Suppose that D is a simple region (a region which is both x-simple and y-simple) and that F = P(x,y)i + Q(x,y)j where P(x,y) and Q(x,y) are each functions with continuous first partial derivatives. y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) We shall let C + represent the boundary of the region D traversed in the counterclockwise (positive) direction, that is, so that the region is on the left as the path is traversed. Observe that  Q — dA =  x D  Q — dx dy =  x c d 1(y)1(y) 2(y)2(y)

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) Observe that  Q — dA =  x D  Q — dx dy =  x c d 1(y)1(y) 2(y)2(y) c d Q(x,y) dy = x =  1 (y) 2(y)2(y) c d Q(  2 (y), y) – Q(  1 (y), y) dy = F = P(x,y)i + Q(x,y)j

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) c d Q(  2 (y), y) – Q(  1 (y), y) dy = F = P(x,y)i + Q(x,y)j c d Q(  2 (t), t) dt –Q(  1 (t), t) dt = c d

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) F = P(x,y)i + Q(x,y)j c d Q(  2 (t), t) dt –Q(  1 (t), t) dt = c d c d (Q(  2 (t), t)j) (  2  (t), 1) dt –(Q(  1 (t), t)j) (  1  (t), 1) dt = c d Note that this line integral is over the right half of C in the counterclockwise direction. Note that this line integral is over the left half of C in the clockwise direction.

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) F = P(x,y)i + Q(x,y)j c Q(  2 (t), t) dt –Q(  1 (t), t) dt = c c d (Q(  2 (t), t)j) (  2  (t), 1) dt –(Q(  1 (t), t)j) (  1  (t), 1) dt = c d C+C+ Q(x,y)j ds dd

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) F = P(x,y)i + Q(x,y)j We see then that  Q — dA =  x DC+C+ Q(x,y)j ds. Similarly, we find  P — dA =  y DC+C+ – P(x,y)i ds.

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) F = P(x,y)i + Q(x,y)j  Q  P — – — dA =  x  y DC+C+ (P(x,y)i + Q(x,y)j) ds =  Q — dA –  x D  P — dA =  y DC+C+ Q(x,y)j ds + C+C+ P(x,y)i ds. C+C+ F ds.

y x ab a  x  b  1 (x)  y   2 (x) y x c d c  y  d  1 (y)  x   2 (y) F = P(x,y)i + Q(x,y)j This proves Green’s Theorem (Theorem 1 on page 522) which can be stated as follows:  Q  P — – — dA =  x  y DC+C+ F ds = Note: Green’s Theorem can be extended to a region which is not a simple region but can be partitioned into several simple regions. C+C+ P dx + Q dy. Note that this is what we have called the scalar curl of F

ExampleLet F(x,y) = yi – xj and let C represent the circle of radius a centered at the origin and traversed counterclockwise. C F ds(a) Findusing the definition of a line integral. First, we parametrize C with (a cos t, a sin t) for 0  t  2 . C F ds = 2  0 F(c(t)) c  (t) dt = 22 0 (a sin t, – a cos t) (– a sin t, a cos t ) dt = – 2  a 2

C F ds (b) Find by using Green’s Theorem to write the line integral as a double integral, and evaluating this double integral. Since F(x,y) = yi – xj, then we let P(x,y) = and Q(x,y) = From Green’s Theorem, we have y– x– x C F ds =  Q  P — – — dx dy =  x  y D where D is the disk of radius a centered at the origin (– 1 – 1) dx dy = D – 2 dx dy = D D – 2(area of D) = – 2  a 2

By applying Green’s Theorem with the vector field F = P(x,y)i + Q(x,y)j with P(x,y) = – y /2 and Q(x,y) = x / 2, we can find the area of a region D in R 2 (as stated in Theorem 2 on page 524), because  Q  P — – — dA =  x  y D dx dy = D 1 1 — + — dA = 2 2 D (area of D). In other words, we have that (area of D) = C F ds where C is the boundary of D traversed in the counterclockwise direction, and F = (– y /2)i + (x / 2)j

ExampleConsider the area in the xy plane inside the graph of x 2/3 + y 2/3 = a 2/3 (where a > 0 is a known constant). (a) Sketch a graph of this area. x y (0, a) (0, – a) (– a, 0) (a, 0)

(b) Write a double integral using rectangular coordinates to obtain the area, and write a double integral using polar coordinates to obtain the area. Observe how difficult evaluating each of these integrals will be. – a a (a 2/3 – x 2/3 ) 3/2 dy dx – (a 2/3 – x 2/3 ) 3/2 0 22 a / (cos 2/3  + sin 2/3  ) 3/2 r dr d  0

(c) Use Green’s Theorem to obtain the area. First, we parametrize the curve which is the boundary of D by Then, by letting F = (– y /2)i + (x / 2)j, the area bounded by the curve is C(  ) = ( a cos 3 , a sin 3  ) for 0    2  C F ds = 22 0 (– a sin 3 , a cos 3  ) 1 — 2 (– 3a cos 2  sin , 3a sin 2  cos  ) d  = 22 0 – (a sin 3  )(– 3a cos 2  sin  ) + (a cos 3  )(3a sin 2  cos  ) d  = 1 — 2

22 0 sin 4  cos 2  + sin 2  cos 4  d  = 3 —a 2 2 22 0 sin 2  cos 2  d  = 3 —a 2 2 22 0 sin 2 2  ——— d  = 4 3 —a 2 2 22 0 1 – cos4  ———— d  = 8 3 —a 2 2 22 0  – (1/4)sin4  —————— = 8 3 —a —a  —= 8 3  — a 2 8 Recall the half-angle and double-angle formulas which give us the following trig identities: 1 sin  cos  =— sin(2  ) 2 1 sin 2  = — (1 – cos(2  )) 2

y x Suppose that D is a simple region (a region which is both x-simple and y-simple) and that the closed path along the border of D in the counterclockwise direction, sometimes denoted as  D, is parametrized by c(t) = (x(t), y(t)). Note: If (a, b) is a non-zero vector in R 2, then any vector which is orthogonal to, and to the right of, (a, b) must be a positive multiple of y x (a, b) (, )– ab (b, – a). At any point along the path, an “outward” normal vector must be one which is orthogonal to the velocity vector c / (t) = (x / (t), y / (t)). That is, an “outward” normal vector must be one which is a positive multiple of (x / (t), y / (t)) (y / (t), – x / (t)) (y / (t), – x / (t)). D

Suppose F = P(x,y)i + Q(x,y)j is a vector field defined on D. Recall Green’s Theorem (Theorem 1 on page 522):  Q  P — – — dA =  x  y D DD F ds Define the vector field V = – Q(x,y)i + P(x,y)j on D, and observe that V c / (t) = V (x / (t), y / (t)) = Applying Green’s Theorem with the vector field V, we have – Q(x,y) x / (t) + P(x,y) y / (t) = F (y / (t), – x / (t))  P  Q — + — dA =  x  y D DD V ds div(F) DD = F n ds where n is understood to be the outward normal to  D This proves the Divergence Theorem in the Plane (Theorem 4, page 527)