Section 17.3
The Fundamental Theorem for Line Integrals
“FTC Number 2”
Recall that in single variable calculus, in order to evaluate a definite
integral, we use the fundamental theorem of calculus. Specifically, we
have
integraldisplay
b
a
F
′
(
x
)
dx
=
F
(
b
)
−
F
(
a
)
.
Provided
F
′
(
x
) is a continuous function, an antiderivative always exists,
so FTC can
always
be used to evaluate single variable integrals of
continuous functions. In this section, we consider a generalization of
FTC to line integrals. The difference in this case however is that we
shall not always be able to apply it.
1.
The Fundamental Theorem
In multivariable calculus, it doesn’t make sense to talk about “the
derivative” since a function of more than one variable can be differ
entiated with respect to different variables. In our earlier discussions
on differentiation, we concluded that though we could not define “the
derivative”, we could define a generalization of the derivative  specif
ically, we said that the gradient vector of a function
f
was the closest
generalization to “the derivative” we could get. With this in mind, we
have the following generalization of FTC in single variable.
Result 1.1.
Let
C
be a smooth curve parameterized by
vector
r
(
t
) with
a
lessorequalslant
t
lessorequalslant
b
and suppose
f
is a differentiable function of two or three
variables whose gradient vector is continuous on
C
. Then
integraldisplay
C
∇
f
·
dvector
r
=
f
(
vector
r
(
b
))
−
f
(
vector
r
(
a
))
.
This results tells us that if
vector
F
is a conservative vector field (meaning
it is the gradient vector field of some function
f
(
x, y
)), then rather
than evaluating the integral by hand, we can simply find the potential
function and take its difference at the end points. We illustrate.
Example 1.2.
(
i
) Suppose
vector
F
= 2
x
vector
i
+ 2
y
vector
j
+ 2
z
vector
k
. Evaluate
integraltext
C
vector
F
·
dvector
r
where
vector
r
= 3
t
2
vector
i
+ 3
t
2
vector
j
+ 2
t
vector
k
with 0
lessorequalslant
t
lessorequalslant
1.
Observe that
vector
F
is a gradient vector field with potential func
tion
f
(
x, y, z
) =
x
2
+
y
2
+
z
2
, so we can use FTC to evaluate
the integral. Since the end points are (0
,
0
,
0) and (3
,
3
,
2), we
have
integraldisplay
C
vector
F
·
dvector
r
=
f
(3
,
3
,
2)
−
f
(0
,
0
,
0) = 9 + 9 + 4 = 22
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
(
ii
) Suppose
vector
F
= 2
x
vector
i
+ 2
y
vector
j
+ 2
z
vector
k
. Evaluate
integraltext
C
vector
F
·
dvector
r
where
vector
r
is
drawn below.
(0,0,0)
(3,3,2)
Since this is a conservative vector field, we can use FTC.
Since the end points are (0
,
0
,
0) and (3
,
3
,
2), we have
integraldisplay
C
vector
F
·
dvector
r
=
f
(3
,
3
,
2)
−
f
(0
,
0
,
0) = 9 + 9 + 4 = 22
In the last two examples, observe that the answer was the same. This
is simply because the vector field
vector
F
is conservative, so the integral
can be evaluated using FTC. This observation motivates the following
definition.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Stefanov
 Calculus, Derivative, Fundamental Theorem Of Calculus, Integrals, Vector Calculus, Manifold, Line integral, Vector field

Click to edit the document details