C 3 C 4. R k C 1. (x,y)

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1 Green s Theorem irculation Density (x,y + y) 3 (x+ x,y + y) 4 k 2 (x,y) 1 (x+ x,y) Suppose that F(x,y) M(x,y)i+N(x,y)j is the velocity field of a fluid flow in the plane and that the first partials of M and N are continuous over a region. Now let k be a small rectangle in (as shown above).

2 Suppose that we wish to approximate the circulation around the small rectangle k. If x and y are small, we expect the velocity field to be nearly constant on each of the four sides of A. For example, the flow along 1 is approximately F(x,y) i x M(x,y) x. Similarly, the flow along each of the other three sides is 2 : F(x+ x,y) j y N(x+ x,y) y 3 : F(x,y + y) ( i) x M(x,y + y) x 4 : F(x,y) ( j) y N(x,y) y Now putting the top and bottom sides together we have ( ) M(x,y) M(x,y + y) M(x,y) x M(x,y + y) x x y y M y x y For the left and right sides we have ( ) N(x+ x,y) N(x,y) N(x+ x,y) y N(x,y) y x y x N x x y

3 It follows that the circulation around the rectangle is approximately ( N (1) circulation x M ) x y y Dividing both sides of (1) by the area of k suggests the following definition. Definition. irculation Density at a Point in the Plane The circulation density or k-component of the curl of a vector field F(x,y) M i+n j at a point (x,y) is (2) (curlf) k N x M y emark. ecall that the curl of the vector field F is given by curlf F i j k x y z M N P ( P y N ) i z ( P x M ) ( N j+ z x M ) k y Notice that circulation density in the plane is a scalar.

4 Theorem 1. Green s Theorem (Tangential Form) Let be a piecewise-smooth, simple closed curve in the plane and let be the region bounded by (in the plane). Suppose also that M and N have continuous partial derivatives on an open region that contains. Then the counterclockwise circulation of the field F M i+n j around is equal to the double integral of the circulation density (k-component of the curl) over the region. ˆ (3) F Tds M dx+n dy (4) or, more conveniently, ( N x M ) dxdy y }{{} circulation density ( F) k }{{} circulation density dxdy

5 Proof. We shall prove a special case of the theorem. Suppose that is the boundary of the triangular region shown in the sketch. Let y g(x) d c (x a)+c and b a x g 1 (y) b a (y c)+c. d c d 3 2 Notice that these are equations for 3 and will be used below. c a 1 b

6 We start with the curl integral (HS of (4)). N x dxdy M y dydx ˆ d ˆ xb c xg 1 (y) N x dxdy ˆ b ˆ yg(x) a yc M y dydx or (5) ˆ d c ˆ d c + ( N(b,y) N ( g 1 (y),y )) dy ˆ b a (M (x,g(x)) M(x,c)) dx ( N(b,y) N ( g 1 (y),y )) dy ˆ b a (M(x,c) M (x,g(x))) dx

7 Now we consider the circulation integral (HS of (3)). It is easy to see that the circulation integral can be rewritten as ˆ ˆ ˆ ˆ M dx+n dy M dx+ N dy + M dx+ N dy Now ˆ M dx 1 ˆ N dy 2 ˆ M dx 3 ˆ N dy 3 ˆ b a ˆ d c ˆ a b ˆ c d M(x,c)dx N(b,y)dy M (x,g(x)) dx N ( g 1 (y),y ) dy ˆ b a ˆ d c M (x,g(x)) dx N ( g 1 (y),y ) dy Putting these together we have ˆ b M dx+n dy (M(x,c) M (x,g(x))) dx + a ˆ d c ( N(b,y) N ( g 1 (y),y )) dy which is the HS of (5).

8 emark. Now to prove the theorem for arbitrary triangles, we first consider the following. From this we could prove the theorem for arbitrary polygonal regions, etc.

9 Example 1. Let f(x,y) x 2 +xy. Verify Green s Theorem for the gradient field f (2x+y)i+xj over the region bounded by the circle : r(t) 2 costi+2 sintj, 0 t 2π So by Green s Theorem (2x+y)dx+xdy 0 M dx+n dy ( N x M ) y (1 1) dxdy dxdy As expected since gradient vector fields are conservative.

10 Example 2. Use Green s Theorem to find the counterclockwise circulation for F (x+e x siny)i+(x+e x cosy)j over the righthand loop of the lemniscate : r 2 cos2θ.

11 M x+e x siny, N x+e x cosy N x 1+ex cosy, M y ex cosy It follows that the circulation density is ( F) k N x M y 1+e x cosy e x cosy 1 So by Green s Theorem M dx+n dy... ( N x M ) y dxdy 1 dx dy (area of the lemniscate) 1 2

12 Example 3. Evaluate the circulation integral of the region (3ydx+2xdy) where is boundary 0 x π, 0 y sinx So let F 3yi+2xj. We sketch this vector field below

13 Now apply Green s Theorem. 3ydx+2xdy as we saw in section M dx+n dy ( N x M ) y ˆ π ˆ sinx 0 0 ˆ π 0 cosx 2 dxdy (2 3)dydx sinxdx π 0

14 Example 4. Applying Green s Theorem - Special esults (a) Let F yi. Let be the boundary of the region in the sketch below. Apply Green s Theorem to quickly find the circulation of F around. y g(x) a 4 b

15 Green s Theorem - Tangential Form ˆ F Tds M dx+n dy ydx ( N x M ) y ˆ b ˆ g(x) a 0 ˆ b a (0 1)dydx g(x) dx dxdy You can verify this result by actually computing the line integral using the parameterizations given below for 1-4. Notice that where 1 : r 1 (t) bi+g(b)tj 2 : r 2 (t) (b (b a)t)i+g(b (b a)t)j 3 : r 3 (t) ai+(1 t)g(a)j 4 : r 4 (t) (a+(b a)t)i, Here each parameterization is given for 0 t 1.

16 (b) Let F G(x)j where G is a differentiable function of x. Let be the boundary of a rectangle shown in the sketch below. Apply Green s Theorem to obtain a well known result from an introductory calculus course a 4 b The vector field is constant along vertical lines. In other words, if c, F(c,y) G(c)j, for all y First we manually compute the circulation of F G(x)j around the closed curve. Notice that F T 0 for the line segments 2 and 4 and that F is constant along 1 and 3. Finally, y 1 y 3 1. It follows that circulation (F(b,y) j) y +(F(a,y) ( j)) y (G(b)) y (G(a)) y G(b) G(a)

17 Now, N G(x) N x G (x) g(x) and M 0 M y 0. So by Green s Theorem G(b) G(a) ˆ F Tds M dx+n dy ˆ b ˆ 1 a ˆ b a ˆ b a ( N x M ) y 0 dxdy (G (x) 0)dydx G (x)dx g(x) dx So the Fundamental Theorem of alculus is a special case of (the tangential form) Green s Theorem. b In this case the definite integral a g(x)dx can be viewed as the circulation of F G(x)j around the closed curve (shown earlier) where G(x) is any antiderivative of g(x).

18 Applications Green s Theorem can be used to derive several useful formulas for area. Let be a simple closed curve in the plane enclosing a region. Then Area of dxdy (6) (1 ( 1)) dxdy ( (x) x ( y) ) y ydx+xdy dxdy The above formula can be used to explain how a planimeter works (see the brief discussion and references on page 1111 of the text).

19 (x j+1,y j+1 ) j+1 (x j,y j ) Figure 1: General Polygonal egion Example 5. Let be a polygon with vertices (x 0,y 0 ),(x 1,y 1 ),...(x n,y n ). A typical edge is shown in Figure 1. Use (6) to derive the formula (7) Area of 1 2 n 1 j0 x j y j+1 x j+1 y j So by (6), the area of the polygon is given by Area 1 xdy ydx 2 1 xdy ydx 2 n j1 j 1 ˆ xdy ydx+ 1 ˆ xdy ydx ˆ xdy ydx 2 n We compute the flow integral along an arbitrary side. So let j+1 : r(t) (x j (1 t)+x j+1 t) i+(y j (1 t)+y j+1 t) j, 0 t 1

20 Then x x j (1 t)+x j+1 t, dy (y j+1 y j ) dt y y j (1 t)+y j+1 t, dx (x j+1 x j ) dt It follows that 1 xdy ydx 1 2 j ˆ 1 0 ˆ 1 0 [(x j (1 t)+x j+1 t)(y j+1 y j ) (y j (1 t)+y j+1 t)( (x j y j+1 x j+1 y j ) dt 1 2 x jy j x j+1y j Thus as desired. Area ˆ xdy ydx+ 1 xdy ydx ˆ xdy ydx 2 n ˆ xdy ydx j+1 n 1 j0 n 1 j0 x j y j+1 x j+1 y j ˆ

21 As another application of (6), we derive the well-known formula for the area of an ellipse. ecall that the general equation of an ellipse centered at the origin is given by x 2 a 2 + y2 b 2 1 for some a,b > 0. It is easy to see that this ellipse can be parameterized by : r(t) acosti+bsintj, 0 t 2π Now let be the region enclosed by the ellipse. Then by (6) Area of ellipse 1 xdy ydx ab 2 ˆ 2π 0 ˆ 2π 0 ((acost)(bcost) (bsint)( asint)) dt dt πab How would you calculate the area using techniques from first semester calculus?

22 Example 6. Evaluate the integral equation ffi 4xydx+xdy. Here is defined by the vector (8) : r(t) 2sinti+3costj, 0 t 2π Let F 4xyi+xj. Then the given integral is a circulation integral. So let be the interior of the ellipse defined in (8). Then by (the tangential form of) Green s Theorem ffi 4xydx+xdy ( (x) x (4xy) ) y (1 4x) dxdy Notice that the ellipse has the rectangular equation dxdy (9) x y2 9 1 Thus (1 4x) dxdy ˆ 2 2 ˆ 2 2 (1 4x) ˆ 36 9x 2 /2 36 9x 2 /2 (1 4x) 36 9x 2 dx dydx

23 We try the trig substitution x 2sinθ. Then ˆ 2 2 (1 4x) 36 9x 2 dx ˆ π/2 π/2 ˆ π/2 π/2 ˆ π/2 π/2 (1 8sinθ)cos 2 θdθ (1+cos2θ)dθ 96 (1+cos2θ)dθ 0 ˆ π/2 π/2 6π We leave it as an exercise to evaluate the line integral directly. sinθ cos 2 θdθ

24 Figure 2: Parametric urve: r sin2t i+sintj Example 7. Find the area of the region whose boundary is given by the vector equation r(t) sin2t i+sint j, 0 t π. So by (6), this is Area ˆ π 0 ˆ π 0 ˆ 1 1 ydx+xdy 2sintcos2t+sin2tcostdt sint(2cos 2 t 1)+sintcos 2 tdt u 2 1du 4 3 The solution above is nice and short but it sure seems like overkill (using Green s Theorem) for a simple area calculation. an we use one of the definitions from sections 15.3 or 15.4 to compute the area of?

25 Example 8. edo the previous example by rewriting the given vector equation in polar form. Let s rewrite the parametric equations (10) x sin2t 2sint cost and y sint in polar form and exploit the formula A rdrdθ. Now and or r 2 x 2 +y 2 sin 2 t(4cos 2 t+1) (1 cos 2 t)(4cos 2 t+1) tanθ y x 1 2cost cost cotθ 2 Let θ 0 arctan(1/2). It follows that ( ) r(θ) 1 cot2 θ (1+cot 2 θ), 4 θ 0 θ π θ 0 Now Area rdrdθ 2 ˆ π/2 ˆ r(θ) θ 0 0 ˆ π θ0 ˆ r(θ) θ 0 0 rdrdθ rdrdθ

26 Area ˆ π/2 θ 0 ˆ π/2 θ 0 r 2 (θ)dθ ( ) 1 cot2 θ (1+cot 2 θ)dθ 4 ˆ π/2 θ cot2 θ 4 cot4 θ 4 dθ. 4 3 As we saw above. Here we have relied on the standard trig reductions formulas for the cotangent function. For example, ˆ ˆ cot 4 θdθ cot 2 θ(csc 2 θ 1)dθ ˆ ˆ cot 2 θcsc 2 θdθ cot 2 θdθ ˆ ˆ cot 2 θcsc 2 θdθ (csc 2 θ 1)dθ cot3 θ 3 +cotθ +θ + That s a lot of work for a simple area calculation. Perhaps working in rectangular coordinates would be easier.

27 Example 9. edo Example 7 by working in rectangular coordinates. This turns out to be easier than working in polar coordinates. Notice that x f(y) since by (10) we have x 2sint cost 2sint 1 sin 2 t 2y 1 y 2 Now the rest is easy. We have Area ˆ 1 ˆ 2y 1 y 2 0 ˆ 1 0 ˆ dxdy 2y 1 y 2 dy udu 4 3 u3/