Suppose that D is a simple region a

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 y a x b 1(x) y 2(x) a b x d c c y d 1(y) x 2(y) x 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. d 2(y) Q Q Observe that — d. A = — dx dy = x x D c 1(y)

F = P(x, y)i + Q(x, y)j y a x b d 1(x) y 2(x) y c b x a c y d 1(y) x 2(y) Q — dx dy x 1(y) d Observe that D d dy x = 1(y) = c = d 2(y) Q(x, y) c Q — d. A x = Q( 2(y) , y) – Q( 1(y) , y) c dy =

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

F = P(x, y)i + Q(x, y)j y a x b d 1(x) y 2(x) y a c b x d d Q( 2(t) , t) dt – Q( 1(t) , t) dt = c c y d 1(y) x 2(y) x c d d (Q( 2(t) , t)j) • ( 2 (t) , 1) dt – (Q( 1(t) , t)j) • ( 1 (t) , 1) dt c c 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. =

F = P(x, y)i + Q(x, y)j y a x b d 1(x) y 2(x) y a b x d d Q( 2(t) , t) dt – Q( 1(t) , t) dt = c c y d 1(y) x 2(y) x c d d (Q( 2(t) , t)j) • ( 2 (t) , 1) dt – (Q( 1(t) , t)j) • ( 1 (t) , 1) dt = c Q(x, y)j • ds C+

F = P(x, y)i + Q(x, y)j y a x b d 1(x) y 2(x) y a c b x We see then that D Similarly, we find D Q — d. A x = P — d. A y = – c y d 1(y) x 2(y) x Q(x, y)j • ds. C+ P(x, y)i • ds. C+

F = P(x, y)i + Q(x, y)j y a x b d 1(x) y 2(x) y a D D Q P — – — x y 1(y) x 2(y) c b x Q — d. A – x c y d D P — d. A y d. A = = x Q(x, y)j • ds + C+ P(x, y)i • ds. C+ (P(x, y)i + Q(x, y)j) • ds = C+ F • ds. C+

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

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

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

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 D Q P — – — x y 1 1 — + — 2 2 d. A = D In other words, we have that (area of D) = dx dy = d. A = D F • ds (area of D). where C C is the boundary of D traversed in the counterclockwise direction, and F = (– y /2)i + (x / 2)j

Example Consider 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. y (0 , a) (a , 0) x (– a , 0) (0 , – a)

(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 2/3 – x 2/3)3/2 2 a / (cos 2/3 + sin 2/3 )3/2 dy dx –a – (a 2/3 – x 2/3)3/2 r dr d 0 0

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

2 3 2 —a 2 2 3 sin 4 cos 2 + sin 2 cos 4 d = —a 2 sin 2 cos 2 d = 2 0 2 2 – (1/4)sin 4 3 2 sin 22 3 2 1 – cos 4 3 —a ——— d = —a ———— d = —a 2 —————— = 2 2 4 2 8 8 0 0 0 Recall the half-angle and double-angle formulas which give us the following 3 2 2 3 2 —a — —a = trig identities: 2 8 1 8 sin cos = — sin(2 ) 2 sin 2 1 = — (1 – cos(2 )) 2

Note: If (a, b) is a non-zero vector in R 2, then any vector which is orthogonal to, (a, b) and to the right of, (a, b) must be a positive multiple of (b, – a). y ( b , – a) 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)). At any point along the path, an (y /(t) , – x /(t)) y (x /(t) , y /(t)) “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 D which is a positive multiple of x (y /(t) , – x /(t)).

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): D Q P — – — x y d. A = F • ds D 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)) = – Q(x, y) x /(t) + P(x, y) y /(t) = F • (y /(t) , – x /(t)) Applying Green’s Theorem with the vector field V, we have D P Q — + — x y d. A = V • ds = D F • n ds D where n is understood to be the outward normal to D div(F) This proves the Divergence Theorem in the Plane (Theorem 4, page 527)