f (x)dx = f(b) f(a). a b f (x)dx is the limit of sums

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1 Green s Theorem If f is funtion of one vrible x with derivtive f x) or df dx to the Fundmentl Theorem of lulus, nd [, b] is given intervl then, ording This is not trivil result, onsidering tht b b f x)dx = fb) f). f x)dx is the limit of sums n f x i ) [x i+ x i ]. The theorem sserts tht the vlue of this limit my be obtined by simply evluting f t the two "boundry" points b nd of the intervl [, b]. Green s theorem sys something similr bout funtions of two vribles. It sserts tht the integrl of ertin prtil derivtives over suitble region in the plne is equl to some line integrl long the "boundry" of. In order to stte it more preisely, it is neessry to introdue some new onepts. Definition A urve = {xt), yt), zt)) : t b} is sid to be simple if there re no numbers s nd r suh tht < r < s < b nd r) = s). Thus is simple if it does not ross itself.) i= A simple urve A urve tht is NOT simple Definition 2 A simple losed urve = {xt), yt)) : t b} is sid to be positively oriented if, s the prmeter t inreses from to b, one tres the urve in suh wy tht the region enlosed by is to one s left. It is negtively oriented if is to one s right. A positively oriented urve Nottion 3 The symbol A negtively oriented urve F dl is sometimes used to denote the line integrl of vetor field F over pieewisesmooth, simply onneted, positively oriented urve. Now we hve the tools to stte nd prove Green s theorem for funtion of two vribles.
2 Theorem 4 Let be pieewisesmooth, simple, positively oriented urve in the plne. Let be the region enlosed by. Suppose tht Mx, y) nd Nx, y) nd their first order prtil derivtives re ontinuous on some open set W ontining nd. Then Mx, y)dx + Nx, y)dy = Proof. se i): is retngle {x, y) : x b nd y d}, whose sides re prllel to the oordinte xes.,d) 3 b,d) 4 2,) b,) It is enlosed by urve = where is the line segment joining, ) to b, ), et. is the set of points {t, ) : t b} = {xt), yt)) : t b} where xt) = t nd yt) =. It follows tht b b Mx, y)dx = Mt, )x t)dt = Mt, )dt For onveniene, write b Mt, )dt s b Mx, )dx. Then Mx, y)dx = b Mx, )dx. 2 is the set {b, t) : t d} = {xt), yt)) : t d} where xt) = b nd yt) = t. Sine x t) =, Mx, y)dx = 2 We determine Mx, y)dx indiretly by determining 3 Mx, y)dx then use the ft tht 3 Mx, y)dx = 3 Mx, y)dx. 3 3 is the set {t, d) : t b} = {xt), yt)) : t b} where xt) = t nd yt) = d. Therefore b b Mx, y)dx = Mx, d)dx = Mt, d)x t)dt = Mt, d)dt 3 3 As we did bove, we write b Mt, d)dt s b Mx, d)dx, therefore 3 Mx, y)dx = 2 b Mx, d)dx
3 Finlly, 4 is the set {, + d t) : t d} = {xt), yt)) : t d} where xt) = nd yt) = +d t. Sine x t) =, 4 Mx, y)dx = Putting the seprte piees together gives Similr omputtions yield Mx, y)dx = Nx, y)dy = b d Mx, ) Mx, d)) dx Nb, y)dy N, y)) dy It follows tht d ) b ) Mx, y)dx + Nx, y)dy = Nb, y) N, y)) dy + Mx, ) Mx, d)) dx Turning to nd Therefore = N x = M = d b b d = N x N x dxdy = d M dydx = d M, we integrte by itertion to get d [Nx, y)] b dy = Nb, y) N, y)) dy b b [Mx, y)] d dx = Mx, ) Mx, d)) dx Nb, y) N, y)) dy + b ) Mx, ) Mx, d)) dx 2) A omprison of ) 2) revels tht Mx, y)dx + Nx, y)dy = se ii): is region bounded by finite number of stright line segments prllel to the oordinte xes, see the figure below). Suh region n be written s union of finite number of retngles, 2,..., n whih interset only in their ommon sides nd hve sides prllel to the oordinte xes. In the exmple below, is union of retngles, 2 nd 3.) 3 2 3
4 The line integrl of given funtion over the boundry of suh set is the sum of its integrl over the boundries of the retngles, 2,..., n. Also, the integrl of given funtion over is the sum of its integrls over the n retngles. It follows tht Mx, y)dx + Nx, y)dy = se iii): is n rbitrry region enlosed by pieewisesmooth, simple, positively oriented urve. This se my be hndled by showing tht is limit of urves onsisting of line segments prllel to the oordinte xes. This requires some hevy lifting tools we hve not introdued. Nottion 5 A urve tht enloses given region is lled the boundry of nd is denoted by. Therefore Green s theorem sttes tht if hs pieewisesmooth, simple, positively oriented boundry nd Mx, y), Nx, y) nd their first order prtil derivtives re ontinuous on then Mx, y)dx + Nx, y)dy =. Exmple 6 Let be the region in the first qudrnt enlosed by the line y = x nd the urve y = x, nd be the boundry of In question?? on pge??, you were sked to show tht question?? on pge??, you were sked to show tht Nx, y) = 2x 2. Then Mx, y)dx + Nx, y)dy = = 6 = 3xy 2 dx = 4 nd 2x 2 dy = 4, nd in 5 4x 6xy) = 6. Tke Mx, y) = 3xy2 nd 4x 6xy) d = Exmple 7 Let be the line segment from, ) to 6, ), 2 the line segment from 6, ) to 2, 4), 3 the line segment from 2, 4) to, ) nd = Sy you hve to evlute 2x 2 + y ) dx 4xydy. A diret evlution will involve lulting totl of six diff erent line integrls. However, use of Green s theorem involves lulting one double integrl. More 4
5 preisely, if is the region enlosed by then 2x 2 + y ) dx 4xydy = = y ) = 4 6 y 4y 2 7y 2 ) dy = 3 2 4y ) dxdy 2 y [ 4y 3 3 7y2 2 2y emrk 8 If we introdue vetor field Fx, y, z) = Mx, y)i + Nx, y)j + k then Green s theorem sttes tht F dl = F) k. To see this, simply evlute F) k. The result, sine N nd M re not funtions of z), is F) k = N z i + M z j + ) k k = emrk 9 Let Fx, y, z) = Mx, y)i+nx, y)j+k. Another form of Green s theorem onnets the integrl of Div F over nd the line integrl of some funtion of F. To derive it, we first rell the definition of tngent vetor to urve given by t) =< xt), yt) >. It is the vetor t) =< x t), y t) >. A unit tngent vetor is obtined by dividing the bove tngent vetor by its norm t) = Verify diretly, by tking dot produt), tht nt) given by nt) = t) < y t), x t) > is orthogonl to the unit tngent vetor. It is lled unit norml nd By formul??) on pge??), F ndl = When we pply Green s theorem to Fxt), yt)) n = Mxt), yt)) y t) t) Nxt), yt)) x t) t) Mdy Ndx) = Mxt), yt))y t) Nxt), yt))x t)) dt Mdy Ndx) Mdy Ndx) we get M x N )) = x ] 4 = 32 M x + N ) d = F x x t)) 2 + y t)) 2. This gives the third form of Green s theorem: F ndl = F Exerise. Let be the region enlosed by the irle entered t, ) with rduis. 5
6 ) Determine ydx nd 4xdy. b) Tke Mx, y) = y, Nx, y) = 4x nd verify tht Mdy + Ndx) =. 2. Let be the urve onsisting of the the sides of the tringle with verties t, ),, 3) nd 3, 2). Use Green s theorem to evlute x 2 y 2 dy + y 3 xdx. 3. Let be simple losed urve in the plne. Use Green s theorem to show tht the region enlosed by. ydx is the re of 6
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