18.02A Topic 39: Green’s theorem.Author: Jeremy OrloffRead: TB: 21.3 to middle p. 768Ingredients: C a simple closed curve (i.e. no self-intersection), R the interior of C.C must be positively oriented (traversed so interior is on left) and piecewise smooth(a few corners are okay).xyR CR CR CGreen’s Theorem:ICM dx + N dy =ZZRNx− MydA.If F = hM, Ni then we call (Nx−My) k = curlF. (The reason for the name is given inthe reading –section V.4.) We can write Green’s theorem asICF · dr =ZZRcurlF · k dA.(In the notes, they drop the k from the curl. This is confusing since when we moveto 3D integrals the curl will have to be a vector.)Example 1: (use the RHS to find the LHS) Use Green’s Theorem to computeI =IC3x2y2dx + 2x2(1 + xy) dy where C is the circle shown.By Green’s Theorem I =ZZR6x2y + 4x − 6x2y dA = 4ZZRx dA.We could compute this directly, but we know xcm=1AZZRx dA = a⇒ZZRx dA = πa3⇒ I = 4πa3.xyCaPath for example 1Example 2: (Use the LHS to find the RHS.)Use Green’s Theorem to find the area under one arch of the cycloidx = a(θ − sin θ), y = a(1 − cos θ).The picture shows the curve C = C1+ C2surrounding the area.By Green’s Theorem,IC−y dx =ZZRdA = area.⇒ area =IC1+C2−y dx =ZC10 · dx +ZC2−y dx=Z2π0a2(1 − cos θ)2dθ = 3πa2.xyC2C12aπa2πaPath for example 2118.02A topic 39 2Other ways to compute area:ZZRdA =IC−y dx =ICx dy =12IC−y dx + x dy.’Proof’ of Green’s Theorem (also see the reading)i) First we’ll work on a rectangle. Later we’ll use a lot of rectangles toapproximate an arbitrary region.ii) We’ll only doICM dx (ICN dy is similar).By direct calculation the right hand side of Green’s TheoremZZR−∂M∂ydA =ZbaZdc−∂M∂ydy dx.Inner integral: −M(x, y)|dc= −M(x, d) + M(x, c)Outer integral:ZZR−∂M∂ydA =ZbaM(x, c) − M (x, d) dx.For the LHS we haveICM dx =ZbottomM dx +ZtopM dx (since dx = 0 along the sides)=ZbaM(x, c) dx +ZabM(x, d) dx =ZbaM(x, c) − M (x, d) dx.So, for a rectangle, we have proved Green’s Thm. by showing the two sides are the same.xyabcdFor line integrals when adding two rectangles with a commonedge the common edges are traversed in opposite directions sothe sum is just the line integral over the outside boundary. Sim-ilarly when adding a lot of rectangles: everything cancels exceptthe outside boundary. This extends Green’s Theorem on a rect-angle to Green’s Theorem on a sum of rectangles. Since anyregion can be approximated as close as we want by a sum ofrectangles, Green’s Theorem must hold on arbitrary regions.=≈End of topic 39
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