VECTOR CALCULUSVECTOR CALCULUS

16

16.7Surface Integrals

In this section, we will learn about:

Integration of different types of surfaces.

VECTOR CALCULUS

SURFACE INTEGRALS

The relationship between surface integrals

and surface area is much the same as

the relationship between line integrals and

arc length.

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.

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.

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∈

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.

PARAMETRIC SURFACES

Then, the surface S

is divided into

corresponding

patches Sij.

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

*

1 1

( )m n

ij iji j

f P S= =

Δ∑∑

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:

*

,1 1

( , , ) lim ( )m n

ij ijm n

i jS

f x y z dS f P S→ ∞

= =

= Δ∑∑∫∫

Equation 1

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)

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.

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.

u v

x y z x y z

u u u v v v

∂ ∂ ∂ ∂ ∂ ∂= + + = + +

∂ ∂ ∂ ∂ ∂ ∂r i j k r i j k

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

Formula 2

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

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)

SURFACE INTEGRALS

Compute the surface integral ,

where S is the unit sphere

x2 + y2 + z2 = 1.

Example 1

2

S

x dS∫∫

SURFACE INTEGRALS

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

0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π That is,

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

Example 1

SURFACE INTEGRALS

As in Example 10 in Section 16.6,

we can compute that:

|rΦ x rθ| = sin Φ

Example 1

SURFACE INTEGRALS

Therefore, by Formula 2,

2

2

2 2 2

0 0

(sin cos ) | |

(sin cos sin

S

D

x dS

dA

d d

φ θ

π π

φ θ

φ θ φ φ θ

= ×

=

∫∫

∫∫

∫ ∫

r r

Example 1

SURFACE INTEGRALS

[ ]

2 2 3

0 0

2 2120 0

2 31 1 12 2 3 0

cos sin

(1 cos 2 ) (sin sin cos )

sin 2 cos cos

4

3

d d

d d

π π

π π

ππ

θ θ φ φ

θ θ φ φ φ φ

θ θ φ φ

π

=

= + −

⎡ ⎤= + − +⎣ ⎦

=

∫ ∫

∫ ∫

0

Example 1

APPLICATIONS

Surface integrals have applications

similar to those for the integrals we have

previously considered.

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).

MASS

Then, the total mass of the sheet

is:

( , , )S

m x y z dSρ=∫∫

CENTER OF MASS

The center of mass is:

where

( ), ,x y z

1( , , )

1( , , )

1( , , )

S

S

S

x x x y z dSm

y y x y z dSm

z z x y z dSm

ρ

ρ

ρ

=

=

=

∫∫

∫∫

∫∫

MOMENTS OF INERTIA

Moments of inertia can also be

defined as before.

See Exercise 39.

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

GRAPHS

Thus,

and

x y

g g

x x

∂ ∂× =− − +

∂ ∂r r i j k

Equation 3

22

| | 1x y

z z

x y

⎛ ⎞∂ ∂⎛ ⎞× = + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠r r

GRAPHS

Therefore, in this case, Formula 2

becomes:

Formula 4

GRAPHS

Similar formulas apply when it is

more convenient to project S onto

the yz-plane or xy-plane.

GRAPHS

For instance, if S is a surface with

equation y = h(x, z) and D is its projection

on the xz-plane, then

GRAPHS

Evaluate where S is the surface

z = x + y2, 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

Example 2

S

y dS∫∫

1

and

2

z

x

zy

y

∂=

∂

∂=

∂

GRAPHS

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

⎛ ⎞∂ ∂⎛ ⎞= + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

= + +

= +

⎤= + =⎦

∫∫ ∫∫

∫ ∫

∫ ∫

Example 2

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= +⋅⋅⋅⋅⋅⋅+

∫∫

∫∫ ∫∫

GRAPHS

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∫∫Example 3

GRAPHS

The surface S is shown.

We have changed the usual position of the axes to get a better look at S.

Example 3

GRAPHS

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

Example 3

GRAPHS

Therefore,

and

sin cos 0 cos sin

0 0 1zθ θ θ θ θ× = − = +

i j k

r r i j

2 2| | cos sin 1zθ θ θ× = + =r r

Example 3

GRAPHS

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

θ

π θ

π

π

π

θ

θ θ

θ θ θ

πθ θ θ

+

= ×

=

= +

= + + +

= + + =⎡ ⎤⎣ ⎦

∫∫ ∫∫

∫ ∫

∫

∫

r r

Example 3

GRAPHS

Since S2 lies in the plane z = 0,

we have:

Example 3

z dSS2

∫∫

= 0dSS2

∫∫=0

GRAPHS

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.

Example 3

GRAPHS

( )

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

π

π

π

π

θ θ

θ θ

θ θ

θ θπ

⎛ ⎞∂ ∂⎛ ⎞= + + +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

= + + +

= +

= +

⎡ ⎤= + =⎢ ⎥⎣ ⎦

∫∫ ∫∫

∫ ∫

∫ ∫

∫

Example 3

GRAPHS

Therefore,

( )

1 2 3

32

30 2

2

2

S S S S

z dS z dS z dS z dS

ππ

π

= + +

= + +

= +

∫∫ ∫∫ ∫∫ ∫∫

Example 3

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).

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.

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.

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.

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.

ORIENTED SURFACES

From now on, we consider

only orientable (two-sided)

surfaces.

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).

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.

POSSIBLE ORIENTATIONS

There are

two possible

orientations for

any orientable

surface.

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

Equation 5

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

Equation 6

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

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

ORIENTATION

So, the orientation induced by r(Φ, θ) is

defined by the unit normal vector

| |

sin cos sin sin cos

1( , )

a

φ θ

φ θ

φ θ φ θ φ

φ θ

×=

×

= + +

=

r rn

r r

i j k

r

POSITIVE ORIENTATION

Observe that n points in the same

direction as the position vector—that is,

outward from the sphere.

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

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.

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.

SURFACE INTEGRALS OF VECTOR FIELDS

Then, the rate of flow

(mass per unit time) per unit

area is ρv.

SURFACE INTEGRALS OF VECTOR FIELDS

If we divide S into small patches Sij ,

then Sij is nearly planar.

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.

VECTOR FIELDS

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.

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

S

S

dS

x y z x y z x y z dS

ρ

ρ

⋅

= ⋅

∫∫

∫∫

v n

v n

Equation 7

VECTOR FIELDS

If we write F = ρv, then F is also a vector

field on .

Then, the integral in Equation 7

becomes:

S

dS⋅∫∫F n

°3

FLUX INTEGRAL

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.

FLUX INTEGRAL

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.

Definition 8

S S

d dS⋅ = ⋅∫∫ ∫∫F S F n

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).

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

FLUX INTEGRAL

Thus, we have:

( )u v

S D

d dA⋅ = ⋅ ×∫∫ ∫∫F S F r r

Formula 9

FLUX INTEGRALS

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

Example 4

FLUX INTEGRALS

Using 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

we have:

F(r(Φ, θ))

= cos Φ i + sin Φ sin θ j + sin Φ cos θ k

Example 4

FLUX INTEGRALS

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

Example 4

FLUX INTEGRALS

Therefore,

F(r(Φ, θ)) · (rΦ x rθ)

= cos Φ sin2 Φ cos θ

+ sin3 Φ sin2 θ

+ sin2 Φ cos Φ cos θ

Example 4

FLUX INTEGRALS

Then, by Formula 9, the flux is:

2 2 3 2

0 0

( )

(2sin cos cos sin sin )

S

D

d

dA

d d

φ θ

π πφ φ θ φ θ φ θ

⋅

= ⋅ ×

= +

∫∫

∫∫

∫ ∫

F S

F r r

Example 4

FLUX INTEGRALS

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

π π

π π

π π

φ φ φ θ θ

φ φ θ θ

φ φ θ θ

π

=

+

= +

=

∫ ∫∫ ∫

∫ ∫

Example 4

FLUX INTEGRALS

The figure shows the vector field F in

Example 4 at points on the unit sphere.

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.

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

VECTOR FIELDS

Thus, Formula 9 becomes:

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

S D

g gd P Q R dA

x y

⎛ ⎞∂ ∂⋅ = − − +⎜ ⎟∂ ∂⎝ ⎠

∫∫ ∫∫F S

Formula 10

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.

VECTOR FIELDS

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.

Example 5

S

d⋅∫∫F S

VECTOR FIELDS

S consists of: A parabolic top surface S1.

A circular bottom surface S2.

Example 5

VECTOR FIELDS

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.

Example 5

VECTOR FIELDS

On S1,

P(x, y, z) = y

Q(x, y, z) = x

R(x, y, z) = z = 1 – x2 – y2

Also,

2 2g g

x yx y

∂ ∂=− =−

∂ ∂

Example 5

VECTOR FIELDS

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

Example 5

VECTOR FIELDS

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

π

π

π

θ θ θ

θ θ θ

θ θ θ

ππ

= + −

= − +

=

= +

=

∫ ∫∫ ∫∫

Example 5

VECTOR FIELDS

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 0

S S D

D

d dS z dA

dA

⋅ = ⋅− = −

= =

∫∫ ∫∫ ∫∫

∫∫

F S F k

Example 5

VECTOR FIELDS

Finally, we compute, by definition,

as the sum of the surface integrals

of F over the pieces S1 and S2:

S

d⋅∫∫F S

1 2

02 2

S S S

d d d

π π

⋅ = ⋅ = ⋅

= + =

∫∫ ∫∫ ∫∫F S F S F S

Example 5

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.

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.

S

d⋅∫∫E S

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 ε0 is a constant (called the permittivity

of free space) that depends on the units used. In the SI system, ε0 ≈ 8.8542 x 10–12 C2/N · m2

0

S

Q dε= ⋅∫∫E S

Equation 11

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

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).

HEAT FLOW

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.

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

HEAT FLOW

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.

Example 6

HEAT FLOW

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.

Example 6

HEAT FLOW

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.

Example 6

HEAT FLOW

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, 2 2 22

( )KC

x y za

⋅ =− + +F n

Example 6

HEAT FLOW

However, on S, we have:

x2 + y2 + z2 = a2

Thus,

F · n = –2aKC

Example 6

HEAT FLOW

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

ππ

⋅ = ⋅ =−

=−

=−

=−

∫∫ ∫∫ ∫∫F S F n

Example 6