III.c Green s Theorem As mentioned repeatedly, if F is not a gradient field then F dr must be worked out from first principles by parameterizing the path, etc. If however is a simple closed path in the (x, y) plane, we can evaluate F dr in terms of a double integral by means of Green s Theorem. We first note the following: A path is a simple closed path if and only if the starting point the end point and does not intersect itself otherwise. For example, (a) Simple closed path: (b) Not a simple closed path: (c) Not a simple closed path: (d) the circle: { x cos θ y sin θ θ π is a simple closed path, but the circle: { x cos θ y sin θ 185 θ 4π
is not. We also need the following observation. Remark. xb xa [ yh(x) yg(x) f(x, y) ] xb [ dy dx f(x, y) xa b xa yh(x) yg(x) ] dx {f(x, h(x)) f(x, g(x))} dx. We can now state the Theorem. Green s Theorem. Let be a simple closed path and F M i + N j, as found in practice. Let region bounded by as shown, and suppose M/, N/ x exist on and inside. Then F dr { N x M } dx dy where means is traversed counterclockwise (also termed: positively oriented). Proof: We sketch a proof only for a special,. Suppose, are as shown. 186
y y h (x) x a 1 x b x y g (x) Note that 1 +. Now on 1 we have y g(x) for a x b. This can be easily parametrized as follows: { x x 1 y g(x) a x b. Here we denote t by x for convenience in what follows. In the same way, traversed backwards, i.e., is given by: { x x y h(x) a x b. So M dx M dx + M dx 1 M dx M dx 1 b xa b [M(x, g(x)) M(x, h(x))] dx xa [M(x, h(x)) M(x, g(x))] dx. By our Remark, this is b [ g(x) ] M (x, y) dy dx xa yh(x) 187
In the same way, we can show [ ] M da. N dy N x da and the result. Before giving examples, we look at some consequences. (1) Note that if N/ x M/, then F is a gradient field, and we just get the old result: F dr. Otherwise, the theorem gives another way to calculate F dr (by a double integral) and another way to calculate { N/ x M/} (by a path integral). Often there is no advantage to be had from this, but sometimes there is. See the examples. { N () Note F dr x M } dx dy. (3) Note also this: We know area of 1 da. Suppose we choose M, N so that N/ x M/ 1. learly there are many choices, so to keept it simple, we pick M and N x. Then Area 1 da x dy. So we get the area as a path integral. In the same way, if density is constant, Now choose M, N so that M x y da. N x M y. Again simple choices are N, M y /, but M and N xy would 188
also work. We then get M x y dx. M y, I x, I y and even M (if the density is simple) can be found by path integrals in this way. (4) Suppose is a region with holes H 1, H as shown. H 1 H We obtain a Green s Theorem for this case as follows: ut the region as shown, so that the holes are removed. 3 5 6 4 7 1 Then { N x M } dx dy F dr + F dr + F dr 1 3 + F dr + F dr + F dr + F dr. 4 5 6 7 Note that we have and 3 since, 3 are traversed clockwise. Furthermore, 189
5 6, 4 7 and so these integrals cancel. We conclude: { N x M } dx dy F dr F dr F dr. 1 3 We now pass to examples. Example 1. Verify Green s Theorem for F y i + x j if is the path consisting of the semicircle x + y 1, y and the straight line from ( 1, ) to (, 1). Answer. To verify Green s Theorem means to calculate both F dr and { N/ x M/} da and show they are identical. In this case, y -1 1 x { N x M } da [1 ( 1)] da area of π, while, we decompose into 1 and and parametrize each. Be careful to traverse these in the right direction. 1 19
Now { x 1 + t, t 1 1 y { x cos θ y sin θ θ π. Then 1 F dr F dr 1 t 1 t π θ [ M dx dt + N dy ] dt dt [ 1 + t] dt +, [( sin θ)( sin θ) + (cos θ)(cos θ)] dθ π 1 dθ π. So { N F dr + π π x M } da. Example. Use Green s Theorem to evaluate F dr if F [e x + tan 1 x 3 + y ] i + [sin 5 (y 3 ) + y 1 x] j and is the region determined by the curves x y, y x. Answer. Here we must evaluate F dr by actually working out { N/ x M/} dx dy. Note that here M, N are real dogs, but N/ x M/ 1 y which is simple. 191
y y x x x y So F dr 1 y 1 y xy xy ( 1 y) dx dy (y + y y 3 ) dy 1 y [ 1 + 1 3 1 ] (1 + y)(y y ) dy 1 3. Example 3. Use Green s Theorem to evaluate the area of the region bounded by the cardioid r 1 + sin θ. Answer. As before, we need to choose M, N so that N/ x M/ 1. So we choose N x and M. Then Area x dy. We need to parametrize r 1 + sin θ. r 1 + sinθ Now x r cos θ (1 + sin θ) cos θ y r sin θ (1 + sin θ) sin θ 19 θ π.
Thus Now A x dy [(1 + sin θ) cos θ][cos θ + sin θ cos θ] dθ [1 + sin θ][1 + sin θ](cos θ) dθ [1 + 3 sin θ + sin θ](cos θ) dθ (cos θ) dθ + cos θ dθ 1 [1 + cos θ] dθ π [ ] π sin θ cos θ dθ cos3 θ 3 sin θ cos θ dθ 1 4 3 sin θ cos θ dθ + sin θ cos θ dθ. (1 cos θ)(1 + cos θ) dθ 4 1 4 1 π π 4. (1 cos θ) dθ 1 4 [ 1 ] (1 + cos 4θ) dθ And so ( π ) A π + 3π 4. Just to check (by ouble Integrals): A 1 1 1+sin θ θ r r dr dθ (1 + sin θ) dθ 1 [ 1 + sin θ + ] 1 cos θ 193 [1 + sin θ + sin θ] dθ dθ 1 3 π 3π.
Example 4. Use Green s Theorem to calculate I x for the region between the circles x + y 1, x + y 4 if the density is constant. Answer. Since I x y da, we need M, N such that N/ x M/ y. So we choose N, M y 3 /3. Next, has a hole, y 1 x so y da 1 ) ( y3 dx 3 ) ( y3 dx. 3 Now while 1 ) ( y3 dx 1 3 3 { x cos θ 1 y sin θ 4 3 { x cos θ y sin θ y 3 3 dx 1 3 θ π, (8 sin 3 θ)( sin θ) dθ 16 3 ( 1 cos θ + 1 + cos 4θ θ π, (sin 3 θ)(sin θ) dθ 1 3 194 ) [ ] 1 cos θ dθ dθ 4 3 3 π 4π, sin 4 θ dθ.
We evaluated sin 4 θ dθ in the previous step, and got and so sin 4 θ dθ 4π 3 8 3π 4, I x 4π π 4 15π 4. Again, to practice we check I x by ouble Integrals. We have r r 1 I x y da θ 15 4 θ r1 (r sin θ) r dr dθ sin θ (4 1) dθ 4 1 cos θ dθ 15π 4. 195
Further Exercises: Verify Green s Theorem if: 1) F yi xj, R is the circle of radius 3 centered at (, ) and is its boundary. ) F yi xj, R is the disc between the circles of radius 1 and 3 and is its boundary. Find the area of a region R by means of path integrals if: 3) R is the ellipse x 4 + y 1. 4) R is the region bounded by r 1 sin θ. 5) R is the region in the upper 1/ plane bounded by y 1 x, x + y 4, y. 6) R is the region bounded by r + cos θ and r 1. Evaluate the following path integrals. 7) if (e x sin x y)dx + (tan 1 y + x)dy if is the path which bounds the region limited below by the parabola y x 1 and above by the parabola y 1 x. 8) (sin 1 (x ) + 3)dx + x dy if is the boundary of the square with vertices (, ), (1, ), (1, 1), (, 1). 9) (sin 1 (x ) + x y)dx + (cos 1 (y ) + (x 3 /3) + y)dy if is the boundary of the triangle with vertices (, ), (1, 1), (, 4). 1) Employ a path integral to find the x coordinate of the center of mass of the region within x + y y and x + y 4 if the density is ρ(x, y) 4 + x. 196