VECTOR CALCULUS

1. 16 VECTOR CALCULUS

2. VECTOR CALCULUS 16.4 Green’s Theorem • In this section, we will learn about: • Green’s Theorem for various regions and its application in evaluating a line integral.

3. INTRODUCTION • 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 bounded by C. • We assume that Dconsists of all points inside C as well as all points on C.

4. INTRODUCTION • In stating Green’s Theorem, we use the convention: • The positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C.

5. 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.

6. 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

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

8. NOTATIONS Note—Equation 1 • Another notation for the positively oriented boundary curve of D is∂D. • So,the equation in Green’s Theorem can be written as:

9. GREEN’S THEOREM • Green’s Theorem should be regarded as the counterpart of the Fundamental Theorem of Calculus (FTC) for double integrals.

10. GREEN’S 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.

11. GREEN’S THEOREM • In the one-dimensional case, the domain is an interval [a, b] whose boundary consists of just two points, a and b.

12. 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 15.3). • Let’s call such regions simple regions.

13. GREEN’S TH. (SIMPLE REGION) Proof—Eqns. 2 & 3 • Notice that the theorem will be proved if we can show that: • and

14. GREEN’S 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.

15. GREEN’S TH. (SIMPLE REGION) Proof—Equations 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.

16. GREEN’S 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.

17. GREEN’S 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,

18. GREEN’S TH. (SIMPLE REGION) Proof • Observe that C3 goes from right to left but –C3 goes from left to right.

19. GREEN’S TH. (SIMPLE REGION) Proof • So, we can write the parametric equations of –C3 as: x =x, y =g2(x), a ≤ x ≤b • Therefore,

20. GREEN’S 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

21. GREEN’S TH. (SIMPLE REGION) Proof • Hence,

22. GREEN’S TH. (SIMPLE REGION) Proof • Comparing this expression with the one in Equation 4, we see that:

23. GREEN’S 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 Green’s Theorem. • See Exercise 28.

24. GREEN’S 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)

25. GREEN’S 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, let’s use Green’s Theorem instead.

26. GREEN’S TH. (SIMPLE REGION) Example 1 • Notice that the region D enclosed by Cis simple and C has positive orientation.

27. GREEN’S TH. (SIMPLE REGION) Example 1 • If we let P(x, y) = x4 and Q(x, y) = xy, then

28. GREEN’S THEOREM Example 2 • Evaluate • where C is the circle x2 + y2 = 9. • The region D bounded by C is the disk x2 + y2≤ 9.

29. GREEN’S THEOREM Example 2 • So, let’s change to polar coordinates after applying Green’s Theorem:

30. GREEN’S 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 you’ll soon be convinced!

31. REVERSE DIRECTION • Sometimes, though, it’s easier to evaluate the line integral, and Green’s 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.

32. 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:

33. 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

34. REVERSE DIRECTION Equation 5 • Then, Green’s Theorem gives the following formulas for the area of D:

35. 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 ≤ 2π

36. REVERSE DIRECTION Example 3 • Using the third formula in Equation 5, we have:

37. UNION OF SIMPLE REGIONS • We have proved Green’s 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.

38. UNION OF SIMPLE REGIONS • For example, if D is the region shown here, we can write: D =D1D2where D1 and D2are both simple.

39. UNION OF SIMPLE REGIONS • The boundary of D1 is C1C3. • The boundary of D2 is C2 (–C3).

40. UNION OF SIMPLE REGIONS • So, applying Green’s Theorem to D1 and D2 separately, we get:

41. 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 = C1C2 . • Thus, this is Green’s Theorem for D =D1D2.

42. UNION OF NONOVERLAPPING SIMPLE REGIONS • The same sort of argument allows us to establish Green’s Theorem for any finite union of nonoverlapping simple regions.

43. 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.

44. 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, θ) | 1 ≤r ≤ 2, 0 ≤ θ ≤π}

45. UNION OF SIMPLE REGIONS Example 4 • So, Green’s Theorem gives:

46. REGIONS WITH HOLES • Green’s Theorem can be extended to regions with holes—that is, regions that are not simply-connected.

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

48. 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.

49. REGIONS WITH HOLES • Let’s divide D into two regions D’ and D”by means of the lines shown here.

50. REGIONS WITH HOLES • Then, applying Green’s Theorem to each of D’ and D” , we get: • As the line integrals along the common boundary lines are in opposite directions, they cancel.