Multivariable MA1104 Calculus Lecture 17
Dr KU Cheng Yeaw Monday March 23, 2009
Recap
Last Lecture ... we considered integral of a vector field. This type of line integrals appear frequently in physics. It turns out that the computation of this line integral simplifies dramatically (in particular, it depends only on the initial and terminal points of a curve) provided the vector field is conservative. This is the Fundamental Theorem for Line Integrals.
Overview
In this lecture ... we study Green's Theorem and see how to apply it to compute line integrals. Green's Theorem can be regarded as the counterpart of the Fundamental Theorem of Calculus for double integral at first glance, you might think Green's Theorem is strange and abstract, one that only mathematician could invent to torture students of Calculus. Regardless of what you think, it is of fundamental importance in the analysis of fluid flows and in the theories of electricity and magnetism.
Green's Theorem
Whenever we mention a line integral C F dr = C P dx + Q dy along a smooth curve C, C must be given by a parametrization. In particular, C has an orientation. This line integral depends on the orientation of C. However, the line integral is the same for any two parametrization of C with the same orientation.
If this parametrization is explicit, then we can evaluate the line integral easily. But if it is not given, then we have to write down one (which is not easy at times) We use -C to denote the curve consisting the same point of C but with opposite orientation, that is, from the terminal point of C to the initial point of C.
In the case when F is (continuous) conservative, that is f = F for some scalar field F, the Fundamental Theorem for Line Integral simplifies the evaluation of a line integral: F dr = f (r(b)) - f (r(a))
C
where C is given by r(t), a t b. That is really amazing because it says that for conservative vector field, the line integral depends just on the initial and terminal points!
Moreover, if F is continuous conservative with domain D, then F dr = 0
C
for every closed path C in D.
Now, let us restrict our discussion to oriented, piecewise-smooth, simple closed curve on R2 and two-dimensional vector fields. The question remains: Given F = P i + Qj, how can we evaluate C F dr along an oriented, piecewise-smooth, simple closed curve C? Green's Theorem helps us to achieve this.
To state Green's Theorem, we require to use the convention that the positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C. We can think of this being given by the right-hand-rule.
Thus, if the positively oriented C is given by the vector function r(t), a t b, then the region D enclosed by C is always on the left as the point r(t) traverses C.
Green's Theorem Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then P dx + Q dy =
C D
Q P - x y
dA.
Notation 1. We also use the notation
P dx + Q dy
C
to indicate that the line integral is calculated using the positive orientation of the closed curve c.
Notation 2. Notice the closed curve C and D in Green's Theorem are closely related: C is the positively oriented boundary of the region D, so the equation in Green's Theorem can be written as Q P - x y
dA =
D
P dx + Q dy.
D
There several interesting remarks about Green"s Theorem: Remark 1. Green's Theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D. This is good news! Because by now we are already experts in multiple integrations!
Remark 2. Green's Theorem should be regarded as the counterpart of the Fundamental Theorem of Calculus (FTC) for double integrals. Compare the Green's Equation: Q P - x y
dA =
D
P dx + Q dy.
D
with the statement of the FTC Part 2:
b
F (x) dx = F (b) - F (a).
a
In both cases there is an integral involving derivatives (F , left side.
Q x
and
P y )
on the
the right side involves the values of the original functions (F , Q, and P ) only on the boundary of the domain.
Proof of Green's Theorem
The theorem is not easy to prove in general. Still, we can give a proof for the special case where the region is both of type I and type II Lets call such regions simple regions.
Notice that the theorem will be proved if we can show that: P dx = -
C D
P dA y
and Q dy =
C D
Q dA. x
Let D be expressed as a type I region: D = {(x, y) : a x b, g1 (x) y g2 (x)} where g1 (x) and g2 (x) are continuous functions. We shall only prove P dx = -
C D
P dA. y
Q dy =
C D
Q dA x
can be proved in the same way by expressing D as a type II region.
OK, to show P dx = -
C D
P dA y
we consider P dA = y =
a b a b g2 (x) g1 (x)
D
P (x, y) dy dx y
[P (x, g2 (x)) - P (x, g1 (x))] dx
where the last step follows from the Fundamental Theorem of Calculus.
On the other hand, we compute as the union of the four curves
P dx + Q dy by breaking up C C3 , and C4 .
C C1 , C2 ,
On C1 : taking x as a parameter, we have a parametrization of C1 : y = g1 (x), a x b. Thus
b
P (x, y) dx =
C1 C1
P (x, g1 (x)) dx =
a
P (x, g1 (x)) dx
by Fundamental Theorem of Calculus.
On C3 , we notice that -C3 goes from left to right, so we write -C3 as y = g2 (x), a x b. Thus
b
P (x, y) dx = -
C3 -C3
P (x, g2 (x)) dx = -
a
P (x, g2 (x)) dx.
On C2 and C3 , x is constant. So P (x, y) dx =
C2
dx dt
= 0. So dx dt = 0. dt dx dt = 0. dt
P (r(t))
C2
P (x, y) dx =
C4 C2
P (r(t))
Combining, we have
P dx =
C C1
P dx +
C2
P dx +
C3
P dx +
C4
P dx
=
C1 b
P dx +
C3
P dx
b
=
a
P (x, g1 (x)) dx -
a
P (x, g2 (x)) dx.
Comparing this expression with the one we got earlier from the double integral, we conclude that P (x, y) dx = -
C D
P dA. y
Example 1. Evaluate x4 dx + xy dy
C
where C is the triangular curve consisting of the line segments from (0, 0) to (1, 0) from (1, 0) to (0, 1) from (0, 1) to (0, 0).
Solution. We can evaluate the line integral the methods in the last lectures: this involve setting up three separate integrals along the three sides of the triangle together with appropriate parametrization. This sounds complicated! Why not use Green's Theorem since the curve is positively oriented, piecewise-smooth, simple closed curve. So, let P (x, y) = x4 and Q(x, y) = xy.
Then, by Green's Theorem, the line integral is just a double integral: Q P - x y
1-x 0 1
x4 dx + xy dy =
C D 1
dA
=
0
(y - 0) dy dx 1 2 y 2
1 y=1-x
=
0
dx
y=0
= =
1 2 1 . 6
(1 - x2 ) dx
0
Wow! Green's is Theorem really cool. The double integral was easier to compute than the line integral. I like it! Lets try it one more time. Example 2. Evaluate (3y - esin x ) dx + (7x +
C
y 4 + 1) dy
where C is the circle x2 + y 2 = 9. Solution. Hold on! It seems that there is no orientation given to C. Actually, the notation oriented, so no worry.
C
already tells us that C is positively
The region D enclosed by C is the disk x2 + y 2 9, so changing to polar coordinates and applying Green's Theorem: (3y - esin x ) dx + (7x +
C
y 4 + 1) dy
=
D 2
(7x + x
3 0
y 4 + 1) -
(3y - esin x ) dA y
=
0 2
(7 - 3)r dr d
3
= =
4
0
d
0
r dr
36.
Application of Green's Theorem
Though we use Green's Theorem to evaluate a line integral by turning it into a double integral, sometimes we can also use it to compute a double integral by turning it into a line integral. One of such application is to compute areas of plane regions. Suppose we want to compute an area A of D. We know that it is given by 1 dA.
D
To use Green's Theorem, we need to find P and Q so that Q P - = 1. x y There are several possibilities: P (x, y) = 0, Q(x, y) = x. or P (x, y) = -y, Q(x, y) = 0. or 1 1 P (x, y) = - y, Q(x, y) = x. 2 2
The Green's Theorem gives the following formula for the area of D: 1 2
A-
C
x dy = -
C
y dx =
x dy - y dx
C
Example 3. Find the area enclosed by the ellipse Solution. The ellipse has parametric equations: x = a cos t, So, we have 1 2 1 2 ab 2
x2 a2
+
y2 b2
= 1.
y = b sin t, 0 t 2.
A = = =
x dy - y dx
C 2
(a cos t)(b cos t) dt - (b sin t)(-a sin t) dt
0 2
dt = ab.
0
Extending Green's Theorem
We have proved Greens Theorem only for the case where D is simple (both Type I and Type II). Still, we can now extend it to the case where D is a finite union of simple regions. D is a region which is not simply connected, that is, it has one or more holes.
For example, if D is the region shown here
Notice D is of Type I but not Type II. So D is not simple.
But we can write D = D1 D2 where D1 and D2 are simple.
The boundary of D1 is C1 C3 . The boundary of D2 is C2 (C3 ).
So, applying Greens Theorem to D1 and D2 separately, we get:
P dx + Q dy =
C1 C3 D1
Q P - x y Q P - x y
dA dA
P dx + Q dy =
C2 (-C3 ) D2
If we add these two equations, the line integral along C3 and -C3 cancel, so we get
P dx + Q dy =
C1 C2 D
Q P - x y
dA
which is Green's Theorem for D = D1 D2 , since its boundary is C = C1 C2 .
The same sort of argument allows us to establish Greens Theorem for any finite union of nonoverlapping simple regions.
Example 4. Evaluate y 2 dx + 3xy dy
C
where C is the boundary of the semiannular region D in the upper half-plane between the circles x2 + y 2 = 1 and x2 + y 2 = 4.
Notice that, though D is not simple, the y-axis divides it into two simple regions.
In polar coordinates we can write D = {(r, ) : 1 r 2, 0 }.
Therefore, Green's Theorem gives 2 (3xy) - (y ) dA x y
2
y 2 dx + 3xy dy =
C D
=
D
y dA =
0 2 1 1
(r sin )r dr d r2 dr
=
0
sin d 14 . 3
=
Observe that the boundary C of the region D here consists of two simple closed curves C1 and C2 .
We assume that these boundary curves are oriented so that the region D is always on the left as the curve C is traversed.
We extend the definition of positive orientation for such a region: counterclockwise for C1 but clockwise for C2 .
Now, divide D into two regions D and D as shown below:
Applying Green's Theorem to each D and D , we get Q P - x y dA
D
=
D
Q P - x y P dx + Q dy +
dA +
D2
Q P - x y
dA
=
D
P dx + Q dy
D
Since the line integrals along the common boundary lines are in opposite directions, they cancel and we get Q P - x y dA
D
=
C1
P dx + Q dy +
C2
P dx + Q dy
=
C
P dx + Q dy
which is Green's Theorem for the region D.
Example 5. If F(x, y) = (-yi + xj)/(x2 + y 2 ) show that F dr = 2
C
for every positively oriented, simple closed path that encloses the origin. Solution. It's difficult to compute the line integral directly because C is an arbitrary closed path that encloses the origin.
So, lets consider a counterclockwise-oriented circle C with center the origin and radius a, where a is chosen to be small enough that C lies inside C.
Let D be the region bounded by C and C . Then, its positively oriented boundary is C (-C ).
So, the general version of Greens Theorem gives: P dx + Q dy +
C -C
P dx + Q dy
=
D
=
D
Q P - dA x y y 2 - x2 y 2 - x2 - 2 (x2 + y 2 )2 (x + y 2 )2
dA
=
0.
Therefore, P dx + Q dy =
C C
P dx + Q dy
that is F dr =
C C
F dr.
We now easily compute this last integral using the parametrization of C given by r(t) = a cos ti + a sin tj, 0 t 2.
Thus
F dr =
C
F dr
C 2
=
0 2
F(r(t)) r (t) dt (-a sin t)(-a sin t) + (a cos t)(a cos t) dt a2 cos2 t + a2 sin2 t dt
0
=
0 2
= = 2.
We end today's lecture by revisiting a question we asked in the last lecture. First, recall a fact which motivates our question: If F(x, y) = P (x, y)i + Q(x, y)j is a conservative vector field where P and Q have continuous first partial derivatives on a domain D, then throughout D we have P Q = . y x
Question. Suppose F = P i + Qj is a vector field with domain D, P and Q have continuous first partial derivative. If P = Q , can y x we conclude that F is conservative? That is, is the converse of the preceding theorem true? Answer. Yes, provided D is open simply-connected region. Recall that a simply-connected region D in the plane is a connected region D such that every simple closed curve in D encloses only points that are in D.
Let F = P i + Qj be a vector field on an open simply-connected D, where P and Q have continuous first-order partial derivatives. Suppose Q P = throughout D. y x Then F is conservative.
Proof. Let C be a simple closed path in D. Let R be the region that C encloses. By Green's Theorem
F dr =
C C
P dx + Q dy Q P - x y 0 dA
R
=
R
dA
= = 0.
A curve that is not simple crosses itself at one or more points and can be broken up into a number of simple curves. We have shown that the line integrals of F around these simple curves are all 0. Adding these integrals, we see that curve C. F dr = 0 for any closed
C
Thus, C F dr is independent of path in D by the equivalence theorem we proved in Slide 16. It follows that F is a conservative vector field.
Example 6. Determine whether or not the vector field F(x, y) = (3 + 2xy)i + (x2 - 3y 2 )j is conservative. Solution. Let P (x, y) = 3 + 2xy and Q(x, y) = x2 - 3y 2 . Then Q P = 2x = . y x Also, the domain of F is the entire plane D = R2 which is open and simply-connected. Therefore, by the preceding Theorem , we conclude that F is conservative.