15 FTLI and Greens Thm

The Fundamental Theorem of Line Integralsand Green’s Theorem

Lucky Galvez

Institute of MathematicsUniversity of the Philippines

Diliman

Math 55 FTLI and Green’s Theorem

Recall: FTOC

Recall from Math 53:

Theorem

Let f(x) be a function that is continuous on [a, b]. Then∫ b

af(x) dx = F (b)− F (a)

where F ′(x) = f(x).

Question: Is there an analog of this theorem for line integrals?


An analog of FTOC for Line Integrals

Theorem

Let C be a smooth curve given by a vector function ~R(t),a ≤ t ≤ b and f be a differentiable function whose gradientvector ∇f is continuous on C. Then∫

C∇f · d~R = f(~R(b))− f(~R(a))

Proof. By the definition of line integrals,∫C

∇f · d~R =

∫ b

a

∇f(~R(t)) · ~R′(t) dt

=

∫ b

a

(∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt

)dt

=

∫ b

a

d

dtf(~R(t)) dt (by the Chain Rule)

= f(~R(b))− f(~R(a)) (by FTOC) �


Notation

If f is a function of two variables and C is a plane curve withinitial point A(x1, y1) and terminal point B(x2, y2), then∫

C∇f · d~R = f(x2, y2)− f(x1, y1).

If f is a function of three variables and C is a plane curve withinitial point A(x1, y1, z1) and terminal point B(x2, y2, z2), then∫

C∇f · d~R = f(x2, y2, z2)− f(x1, y1, z1).

In other words, we evaluate f at the endpoints.


Independence of Path

Suppose C1 and C2 are two curves (paths) having the sameinitial and terminal points. We know in general that∫

C1

~F · d~R 6=∫C1

~F · d~R

If the equality holds for any two paths C1 and C2 having the

same initial and terminal points, then we say that

∫C

~F · d~R is

independent of path.


Independence of Path

Theorem

Let ~F be a continuous vector field with domain D. The line

integral

∫C

~F · d~R is independent of path in D if and only if∫C

~F · d~R = 0 for every closed path C in D.

Proof. Let

∫C

~F · d~R be independent of path and consider a

closed path C in D. We can regard C = C1 ∪ C2. Note that C1

and −C2 have the same initial and terminal points. Hence,∫C

~F · d~R =

∫C1

~F · d~R+

∫C2

~F · d~R

=

∫C1

~F · d~R−∫−C2

~F · d~R

= 0 �


Fundamental Theorem of Line Integrals

We have seen that

∫C∇f · d~R is independent of path.

If we let ~F = ∇f , i.e., ~F is a conservative vector field withpotential function f , then we have the following theorem:

Theorem (FTLI)

The line integral of a conservative vector field ~F is independentof path. That is, if C is a smooth curve given by ~R(t), a ≤ t ≤ bwith initial point A(x1, y1, z1) and terminal point B(x2, y2, z2)and ~F is a conservative vector field which is continuous on C,then ∫

C

~F · d~R = f(x2, y2, z2)− f(x1, y1, z1)

where f is a potential function for ~F .



The following follows immediately from the previous theorem.

Corollary

Let ~F be a conservative vector field with domain D. Then∫C

~F · d~R = 0

for any closed curve C in D.



Example

Show that the vector field ~F (x, y) =⟨y2, 2xy

⟩is conservative

and evaluate

∫C

~F · d~R where C is the unit circle.

Solution. Recall that ~F = 〈P,Q〉 is conservative iff Py = Qx.Since

∂

∂yy2 = 2y =

∂

∂x2xy,

F (x, y) =⟨y2, 2xy

⟩is conservative.

Now, C is a closed curve so by the previous corollary,∫C

~F · d~R = 0.



Example

Evaluate

∫C

~F · d~R where ~F (x, y) =⟨4x3y4 + 2x, 4x4y3 + 2y

⟩and C is given by ~R(t) =

⟨tπ cos t− 1, sin

(t2

)⟩, 0 ≤ t ≤ π.

Solution. We can use the definition of line integrals but thatwould give a nasty solution.Instead, we can show that ~F is conservative with potentialfunction

f(x, y) = x4y4 + x2 + y2 + c.

To use FTLI, we find the endpoints of C. Note that

~R(0) = 〈−1, 0〉 and ~R(π) = 〈−2, 1〉 .

Hence,

∫C

~F · d~R = f(−2, 1)− f(−1, 0) = 21− 1 = 20.


Exercises

1 Given ~F (x, y) =

⟨y2

1 + x2, 2y tan−1 x

⟩,

a. Show that ~F is conservative and find a potential function.

b. Evaluate

∫C

~F · d~R for any path C from (0, 0) to (1, 2).

2 Evaluate

∫C

sin y dx+ x cos y dy, where C is the ellipse

x2 + xy + y2 = 1.

3 Find the work done by the force field~F (x, y) = 〈e−y,−xe−y〉 in moving an object from the pointA(0, 1) to the point B(2, 0).


Orientation of a Curve

Let C be a simple closed curve. The positive orientation ofC refers to the single counterclockwise traversal of C.

Figure: positive orienation Figure: negative orienation


Green’s Theorem

Theorem

Let C be a positively oriented, piecewise-smooth, simple closedcurve in the plane and let D be the region bounded by C. If Pand Q have continuous partial derivatives on an open regionthat contains D, then∫

CP dx+Qdy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA


Remarks and Notation

1 The Green’s Theorem relates a line integral along a curveC and the double integral over the plane region D boundedby C.

2 The notation

∮CP dx+Qdy is sometimes used to indicate

that the line integral is calculated using the positiveorientation of C.

3 Other texts denote the positively oriented boundary curveof D as ∂D and hence∫∫

D

(∂Q

∂x− ∂P

∂y

)dA =

∫∂D

P dx+Qdy

4 The Green’s Theorem can be regarded, in some sense, asthe analog of FTOC for double integrals.


Green’s Theorem

Green’s Theorem will also hold if the region D is not simplyconnected, i.e, the boundary of D is C = C1 ∪ C2 (assume theyare positively oriented).

∫∫D

(∂Q

∂x− ∂P

∂y

)dA =

∫∫D′

(∂Q

∂x− ∂P

∂y

)dA+

∫∫D′′

(∂Q

∂x− ∂P

∂y

)dA

=

∫∂D′

P dx+Qdy +

∫∂D′′

P dx+Qdy


Green’s Theorem

Green’s Theorem will also hold if the region D is not simplyconnected, i.e, the boundary of D is C = C1 ∪ C2 (assume theyare positively oriented).

The line integrals along each of the common boundary pointsare on opposite directions, hence they cancel and we get∫∫D

(∂Q

∂x− ∂P

∂y

)dA =

∫C1

P dx+Qdy +

∫C2

P dx+Qdy

=

∫CP dx+Qdy


Green’s Theorem

Example

Evaluate

∫Cx2y dx+ xy2 dy where C is the boundary of the

region D between the circles x2 + y2 = 4 and x2 + y2 = 1.

Solution. Note that the region D can be expressedconveniently in polar coordinates, i.e.,

D = {(r, θ) : 1 ≤ r ≤ 2, 0 ≤ θ2π} .Hence, by Green’s Theorem,∫

C

x2y dx+ xy2 dy =

∫∫D

(∂(xy2)

∂x− ∂(x2y)

∂y

)dA

=

∫∫D

y2 − x2 dA

=

∫ 2π

0

∫ 2

1

r3(sin2 θ − cos2 θ) dr dθ = 0


Exercises

1 Evaluate

∫C

cos y dx+ x2 sin y dy where C is the rectangle

from (0, 0) to (0, π) to (2, π) to (2, 0) to (0, 0).

2 Evaluate

∫Cx2y dx+ xy2 dy where C is the positively

oriented triangle with vertices at (0, 0), (1, 0) and (0, 1).

3 Evaluate

∫C

(y + e√x) dx+ (2x+ cos y2) dy, where C is the

positively oriented boundary of the region enclosed by theparabolas y = x2 and x = y2.

4 Use Green’s Theorem to evaluate

∫C

~F · d~R where

~F (x, y) =⟨√

x+ y3, x2 +√y⟩

and C consists of the curvey = sinx from (0, 0) to (π, 0) and the line segment from(π, 0) to (0, 0).


References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/


15 FTLI and Greens Thm

Documents

15 FTLI and Greens Thm