Green’s Theorem Green’s Theorem We start with the ingredients for Green’s theorem. (i) C a simple closed curve (simple means it never intersects itself) (ii) R the interior of C. We also require that C must be positively oriented, that is, it must be traversed so its interior is on the left as you move in around the curve. Finally we require that C be piecewise smooth. This means it is a smooth curve with, possibly a ﬁnite number of corners. Here are some examples. y R C R R C C x Green’s Theorem With the above ingredients for a vector ﬁeld F = (M, N ) we have Nx − My dA. M dx + N dy = C R We call Nx − My the two dimensional curl and denote it curl F. We can write also Green’s theorem as F · dr = C curl F dA. R Example 1: (use the right hand side (RHS) to ﬁnd the left hand side (LHS)) Use Green’s Theorem to compute 3x2 y 2 dx + 2x2 (1 + xy) dy where C is the circle shown. I= y C C 6x2 y + 4x − 6x2 y dA = 4 By Green’s Theorem I = R We could compute this directly, but we know xcm x dA. R 1 = A x dA = a R x dA = πa3 ⇒ I = 4πa3 . ⇒ R Example 2: (Use the LHS to ﬁnd the RHS.) Use Green’s Theorem to ﬁnd the area under one arch of the cycloid x = a(θ − sin θ), y = a(1 − cos θ). a x The picture shows the curve C = C1 − C2 surrounding the area we want to ﬁnd.(Note the minus sign on C2 .) By Green’s Theorem, I ZZ −y dx = dA = area. C R Thus, I Z −y dx = area = C1 −C2 Z Z 0 · dx − C1 2π −y dx = C2 a2 (1 − cos θ)2 dθ = 3πa2 . 0 y 2a C2 πa C1 x 2πa