientation of the surface.

Oriented Surfaces

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

and the opposite orientation is given by –n.

For instance, the parametric representation is

r(φ, θ) = a sin φ cos θ i + a sin φ sin θ j + a cos φ k

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Oriented Surfaces

We know that

r_φ × r_θ = a² sin² φ cos θ i + a² sin² φ sin θ j + a² sin φ cos φ k and | r_φ × r_θ| = a² sin φ

So the orientation induced by r(φ, θ) is defined by the unit normal vector

Observe that n points in the same direction as the position vector, that is, outward from the sphere (see Figure 8).

Positive orientation

Figure 8

The opposite (inward) orientation would have been

obtained (see Figure 9) if we had reversed the order of the parameters because r_θ × r_φ = –r_φ × r_θ.

For a closed surface, that is, 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

Oriented Surfaces

Negative orientation

Figure 9

Positive orientation

Figure 8

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Surface Integrals of Vector Fields

Surface Integrals of Vector Fields

Suppose that S is an oriented surface with unit normal vector n, and 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.)

Then the rate of flow (mass per unit time) per unit area is ρv.

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Surface Integrals of Vector Fields

If we divide S into small patches S_ij, as in Figure 10 (compare with Figure 1).

Figure 1 Figure 10

Surface Integrals of Vector Fields

Then S_ij is nearly planar and so we can approximate the

mass of fluid per unit time crossing S_ij in the direction of the normal n by the quantity

(ρv  n)A(S_ij)

where ρ, v, and n are evaluated at some point on S_ij.

(Recall that the component of the vector ρv in the direction of the unit vector n is ρv  n.)

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Surface Integrals of Vector Fields

By summing these quantities and taking the limit we get, according to Definition 1, the surface integral of the function ρv  n over S:

and this is interpreted physically as the rate of flow through S.

If we write F = ρv, then F is also a vector field on and the integral in Equation 7 becomes

Surface Integrals of Vector Fields

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

In words, Definition 8 says that the surface integral of a vector field over S is equal to the surface integral of its

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Surface Integrals of Vector Fields

If S is given by a vector function r(u, v), then n is given by Equation 6, and from Definition 8 and Equation 2 we have

where D is the parameter domain. Thus we have

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.

Solution:

As in Example 1, we use the parametric representation r(φ, θ) = sin φ cos θ i + sin φ sin θ j + cos φ k

0 ≤ φ ≤ π 0 ≤ θ ≤ 2π Then

F(r(φ, θ)) = cos φ i + sin φ sin θ j + sin φ cos θ k and,

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Example 4 – Solution

Therefore F(r(φ, θ))  (r_φ × r_θ) = cos φ sin² φ cos θ + sin³ φ sin² θ + sin² φ cos φ cos θ and, by Formula 9, the flux is

by the same calculation as in Example 1.

cont’d

Surface Integrals of 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.

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

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Surface Integrals of Vector Fields

Thus Formula 9 becomes

This formula assumes the upward orientation of S; for a downward orientation we multiply by –1.

Similar formulas can be worked out if S is given by y = h(x, z) or x = k(y, z).

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

Surface Integrals of Vector Fields

For instance, if E is an electric field, then the surface integral

is called the electric flux of E through the surface S. 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,

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Surface Integrals of Vector Fields

Therefore, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is Q = πε₀.

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

Surface Integrals of Vector Fields

The rate of heat flow across the surface S in the body is then given by the surface integral