SECTION 13.7

SURFACE INTEGRALS

P213.7

SURFACE INTEGRALS

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

P313.7

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.

We start with parametric surfaces.Then, we deal with the special case where S is

the graph of a function of two variables.

P413.7

PARAMETRIC SURFACES

Suppose a surface S has a vector equationr(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 Rij with dimensions ∆u and ∆v.

P513.7

PARAMETRIC SURFACES

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

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

*

1 1

( )m n

ij iji j

f P S

P613.7

SURFACE INTEGRAL

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

*

max , 01 1

( , , ) lim ( )i j

m n

ij iju vi jS

f x y z dS f P S

P713.7

SURFACE INTEGRALS

Notice the analogy with: The definition of a line integral (Definition 2 in

Section 13.2) The definition of a double integral (Definition 5 in

Section 12.1)

P813.7

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.

P913.7

SURFACE INTEGRALS

In our discussion of surface area in Section 13.6, we made the approximation

∆Sij ≈ |ru × rv| ∆u ∆v

where:

are the tangent vectors at a corner of Sij.

u v

x y z x y z

u u u v v v

r i j k r i j k

P1013.7

SURFACE INTEGRALS

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:

( , , ) ( ( , )) | |u v

S D

f x y z dS f u v dA r r r

P1113.7

SURFACE INTEGRALS

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

Observe also that:

( , , ) ( ( )) | '( ) |b

C af x y z ds f t t dt r r

1 | | ( )u v

S D

dS dA A S r r

P1213.7

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)

P1313.7

Example 1

Compute the surface integral ,

where S is the unit sphere x2 + y2 + z2 = 1.

2

S

x dS

P1413.7

Example 1 SOLUTION

As in Example 4 in Section 13.6, we use the parametric representation

x = sin cos , y = sin sin , z = cos 0 ≤ ≤ , 0 ≤ ≤ 2

That is, r(, ) = sin cos i + sin sin j + cos k

P1513.7

Example 1 SOLUTION

As in Example 9 in Section 13.6, we can compute that:

|r × r| = sin

P1613.7

Example 1 SOLUTION

Therefore, by Formula 2,2 2

2 2 2

0 0

(sin cos ) | |

sin cos sin

S D

x dS dA

d d

r r

2 2 3

0 0

2 2120 0

2 31 1 12 2 3 0

cos sin

(1 cos 2 ) (sin sin cos )

4sin 2 cos cos

3

d d

d d

0

P1713.7

APPLICATIONS

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

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 (x, y, z).

P1813.7

MASS

Then, the total mass of the sheet is:

( , , )S

m x y z dS

P1913.7

CENTER OF MASS

The center of mass is , where

Moments of inertia can also be defined as before. See Exercise 35.

, , x y z

1 1( , , ) ( , , )

1( , , )

S S

S

x x x y z dS y y x y z dSm m

z z x y z dSm

P2013.7

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:

x y

g g

x y

r i k r j k

P2113.7

GRAPHS

Thus,

and

x y

g g

x x

r r i j k

22

| | 1x y

z z

x y

r r

P2213.7

GRAPHS

Therefore, in this case, Formula 2 becomes:

22

( , , )

( , , ( , )) 1

S

D

f x y z dS

z zf x y g x y dA

x y

P2313.7

GRAPHS

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

P2413.7

GRAPHS

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

2 2

( , , )

( , ( , ), ) 1

S

D

f x y z dS

y yf x h x z z dA

x z

P2513.7

Example 2

Evaluate where S is the surface

z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2See Figure 2.

SOLUTION

S

y dS

1 and 2z z

yx y

P2613.7

Example 2 SOLUTION

So, Formula 4 gives:

22

1 2 2

0 0

1 2 2

0 0

22 3/ 21 2

4 30

1

1 1 4

2 1 2

13 22 (1 2 )

3

S D

z zy dS y dA

x y

y y dy dx

dx y y dy

y

P2713.7

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:

1

( , , )

( , , ) ( , , )n

S

S S

f x y z dS

f x y z dS f x y z dS

P2813.7

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.

S

z dS

P2913.7

Example 3 SOLUTION

The surface S is shown in Figure 3. We have changed the usual position of the axes to

get a better look at S.

P3013.7

Example 3 SOLUTION

For S1, we use and z as parameters (Example 5 in Section 13.6) and write its parametric equations as:

x = cos y = sin z = z

where: 0 ≤ ≤ 2 0 ≤ z ≤ 1 + x = 1 + cos

P3113.7

Example 3 SOLUTION

Therefore,

and

sin cos 0 cos sin

0 0 1z

i j k

r r i j

2 2| | cos sin 1z r r

P3213.7

Example 3 SOLUTION

Thus, the surface integral over S1 is:

1

2 1 cos

0 0

2 2120

21 12 20

231 102 2 4

| |

(1 cos )

1 2cos (1 cos 2 )

32sin sin 2

2

z

S D

z dS z dA

z dz d

d

d

r r

P3313.7

Example 3 SOLUTION

Since S2 lies in the plane z = 0, we have:

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.

2 2

0 0S S

z dS dS

P3413.7

Example 3 SOLUTION

3

22

2 1

0 0

2 1 2

0 0

21 12 30

2

0

(1 ) 1

(1 cos ) 1 1 0

2 ( cos )

2 cos

sin2 2

2 3

S D

z zz dS x dA

x y

r r dr d

r r dr d

d

P3513.7

Example 3 SOLUTION

Therefore,

1 2 3

32

30 2

2

2

S S S S

z dS z dS z dS z dS

P3613.7

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).

P3713.7

MOBIUS STRIP

You can construct one for yourself by: 1. Taking a long rectangular strip of paper.

2. Giving it a half-twist.

3. Taping the short edges together.

P3813.7

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.

P3913.7

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.

P4013.7

MOBIUS STRIP

Therefore, a Möbius strip really has only one side. You can graph the Möbius strip using the parametric

equations in Exercise 28 in Section 13.6.

P4113.7

ORIENTED SURFACES

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

P4213.7

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.

P4313.7

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.

P4413.7

POSSIBLE ORIENTATIONS

There are two possible orientations for any orientable surface.

P4513.7

UPWARD ORIENTATION

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.

22

1

g gx y

g gx y

i j kn

P4613.7

ORIENTATION

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.

| |u v

u v

r r

nr r

P4713.7

ORIENTATION

For instance, in Example 4 in Section 13.6, we found the parametric representation

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

for the sphere x2 + y2 + z2 = a2

P4813.7

ORIENTATION

Then, in Example 9 in Section 13.6, we found that:

r × r = a2 sin2 cos i + a2 sin2 sin j

+ a2 sin cos kand

|r × r | = a2 sin

P4913.7

ORIENTATION

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

| |

sin cos sin sin cos

1( , )

a

r rn

r r

i j k

r

P5013.7

POSITIVE ORIENTATION

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

See Figure 8.

Fig. 17.7.8, p. 1122

P5113.7

NEGATIVE ORIENTATION

The opposite (inward) orientation would have been obtained (see Figure 9) if we had reversed the order of the parameters because r × r = –r × r

P5213.7

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.

P5313.7

SURFACE INTEGRALS OF VECTOR FIELDS

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

Then, imagine a fluid with density (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.

P5413.7

SURFACE INTEGRALS OF VECTOR FIELDS

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

If we divide S into small patches Sij , as in Figure 10 (compare with Figure 1), then Sij is nearly planar.

P5513.7

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

( v · n)A(Sij)

where , v, and n are evaluated at some point on Sij. Recall that the component of the vector v in the

direction of the unit vector n is v · n.

P5613.7

VECTOR FIELDS

Summing these quantities and taking the limit, we get, according to Definition 1, the surface integral of the function v · n over S:

This is interpreted physically as the rate of flow through S.

( , , ) ( , , ) ( , , )

S

S

dS

x y z x y z x y z dS

v n

v n

P5713.7

VECTOR FIELDS

If we write F = v, then F is also a vector field on .

Then, the integral in Equation 7 becomes:

S

dSF n

3

P5813.7

FLUX INTEGRAL

A surface integral of this form occurs frequently in physics—even when F is not v.

It is called the surface integral (or flux integral) of F over S.

P5913.7

Definition 8

S S

d dS F S F n

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.

P6013.7

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).

P6113.7

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.

( ( , ))

u v

u vS S

u vu v

u vD

d dS

u v dA

r rF S F

r r

r rF r r r

r r

P6213.7

FLUX INTEGRAL

Thus, we have:

( )u v

S D

d dA F S F r r

P6313.7

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

P6413.7

Example 4 SOLUTION

Using the parametric representation

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

0 ≤ ≤ 0 ≤ ≤ 2

we have:F(r(, )) = cos i + sin sin j + sin cos k

P6513.7

Example 4 SOLUTION

From Example 9 in Section 13.6,r × r

= sin2 cos i + sin2 sin j + sin cos kTherefore,

F(r(, )) · (r × r )= cos sin2 cos + sin3 sin2 + sin2 cos cos

P6613.7

Example 4 SOLUTION

Then, by Formula 9, the flux is:

2 2 3 2

0 0

( )

(2sin cos cos sin sin )

S

D

d

dA

d d

F S

F r r

P6713.7

Example 4 SOLUTION

This is by the same calculation as in Example 1.

22

0 0

23 2

0 0

23 2

0 0

2 sin cos cos

sin sin

0 sin sin

4

3

d d

d d

d d

P6813.7

FLUX INTEGRALS

Figure 11 shows the vector field F in Example 4 at points on the unit sphere.

P6913.7

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, 4/3, represents: The rate of flow through the unit sphere in units of

mass per unit time.

P7013.7

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:

( ) ( )x y

g gP Q R

x y

F r r i j k i j k

P7113.7

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). See Exercises 31 and 32.

S D

g gd P Q R dA

x y

F S

P7213.7

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.

S

dF S

P7313.7

Example 5 SOLUTION

S consists of: A parabolic top surface S1.

A circular bottom surface S2. See Figure 12.

P7413.7

Example 5 SOLUTION

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.

P7513.7

Example 5 SOLUTION

On S1,P(x, y, z) = y Q(x, y, z) = xR(x, y, z) = z = 1 – x2 – y2

Also,

2 2g g

x yx y

P7613.7

Example 5 SOLUTION

So, we have:

1

2 2

2 2

[ ( 2 ) ( 2 ) 1 ]

(1 4 )

S

D

D

D

d

g gP Q R dA

x y

y x x y x y dA

xy x y dA

F S

P7713.7

Example 5 SOLUTION

2 1 2 2

0 0

2 1 3 3

0 0

2140

14

(1 4 cos sin )

( 4 cos sin )

( cos sin )

(2 ) 0

2

r r r dr d

r r r dr d

d

P7813.7

Example 5 SOLUTION

The disk S2 is oriented downward.

So, its unit normal vector is n = –k and we have:

since z = 0 on S2.

2 2

( ) ( ) 0 0S S D D

d dS z dA dA F S F k

P7913.7

Example 5 SOLUTION

Finally, we compute, by definition,as the sum of the surface integrals of F over the pieces S1 and S2:

S

dF S

1 2

02 2S S S

d d d

F S F S F S

P8013.7

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.

P8113.7

ELECTRIC FLUX

For instance, if E is an electric field (Example 5 in Section 13.1), the surface integral

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

S

dE S

P8213.7

GAUSS’S LAW

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 0 is a constant (called the permittivity of free space) that depends on the units used. In the SI system, 0 ≈ 8.8542 × 10–12 C2/N · m2

0

S

Q d E S

P8313.7

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 = 40/3

P8413.7

HEAT FLOW

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).

Then, the heat flow is defined as the vector field

F = –K ∇uwhere K is an experimentally determined constant called the conductivity of the substance.

P8513.7

HEAT FLOW

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

S S

d K u d F S S

P8613.7

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.

P8713.7

Example 6 SOLUTION

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.

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.

P8813.7

Example 6 SOLUTION

Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the spherex2 + y2 + z2 = a2 at the point (x, y, z) is:

n = 1/a (x i + y j + z k)Thus,

2 2 22( )

KCx y z

a F n

P8913.7

Example 6 SOLUTION

However, on S, we have:x2 + y2 + z2 = a2

Thus, F · n = –2aKC

P9013.7

Example 6 SOLUTION

Thus, the rate of heat flow across S is:

2

3

= = 2

= 2 ( )= 2 (4 )

= 8

S S S

d dS aKC dS

aKCA S aKC a

KC a

F S F n