SECTION
17.5
Surface Integrals of Vector Fields
(ET Section 16.5)
1141
Using the Surface Integral over a Graph we have:
Area
(
S
)
=
ZZ
S
1
dS
=
D
q
1
+
g
2
x
+
g
2
y
dA
(1)
In parametrizing the surface by
φ
(
x
,
y
)
=
(
x
,
y
,
g
(
x
,
y
))
,
(
x
,
y
)
=
D
,wehave:
T
x
=
∂8
∂
x
= h
1
,
0
,
g
x
i
T
y
=
∂
y
=

0
,
1
,
g
y
®
Hence,
n
=
T
x
×
T
y
=
¯
¯
¯
¯
¯
¯
ij k
10
g
x
01
g
y
¯
¯
¯
¯
¯
¯
=−
g
x
i
−
g
y
j
+
k
=

−
g
x
,
−
g
y
,
1
®
k
n
k=
q
g
2
x
+
g
2
y
+
1
n
k
−
k
There are two adjacent angles between the normal
n
and the vertical, and the cosines of these angles are opposite numbers.
Therefore we take the absolute value of cos
to obtain a positive value for Area
(
S
)
. Using the Formula for the cosine of
the angle between two vectors we get:

cos
=

n
·
k

k
n
kk
k
k
=


−
g
x
,
−
g
y
,
1
®
· h
0
,
0
,
1
i 
q
1
+
g
2
x
+
g
2
y
·
1
=
1
q
1
+
g
2
x
+
g
2
y
Substituting in (1) we get:
Area
(
S
)
=
D

cos

17.5 Surface Integrals of Vector Fields
(ET Section 16.5)
Preliminary Questions
1.
Let
F
be a vector ±eld and
8(
u
,v)
a parametrization of a surface
S
,andset
n
=
T
u
×
T
v
. Which of the following
is the normal component of
F
?
(a) F
·
n(
b
)
F
·
e
n
SOLUTION
The normal component of
F
is
F
·
e
n
rather than
F
·
n
.
2.
The vector surface integral
S
F
·
d
S
is equal to the scalar surface integral of the function (choose the correct
answer):
(a)
k
F
k
(b) F
·
n
,where
n
is a normal vector
(c) F
·
e
n
e
n
is the unit normal vector
The vector surface integral
R
S
F
·
d
S
is de±ned as the scalar surface integral of the normal component of
F
on the oriented surface. That is,
R
S
F
·
d
S
=
R
S
(
F
·
e
n
)
as stated in (c).
3.
S
F
·
d
S
is zero if (choose the correct answer):
(a) F
is tangent to
S
at every point.
(b) F
is perpendicular to
S
at every point.
Since
R
S
F
·
d
S
is equal to the scalar surface integral of the normal component of
F
on
S
, this integral is
zero when the normal component is zero at every point, that is, when
F
is tangent to
S
at every point as stated in (a).