View
15
Download
0
Category
Preview:
Citation preview
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?
Math 55 FTLI and Green’s Theorem
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) �
Math 55 FTLI and Green’s Theorem
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.
Math 55 FTLI and Green’s Theorem
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.
Math 55 FTLI and Green’s Theorem
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 �
Math 55 FTLI and Green’s Theorem
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 .
Math 55 FTLI and Green’s Theorem
Fundamental Theorem of Line Integrals
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.
Math 55 FTLI and Green’s Theorem
Fundamental Theorem of Line Integrals
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.
Math 55 FTLI and Green’s Theorem
Fundamental Theorem of Line Integrals
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.
Math 55 FTLI and Green’s Theorem
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).
Math 55 FTLI and Green’s Theorem
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
Math 55 FTLI and Green’s Theorem
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
Math 55 FTLI and Green’s Theorem
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.
Math 55 FTLI and Green’s Theorem
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
Math 55 FTLI and Green’s Theorem
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
Math 55 FTLI and Green’s Theorem
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
Math 55 FTLI and Green’s Theorem
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).
Math 55 FTLI and Green’s Theorem
Recommended