16.7

Surface Integrals

In this section, we will learn about:

Integration of different types of surfaces.

VECTOR CALCULUS

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.

•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 ij

i j

f P S

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:

Analogues to: The definition of a line integral

(Definition 2 in Section 16.2);The definition of a double

integral (Definition 5 in Section 15.1)

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.

*

,1 1

( , , ) lim ( )m n

ij ijm n

i jS

f x y z dS f P S

Equation 1

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

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

we can compute: |rΦ x rθ| = sin Φ

Example 1

SURFACE INTEGRALS

Therefore, by Formula 2,

2

2

22 2

0 0

(sin cos ) | |

(sin cos sin

S

D

x dS

dA

d d

r r

Example 1

22 3

0 0

221

20 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

0

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).

CENTER OF MASS

Then, the total mass of the sheet

is:

The center of mass is:

where

, ,x y z1

( , , )

1( , , )

1( , , )

S

S

S

x x x y z dSm

y y x y z dSm

z z x y z dSm

( , , )S

m x y z dS

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

•Formula 2 becomes:

x y

g g

x x

r r i j k

Equation 3

22

| | 1x y

z z

x y

r r

22

( , , )

( , , ( , )) 1

S

D

f x y z dS

z zf x y g x y dA

x y

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 22

0 0

1 22

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 dSExample 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 1

z

i j k

r r i j

2 2| | cos sin 1z r r

Example 3

GRAPHS

Thus, the surface integral over S1 is:

1

2 1 cos

0 0

221

20

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

Example 3

GRAPHS

Since S2 lies in the plane z = 0,

we have:

Example 3

z dSS

2

0 dSS

2

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 12

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

Example 3

GRAPHS

Therefore,

1 2 3

32

30 2

2

2

S S S S

z dS z dS z dS z dS

Example 3

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.

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

v n

v n

Equation 7

VECTOR FIELDS

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

field on R3. Then, the integral in Equation 7

becomes:

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.

S

dSF n

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

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 (D is the parameters’ domain):

So,

( ( , ))

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

( )u v

S D

d dA F S F r r

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

•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

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

Therefore, F(r(Φ, θ)) · (rΦ x rθ) = cos Φ sin2 Φ

cos θ + sin3 Φ sin2 θ + sin2 Φ cos Φ cos θ

Then, by Formula 9, the flux is:

Example 4

22 3 2

0 0

( )

(2sin cos cos sin sin )

S

D

d

dA

d d

F S

F r r

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

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

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

dF 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 12 2

0 0

2 13 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

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

dF S

1 2

02 2

S S S

d d d

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

dE 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 d 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.

•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