A three-part examination of the Fundamental Theorem of Line Integrals. Learn how to use this theorem in multivariable calculus. Simplify the process of solving line integrals using the FTLI.
Concepts and Applications of the Fundamental Theorem of Line Integrals.pdf
1. 1
Concepts and Applications of the
Fundamental Theorem of Line Integrals
Samson Axelrod, Jacob Braginsky, Pranav Dwarakanath, Param Thakkar
May 7, 2023
2. The Fundamental Theorem of Line Integrals
Section 1.1
Introduction to the Fundamental Theorem of Line Integrals
Background Information: Gradient Fields and Path Independence
Understanding gradient fields is essential in order to apply the Fundamental Theorem of Line Integrals.
Consider , a continuous vector field that satisfies = โ . This scalar-valued function is known as the
๐น
โ
๐น
โ
๐ ๐
potential function of . This property classifies as a conservative vector field, or a gradient field. Any
๐น
โ
๐น
โ
gradient field defined on an open, connected domain will produce path-independent line integrals. This means
that
d = d
๐ถ1
โซ๐น
โ
ยท ๐
โ
๐ถ2
โซ๐น
โ
ยท ๐
โ
for any two piecewise-smooth curves C1 and C2 with the same start and end points. It is important that these
curves are defined on the domain of .
๐น
โ
Defining the Fundamental Theorem of Line Integrals
Let and = โ , where is continuous on a connected, open region in . On any
๐: โ
๐
โ โ ๐น
โ
๐ ๐น
โ
โ
๐
piecewise-smooth curve from A to B,
๐ถ
.
๐ถ
โซ ๐น
โ
ยท ๐๐
โ
= ๐(๐ต
โ
) โ ๐(๐ด
โ
)
3. 3
Proving the Fundamental Theorem of Line Integrals
Now we will prove the Fundamental Theorem of Line Integrals. Let and . We can use the
๐ด = ๐ฅ
โ
(ฮฑ) ๐ต = ๐ฅ
โ
(ฮฒ)
chain rule to obtain,
๐ถ
โซ ๐น
โ
ยท ๐๐
โ
=
ฮฑ
ฮฒ
โซ โ๐(๐ฅ
โ
(๐ก)) ยท ๐ฅ
โ
'(๐ก)๐๐ก
=
ฮฑ
ฮฒ
โซ
๐
๐๐ก
[๐(๐ฅ
โ
(๐ก))]๐๐ก
= ๐(๐ฅ
โ
(ฮฒ)) โ ๐(๐ฅ
โ
(ฮฑ))
= ๐(๐ต
โ
) โ ๐(๐ด
โ
).
This shows that the line integral only depends on the endpoints, making it path-independent. Path independence
is a corollary of the FTLI.
Section 1.2
Finding the Potential Function
Option 1 - The โSafe Methodโ
We begin by using the โSafe Method.โ Define = for some and . Our goal is
๐น
โ โ๐
โ๐ฅ
,
โ๐
โ๐ฆ
( ) ๐น
โ
: โ
2
โ โ
2
๐: โ
2
โ โ
to find the function such that โ = . Take the antiderivative of with respect to .
๐ ๐ ๐น
โ โ๐
โ๐ฅ
๐ฅ
โซ
โ๐
โ๐ฅ
๐๐ฅ = ๐(๐ฅ, ๐ฆ) = ๐(๐ฅ, ๐ฆ) + ๐ถ(๐ฆ)
4. 4
In the equation above, is a function of and . This โintegration functionโ is the most important
๐ถ ๐ฆ, ๐: โ
2
โ โ
part of this process and can not be overlooked. Seeing that this is a function of y, if we took the derivative of
+ with respect to , would disappear. Letโs move on to the next step. Take the derivative of
๐ ๐ถ(๐ฆ) ๐ฅ ๐ถ(๐ฆ)
+ with respect to . We can cross reference it with the component from the vector field to obtain
๐ ๐ถ(๐ฆ) ๐ฆ
โ๐
โ๐ฆ
= = .
โ
โ๐ฆ
(๐ + ๐ถ(๐ฆ)) โ ๐๐ฆ
+ ๐ถ'(๐ฆ)
โ๐
โ๐ฆ
โ ๐ถ'(๐ฆ)
โ๐
โ๐ฆ
โ ๐๐ฆ
Now, we take the antiderivative of with respect to , obtaining
๐ถ'(๐ฆ) ๐ฆ
.
โซ๐ถ'(๐ฆ) ๐๐ฆ = ๐ถ(๐ฆ)
Recall back to the beginning of our process when we took the antiderivative of with respect to and got
โ๐
โ๐ฅ
๐ฅ
+ . We just discovered what is, so weโve filled in the missing piece. At this point, we have
๐ ๐ถ(๐ฆ) = ๐ ๐ถ(๐ฆ)
fully assembled the potential function. Here is an example with a defined vector field .
๐น
โ
= =
๐น
โ โ๐
โ๐ฅ
,
โ๐
โ๐ฆ
( ) ๐ฆ cos(๐ฅ) + 3๐ฅ
2
+ 1, sin(๐ฅ) + 4๐
4๐ฆ
( )
Integrate with respect to .
โ๐
โ๐ฅ
๐ฅ
=
โซ
โ๐
โ๐ฅ
๐๐ฅ โซ(๐ฆ cos(๐ฅ) + 3๐ฅ
2
+ 1) ๐๐ฅ
= ๐ฆ sin(๐ฅ) + ๐ฅ
3
+ ๐ฅ + ๐ถ(๐ฆ)
= ๐(๐ฅ, ๐ฆ)
Differentiate with respect to .
๐(๐ฅ, ๐ฆ) ๐ฆ
=
โ
โ๐ฆ
(๐(๐ฅ, ๐ฆ))
โ
โ๐ฆ
(๐ฆ sin(๐ฅ) + ๐ฅ
3
+ ๐ฅ + ๐ถ(๐ฆ))
5. 5
= sin(๐ฅ) + ๐ถ'(๐ฆ)
Compare the result above to .
โ๐
โ๐ฆ
=
sin(๐ฅ) + ๐ถ'(๐ฆ)
โ๐
โ๐ฆ
= +
sin(๐ฅ) + ๐ถ'(๐ฆ) sin(๐ฅ) 4๐
4๐ฆ
=
๐ถ'(๐ฆ) 4๐
4๐ฆ
Integrate with respect to to find .
๐ถ'(๐ฆ) = 4๐
4๐ฆ
๐ฆ ๐ถ(๐ฆ)
โซ4๐
4๐ฆ
๐๐ฆ = ๐
4๐ฆ
= ๐ถ(๐ฆ)
Put together the scalar potential function.
๐ = ๐ฆ sin(๐ฅ) + ๐ฅ
3
+ ๐ฅ + ๐
4๐ฆ
Option 2 - The โDangerous Methodโ
We can also use the โDangerous Methodโ to find the potential function. Weโll explain this method using an
example vector field. Define .
๐น
โ
(๐ฅ, ๐ฆ) = 2 cos(2๐ฅ) cos(๐ฆ) โ 1 + ๐ฆ๐
๐ฅ๐ฆ
, โ sin(2๐ฅ) sin(๐ฆ) + ๐ฅ๐
๐ฅ๐ฆ
+ 3๐ฆ
2
( )
First, we take the antiderivative of with respect to and of with respect to .
โ๐
โ๐ฅ
๐ฅ
โ๐
โ๐ฆ
๐ฆ
โซ(2 cos(2๐ฅ) cos(๐ฆ) โ 1 + ๐ฆ๐
๐ฅ๐ฆ
) ๐๐ฅ = sin(2๐ฅ) cos(๐ฆ) โ ๐ฅ + ๐
๐ฅ๐ฆ
+ ๐ถ(๐ฆ)
โซ(โ sin(2๐ฅ) sin(๐ฆ) + ๐ฅ๐
๐ฅ๐ฆ
+ 3๐ฆ
2
)๐๐ฆ = sin(2๐ฅ) cos(๐ฆ) + ๐
๐ฅ๐ฆ
+ ๐ฆ
3
+ ๐ถ(๐ฅ)
Now, we compare the two antiderivatives to see what they have in common. We do this to synthesize a โguessโ
as to what the scalar potential function might be by writing down function components that both
๐
antiderivatives exhibited, followed by each unique component seen in either antiderivative. In this case, both
6. 6
antiderivatives had and with unique components being - and . Combing these gives
๐ ๐๐(2๐ฅ)๐๐๐ (๐ฆ) ๐
๐ฅ๐ฆ
๐ฅ ๐ฆ
3
us
.
๐ = sin(2๐ฅ) cos(๐ฆ) + ๐
๐ฅ๐ฆ
โ ๐ฅ + ๐ฆ
3
Next, we verify we got the correct function by confirming = .
๐ โ๐ ๐น
โ
โ = , )
๐ (
โ
โ๐ฅ
(sin(2๐ฅ) cos(๐ฆ) + ๐
๐ฅ๐ฆ
โ ๐ฅ + ๐ฆ
3
)
โ
โ๐ฆ
(sin(2๐ฅ) cos(๐ฆ) + ๐
๐ฅ๐ฆ
โ ๐ฅ + ๐ฆ
3
)
= ( )
2 cos(2๐ฅ) cos(๐ฆ) + ๐ฆ๐
๐ฅ๐ฆ
โ 1, โ sin(2๐ฅ) sin(๐ฆ) + ๐ฅ๐
๐ฅ๐ฆ
+ 3๐ฆ
2
= ๐น
โ
Section 1.3
FTLI Examples
A โMundaneโ Example
Suppose we have the vector field = We wish to compute where C is any piecewise smooth
๐น
โ
(4๐ฅ
3
, 2).
๐ถ
โซ๐น
โ
ยท๐๐
โ
,
curve from the point (0,0) to (3,4). The first step for us is to integrate with respect to . We solve to obtain
4๐ฅ
3
๐ฅ
=
โซ4๐ฅ
3
๐๐ฅ ๐ฅ
4
+ ๐ถ(๐ฆ).
The next step is to integrate 2 with respect to so we then obtain
๐ฆ,
=
โซ 2 ๐๐ฆ 2๐ฆ + ๐ถ(๐ฅ).
7. 7
Combining our two results, we see that We can double check our answer by computing the
๐(๐ฅ, ๐ฆ) = ๐ฅ
4
+ 2๐ฆ.
gradient of .
๐
โ๐ = (
โ๐
โ๐ฅ
,
โ๐
โ๐ฆ
) = (4๐ฅ
3
, 2)
Now that we have checked our function we are ready to apply the Fundamental Theorem of Line Integrals
๐,
and compute our integral.
๐ถ
โซ๐น
โ
ยท๐๐
โ
= ๐(3, 4) โ ๐(0, 0)
= 3
4
+ 2 ยท 4 โ 0
4
โ 0 ยท 8
= 81 + 8
= 89
For any paths from (0,0) to (3,4) to which the FTLI is applicable, for , the line integral will always be 89
๐น
โ
because of path-independence.
To further our understanding, let's compute an integral using the definition of line integral over a path to show
that the FTLI works and is indeed equal to what we computed. Let our first curve be where
๐ฅ(๐ก) = (3๐ก, 4๐ก),
Plugging everything in,
๐ก ฯต [0, 1].
0
1
โซ (4 ยท (3๐ก)
3
, 2) ยท (3, 4) ๐๐ก =
0
1
โซ 324๐ก
3
+ 8 ๐๐ก
= 81 (1)
3
+ 8 โ 81 (0)
3
+ 8 ยท 0
= 81 + 8
.
= 89
8. 8
Thus, the Fundamental of Line Integrals indeed works as both of our computations were equal, showing that we
may use this shortcut.
Solving the Circulation Integral with FTLI
Define a gradient field on an open, connected space in , with a scalar-valued potential function . Consider
๐น
โ
โ
๐
๐
C, a closed curve with a counterclockwise parameterization defined on the same space. Using the Fundamental
Theorem of Line Integrals, we can prove that d = 0. Points A and B lie on curve C, dividing it into two
๐ถ
โฎ๐น
โ
ยท ๐
โ
open curves. We label the curve traveling from A to B as C1, and the curve traveling from B to A as C2, as shown
in Fig. 1.
Since both C1 and C2 are open paths with initial and terminal points, we can apply the Fundamental Theorem of
Line Integrals to calculate the following line integrals:
9. 9
d = (B) - (A)
๐ถ1
โซ ๐น
โ
ยท ๐
โ
๐ ๐
d = (A) - (B)
๐ถ2
โซ ๐น
โ
ยท ๐
โ
๐ ๐
Our objective is to calculate the circulation integral over C. Seeing that C1 and C2 combined form C, we
accomplish our objective by finding the sum of the line integrals over C1 and C2. Thus, we obtain
d = d + d
๐ถ
โฎ ๐น
โ
ยท ๐
โ
๐ถ1
โซ๐น
โ
ยท ๐
โ
๐ถ2
โซ๐น
โ
ยท ๐
โ
= (B) - (A) + (A) - (B)
๐ ๐ ๐ ๐
= 0.
The relationship between the circulation integral and the FTLI can be conceptualized using path independence.
Since we know that the line integral over a single point is zero, we can deduce that the line integral over a curve
with the same initial and terminal point is also zero, regardless of the curveโs length. Thus, d = 0 for any
๐ถ
โฎ๐น
โ
ยท ๐
โ
closed curve C.