Section 14.7: Surface Integrals
Suppose f is a function of three variables defined on a surface S. Partition S into patches
Sij with area ∆Sij . Evaluate f at a point Pij∗ in each patch and form the sum
m X
n
X
f (Pij∗ )∆Sij .
i=1 j=1
Let ||P || denote the norm of the partition P which is the area of the largest patch.
Definition: The surface integral of f over the surface S is
ZZ
m X
n
X
f (x, y, z)dS = lim
f (Pij∗ )∆Sij
||P ||→0
S
i=1 j=1
provided this limit exists.
Theorem: (Formula for Surface Integrals)
Suppose f is a function of three variables defined on a surface S with equation z = g(x, y)
for (x, y) ∈ D. Then
s
2 2
ZZ
ZZ
∂z
∂z
+
dA.
f (x, y, g(x, y)) 1 +
f (x, y, z)dS =
∂x
∂y
D
S
ZZ
Example: Evaluate the surface integral
xdS, where S is the part of the plane 2x+2y+z =
S
6 that lies in the first octant.
The projection of S onto the xy-plane is the region
D = {(x, y)|0 ≤ x ≤ 2, 0 ≤ y ≤ 3 − x}.
Then zx = −2, zy = −2, and
ZZ
Z
ydS =
S
3
Z
3−x
y
0
p
(−2)2 + (−2)2 + 1dydx
0
3
3−x
3
2
=
y dx
2 0
0
Z
3 3
=
(3 − x)2 dx
2 0
3
1
3
= − (3 − x) 2
Z
0
27
=
.
2
Theorem: (Center of Mass of a Surface)
If a thin sheet has the shape of a surface S and the density at (x, y, z) is ρ(x, y, z), then the
total mass of the sheet is
ZZ
m=
ρ(x, y, z)dS
S
and the center of mass is (x̄, ȳ, z̄), where
ZZ
ZZ
1
1
x̄ =
xρ(x, y, z)dS ȳ =
yρ(x, y, z)dS
m S
m S
1
z̄ =
m
ZZ
Example: Find the mass of a thin funnel in the shape of a cone z =
its density function is ρ(x, y, z) = 10 − z.
zρ(x, y, z)dS.
S
p
x2 + y 2 , 1 ≤ z ≤ 4 if
First,
x
∂z
=p
2
∂x
x + y2
and
∂z
y
=p
.
2
∂y
x + y2
The mass of the funnel is
ZZ
(10 − z)dS
m =
=
=
=
=
=
√
SZ Z
p
(10 − x2 + y 2 )dA
D
√ Z 2π Z 4
(10 − r)rdrdθ
2
0
1
√ Z 4
2 2π
(10r − r2 )dr
1
4
√
1 3 2
2 2π 5r − r 3
1
√
108 2π.
2
Theorem: (Surface Integral for a Parametric Surface)
Suppose f is a function of three variables defined on a surface S. If S has a vector equation
~
R(u,
v) = hx(u, v), y(u, v), z(u, v)i for (u, v) ∈ D, then
ZZ
ZZ
~
~u × R
~ v ||dA.
f (x, y, z)dS =
f (R(u,
v))||R
S
D
ZZ
Example: Evaluate the surface integral
yzdS, where S is the helicoid with vector equaS
~
tion R(u,
v) = hu cos v, u sin v, vi, 0 ≤ u ≤ 1, 0 ≤ v ≤ π.
The partial derivatives are
~ u = hcos v, sin v, 0i and R
~ v = h−u sin v, u cos v, 1i.
R
Then
~k
0
1
~i
~j
~
~
Ru × Rv = cos v
sin v
−u sin v u cos v
Thus
~u × R
~ v || =
||R
√
= hsin v, − cos v, ui
1 + u2 .
Therefore,
ZZ
√
uv sin v u2 + 1dudv
0Z 1 0
Z π
√
2
u u + 1du
v sin vdv
=
0
0
1 !
1 2
=
(u + 1)3/2 (−v cos v + sin v|π0 )
3
0
π √
(2 2 − 1).
=
3
Z
π
Z
1
yzdS =
S
~ that varies
Definition: A surface S is called orientable if it has a normal vector field N
~ that provide S with an
continuously over S. If S is orientable, there are two choices for N
orientation. For the positive orientation, the normal vectors point outward and for the
negative orientation, the normal vectors point inward.
Note: Examples of non-orientable surfaces are the Möbius strip or Klein bottle.
Definition: If F~ is a continuous vector field defined on an orientable surface S with unit
~ , then the surface integral of F~ over S is
normal vector N
ZZ
ZZ
~
~
~ dS.
F · dS =
F~ · N
S
S
This integral is also called the flux of F~ across S.
Theorem: (Formula for Flux)
If F~ (x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i and S is given by z = g(x, y) for (x, y) ∈ D,
then
ZZ
ZZ ∂g
∂g
~=
−Q
+ R dA,
F~ · dS
−P
∂x
∂y
S
D
~ is the upward unit normal. For the downward unit normal, multiply by −1.
where N
Example: Find the flux of F~ (x, y, z) = hey , yex , x2 yi across the part of the paraboloid z =
x2 + y 2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and has upward orientation.
Applying the previous theorem, the flux of F~ across S is
ZZ
ZZ
y
~ =
F~ · dS
−e (2x) − yex (2y) + x2 y dA
S
Z 1DZ 1
=
(x2 y − 2xey − 2y 2 ex )dydx
0
0
1 !
Z 1
2
1 2 2
x y − 2xey − y 3 ex dx
=
2
3
0
0
Z 1
1 2
2
=
x + 2(1 − e)x − ex dx
2
3
0
1
2 1 3
x + (1 − e)x2 − ex =
6
3 0
2
2
1
+ (1 − e) − e +
=
6
3
3
11 − 10e
.
=
6
Note: If S is a piecewise-smooth surface, then the surface integral of F~ over S is defined by
ZZ
ZZ
ZZ
~
~
~
~
~
F · dS =
F · dS + · · · +
F~ · dS.
S
S1
Sn
Example: Find the outward flux of F~ (x, y, z) = hx, y, zi across the boundary of the region
enclosed by the paraboloid z = 4 − x2 − y 2 and the plane z = 0.
The surface is given by S = S1 ∪ S2 , where S1 is the paraboloid z = 4 − x2 − y 2 and S2 is the
disk x2 + y 2 ≤ 4 in the xy-plane. Since S has the positive orientation, S1 is oriented upward
and S2 is oriented downward. On S1 ,
F~ (x, y, z) = hx, y, zi = hx, y, 4 − x2 − y 2 i
and
∂g
= −2x,
∂x
∂g
= −2y.
∂y
Then
ZZ
~ =
F~ · dS
S1
ZZ
(2x2 + 2y 2 + 4 − x2 − y 2 )dA
Z ZD
=
=
=
=
=
(x2 + y 2 + 4)dA
D
Z 2
Z 2π
(r2 + 4)rdrdθ
0
Z 20
(r3 + 4r)dr
2π
0
2
1 4
2 2π
r + 2r 4
0
24π.
~ = h0, 0, −1i and
Since the disk S2 is oriented downward, its normal vector is N
ZZ
ZZ
ZZ
~
~
F · N dS =
(−z)dA =
0dA = 0
S2
D
D
since z = 0 on S2 . Thus,
ZZ
~=
F~ · dS
Flux =
ZZ
S
~+
F~ · dS
S1
ZZ
~ = 24π.
F~ · dS
S2
Theorem: (Flux Integrals for Parametric Surfaces)
~
If S is a parametric surface defined by a vector function R(u,
v) for (u, v) ∈ D, then
ZZ
ZZ
~=
~
~u × R
~ v )dA.
F~ (R(u,
v)) · (R
F~ · dS
S
D
Example: Find the outward flux of F~ (x, y, z) = hz, y, xi across the sphere x2 + y 2 + z 2 = 1.
Using spherical coordinates, the sphere can be parameterized by
~
R(φ,
θ) = hsin φ cos θ, sin φ sin θ, cos φi
for 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Then
~ φ = hcos φ cos θ, cos φ sin θ, − sin φi
R
~ θ = h− sin φ sin θ, sin φ cos θ, 0i
R
F~ (φ, θ) = hcos φ, sin φ sin θ, sin φ cos θi.
Thus,
~k
~i
~j
= cos φ cos θ cos φ sin θ − sin φ
− sin φ sin θ sin φ cos θ
0
~φ × R
~θ
R
= hsin2 φ cos θ, sin2 φ sin θ, sin φ cos φi.
Then
~φ × R
~ θ ) = 2 sin2 φ cos φ cos θ + sin3 φ sin2 θ.
F~ (φ, θ) · (R
Therefore,
ZZ
~ =
F~ · dS
S
=
=
=
=
=
Z
2π
Z
π
(2 sin2 φ cos φ cos θ + sin3 φ sin2 θ)dφdθ
0
Z 2π
Z π
Z 2π
Z π 0
2
3
sin φ cos φdφ
cos θdθ +
sin φdφ
sin2 θdθ
2
0
0
0
Z 2π
Z π0
sin3 φdφ
sin2 θdθ
0
Z0 π
Z 2π
1 − cos 2θ
2
(1 − cos φ) sin φdφ
dθ
2
0
0
π 2π !
1
θ 1
3 − sin 2θ
− cos φ + cos φ
3
2 4
0
0
4π
.
3
Definition: Suppose the temperature at a point (x, y, z) in a body is T (x, y, z). Then the
heat flow is defined as the vector field
F~ = −K∇T,
where K is an experimentally determined constant called the conductivity of the substance.
The rate of heat flow across a surface S in the body is given by the flux integral.
Example: The temperature at the point (x, y, z) in a substance with conductivity K = 6.5
is T (x, y, z) = 2(x2 + y 2 ). Find the rate of heat flow inward across the cylindrical surface
x2 + y 2 = 6 for 0 ≤ z ≤ 4.
Using cylindrical coordinates, the surface S can be parameterized as
√
√
~ z) = h 6 cos θ, 6 sin θ, zi
R(θ,
for 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 4. Then
√
√
~ θ = h− 6 sin θ, 6 cos θ, 0i and R
~ z = h0, 0, 1i.
R
Thus,
~k ~
~j
√
√
√i
~θ × R
~ z = − 6 sin θ √6 cos θ 0 = h 6 cos θ, 6 sin θ, 0i.
R
0
0
1 The heat flow is
F~ (x, y, z) = −K∇T = −6.5h4x, 4y, 0i = h−26x, −26y, 0i.
Therefore,
√
√
F~ (θ, z) = h−26 6 cos θ, −26 6 sin θ, 0i
and
~θ × R
~ z ) = −156(cos2 θ + sin2 θ) = −156.
F~ (θ, z) · (R
Thus, the rate of heat flow inward across the surface S is
ZZ
Z 2π Z 4
~
~
−156dzdθ
−
F · dS = −
S
0
= 1248π.
0