Let R be the region shown above bounded by the curve C C1 C
Solution
Green’s Theorem: M dx + N dy = Nx My dA. C R
Proof:
i) First we’ll work on a rectangle. Later we’ll use a lot of rectangles to y approximate an arbitrary o region. d ii) We’ll only do M dx ( N dy is similar). C C direct calculation the righ o By t hand side of Green’s Theorem b M d M dA = dy dx. c R y a c y x integral: M(x, d a Inner y)| c = M(x, d) + M(x, c) b M b Outer integral: M( c) R dA = x, y a M(x, d) dx. F o or the LHS we have M dx = M dx + M dx (since dx = 0 along the sides) C bottom top b a = b M(x, c) dx + M(x, d) dx = M(x, c) M(x, d) dx. a 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 approximated as closely as we want by a sum of rectangles, Green’s Theorem must hold on arbitrary regions.
