Green’s Theorem: Sketch of Proof o Green’s Theorem: Nx − My dA. M dx + N dy = C R Proof: i) First we’ll work on a rectangle. Later we’ll use a lot of rectangles to approximate an arbitrary region. o o ii) We’ll only do M dx ( N dy is similar). C − R bottom b = a M (x, c) − M (x, d) dx. a M dx (since dx = 0 along the sides) top a M (x, c) dx + a c b ∂M dA = ∂y Foor the LHS we have M dx + M dx = C y C By direct calculation the right hand side of Green’s Theorem b d ∂M ∂M − − dA = dy dx. ∂y ∂y R a c Inner integral: −M (x, y)|dc = −M (x, d) + M (x, c) Outer integral: d b M (x, c) − M (x, d) dx. M (x, d) dx = b a So, for a rectangle, we have proved Green’s Theorem by showing the two sides are the same. In lecture, Professor Auroux divided R into “vertically simple regions”. This proof instead approximates R by a collection of rectangles which are especially simple both vertically and horizontally. For line integrals, when adding two rectangles with a common edge the common edges are traversed in opposite directions so the sum is just the line integral over the outside boundary. Similarly when adding a lot of rectangles: everything cancels except the outside boundary. This extends Green’s Theorem on a rectangle to Green’s Theorem on a sum of rectangles. Since any region can be approxi mated as closely as we want by a sum of rectangles, Green’s Theorem must hold on arbitrary regions. = ≈ x b