VECTOR CALCULUS

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17
VECTOR CALCULUS
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VECTOR CALCULUS
17.4 Greens Theorem
In this section, we will learn about Greens
Theorem for various regions and its application
in evaluating a line integral.
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INTRODUCTION

• Greens Theorem gives the relationship between a
  line integral around a simple closed curve C and
  a double integral over the plane region D bounded
  by C.
• We assume that D consists of all points inside
  C as well as all points on C.

Fig. 17.4.1, p. 1091
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INTRODUCTION

• In stating Greens Theorem, we use the
  convention
• The positive orientation of a simple closed curve
  C refers to a single counterclockwise traversal
  of C.

Fig. 17.4.2a, p. 1091
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INTRODUCTION

• Thus, if C is given by the vector function r(t),
  a t b, then the region D is always on the
  left as the point r(t) traverses C.

Fig. 17.4.2b, p. 1091
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GREENS 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

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NOTATIONS
Note

• The notation
• is sometimes used to indicate that the line
  integral is calculated using the positive
  orientation of the closed curve C.

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NOTATIONS
NoteEquation 1

• Another notation for the positively oriented
  boundary curve of D is ?D.
• So, the equation in Greens Theorem can be
  written as

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GREENS THEOREM

• Greens Theorem should be regarded as the
  counterpart of the Fundamental Theorem of
  Calculus (FTC) for double integrals.

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GREENS THEOREM

• Compare Equation 1 with the statement of the FTC
  Part 2 (FTC2), in this equation
• In both cases,
• There is an integral involving derivatives (F,
  ?Q/?x, and ?P/?y) on the left side.
• The right side involves the values of the
  original functions (F, Q, and P) only on the
  boundary of the domain.

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GREENS THEOREM

• In the one-dimensional case, the domain is an
  interval a, b whose boundary consists of just
  two points, a and b.

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SIMPLE REGION

• 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
  (Section 16.3).
• Lets call such regions simple regions.

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GREENS TH. (SIMPLE REGION)
ProofEqns. 2 3

• Notice that the theorem will be proved if we can
  show that
• and

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GREENS TH. (SIMPLE REGION)
Proof

• We prove Equation 2 by expressing D as a type I
  region
• D (x, y) a x b, g1(x) y g2(x)
• where g1 and g2 are continuous functions.

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GREENS TH. (SIMPLE REGION)
ProofEquations 4

• That enables us to compute the double integral on
  the right side of Equation 2 as
• where the last step follows from the FTC.

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GREENS TH. (SIMPLE REGION)
Proof

• Now, we compute the left side of Equation 2 by
  breaking up C as the union of the four curves C1,
  C2, C3, and C4.

Fig. 17.4.3, p. 1092
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GREENS TH. (SIMPLE REGION)
Proof

• On C1 we take x as the parameter and write the
  parametric equations as x x, y g1(x), a
  x b
• Thus,

Fig. 17.4.3, p. 1092
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GREENS TH. (SIMPLE REGION)
Proof

• Observe that C3 goes from right to left but C3
  goes from left to right.

Fig. 17.4.3, p. 1092
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GREENS TH. (SIMPLE REGION)
Proof

• So, we can write the parametric equations of C3
  as x x, y g2(x), a x b
• Therefore,

Fig. 17.4.3, p. 1092
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GREENS TH. (SIMPLE REGION)
Proof

• On C2 or C4 (either of which might reduce to just
  a single point), x is constant.
• So, dx 0 and

Fig. 17.4.3, p. 1092
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GREENS TH. (SIMPLE REGION)
Proof

• Hence,

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GREENS TH. (SIMPLE REGION)
Proof

• Comparing this expression with the one in
  Equation 4, we see that

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GREENS TH. (SIMPLE REGION)
Proof

• Equation 3 can be proved in much the same way by
  expressing D as a type II region.
• Then, by adding Equations 2 and 3, we obtain
  Greens Theorem.
• See Exercise 28.

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GREENS THEOREM
Example 1

• Evaluate 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)

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GREENS THEOREM
Example 1

• The given line integral could be evaluated as
  usual by the methods of Section 16.2.
• However, that would involve setting up three
  separate integrals along the three sides of the
  triangle.
• So, lets use Greens Theorem instead.

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GREENS TH. (SIMPLE REGION)
Example 1

• Notice that the region D enclosed by C is simple
  and C has positive orientation.

Fig. 17.4.4, p. 1093
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GREENS TH. (SIMPLE REGION)
Example 1

• If we let P(x, y) x4 and Q(x, y) xy, then

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GREENS THEOREM
Example 2

• Evaluate
• where C is the circle x2 y2 9.
• The region D bounded by C is the disk x2 y2
  9.

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GREENS THEOREM
Example 2

• So, lets change to polar coordinates after
  applying Greens Theorem

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GREENS THEOREM

• In Examples 1 and 2, we found that the double
  integral was easier to evaluate than the line
  integral.
• Try setting up the line integral in Example 2
  and youll soon be convinced!

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REVERSE DIRECTION

• Sometimes, though, its easier to evaluate the
  line integral, and Greens Theorem is used in
  the reverse direction.
• For instance, if it is known that P(x, y) Q(x,
  y) 0 on the curve C, the theorem gives
  no matter what values P and Q assume in D.

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REVERSE DIRECTION

• Another application of the reverse direction of
  the theorem is in computing areas.
• As the area of D is , we wish to choose
  P and Q so that

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REVERSE DIRECTION

• There are several possibilities
• P(x, y) 0
• P(x, y) y
• P(x, y) ½y
• Q(x, y) x
• Q(x, y) 0
• Q(x, y) ½x

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REVERSE DIRECTION
Equation 5

• Then, Greens Theorem gives the following
  formulas for the area of D

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REVERSE DIRECTION
Example 3

• Find the area enclosed by the ellipse
• The ellipse has parametric equations x a
  cos t, y b sin t, 0 t 2p

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REVERSE DIRECTION
Example 3

• Using the third formula in Equation 5, we have

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UNION OF SIMPLE REGIONS

• We have proved Greens Theorem only for the case
  where D is simple.
• Still, we can now extend it to the case where D
  is a finite union of simple regions.

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UNION OF SIMPLE REGIONS

• For example, if D is the region shown here, we
  can write D D1 D2 where D1 and D2
  are both simple.

Fig. 17.4.5, p. 1094
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UNION OF SIMPLE REGIONS

• The boundary of D1 is C1 C3.
• The boundary of D2 is C2 (C3).

Fig. 17.4.5, p. 1094
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UNION OF SIMPLE REGIONS

• So, applying Greens Theorem to D1 and D2
  separately, we get

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UNION OF SIMPLE REGIONS

• If we add these two equations, the line integrals
  along C3 and C3 cancel.
• So, we get
• Its boundary is C C1 C2 .
• Thus, this is Greens Theorem for D D1 D2.

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UNION OF NONOVERLAPPING SIMPLE REGIONS

• The same sort of argument allows us to establish
  Greens Theorem for any finite union of
  nonoverlapping simple regions.

Fig. 17.4.6, p. 1094
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UNION OF SIMPLE REGIONS
Example 4

• Evaluate where C is the boundary of the
  semiannular region D in the upper half-plane
  between the circles x2 y2 1 and x2 y2 4.

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UNION OF SIMPLE REGIONS
Example 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 ? p

Fig. 17.4.7, p. 1095
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UNION OF SIMPLE REGIONS
Example 4

• So, Greens Theorem gives

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REGIONS WITH HOLES

• Greens Theorem can be extended to regions with
  holesthat is, regions that are not
  simply-connected.

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REGIONS WITH HOLES

• Observe that the boundary C of the region D here
  consists of two simple closed curves C1 and C2.

Fig. 17.4.8, p. 1095
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REGIONS WITH HOLES

• We assume that these boundary curves are oriented
  so that the region D is always on the left as
  the curve C is traversed.
• So, the positive direction is counterclockwise
  for C1 but clockwise for C2.

Fig. 17.4.8, p. 1095
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REGIONS WITH HOLES

• Lets divide D into two regions D and D by
  means of the lines shown here.

Fig. 17.4.9, p. 1095
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REGIONS WITH HOLES

• Then, applying Greens Theorem to each of D and
  D , we get
• As the line integrals along the common boundary
  lines are in opposite directions, they cancel.

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REGIONS WITH HOLES

• Thus, we get
• This is Greens Theorem for the region D.

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REGIONS WITH HOLES
Example 5

• If F(x, y) (y i x j)/(x2 y2) show that
  ?C F dr 2p for every positively oriented,
  simple closed path that encloses the origin.

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REGIONS WITH HOLES
Example 5

• C is an arbitrary closed path that encloses the
  origin.
• Thus, its difficult to compute the given
  integral directly.

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REGIONS WITH HOLES
Example 5

• 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.

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REGIONS WITH HOLES
Example 5

• Let D be the region bounded by C and C.
• Then, its positively oriented boundary is C
  (C).

Fig. 17.4.10, p. 1095
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REGIONS WITH HOLES
Example 5

• So, the general version of Greens Theorem gives

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REGIONS WITH HOLES
Example 5

• Therefore,
• That is,
• We now easily compute this last integral using
  the parametrization given by r(t) a cos t
  i a sin t j, 0 t 2p

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REGIONS WITH HOLES
Example 5

• Thus,

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GREENS THEOREM

• We end by using Greens Theorem to discuss a
  result that was stated in Section 17.3

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THEOREM 6 IN SECTION 17.3
Proof

• Were assuming that
• F P i Q j is a vector field on an open
  simply-connected region D.
• P and Q have continuous first-order partial
  derivatives.
• throughout D

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THEOREM 6 IN SECTION 17.3
Proof

• If C is any simple closed path in D and R is the
  region that C encloses, Greens Theorem gives

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THEOREM 6 IN SECTION 17.3
Proof

• 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 ?C F dr 0
  for any closed curve C.

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THEOREM 6 IN SECTION 17.3
Proof

• Thus, ?C F dr is independent of path in D y
  Theorem 3 in Section 16.3.
• It follows that F is a conservative vector field.