Line Integrals. Partitioning the Curve. Estimating the Mass

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1 Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to construct model where we pretend curve does hve density nd mss, for exmple, when nlyzing wire. Although we re now looking curve in plne, tht is mere convenience; nlogs of our results will hold for curves in R 3. Prtitioning the urve If we wished to estimte the mss of the curve, we might prtition the curve into smll pieces nd estimte the mss of ech smll piece by multiplying its density times its length. Although the density my vry, it probbly won t vry much in smll piece of the curve, so we could use the density t ny point in the piece of the curve in our estimte. If the curve hs the prmetriztion x = x(t), y = y(t), t b, prtitioning the curve mounts to tking prtition P = {t 0, t 1, t 2,..., t n } of [, b]. Estimting the Mss For ech i, 1 i n, We could choose some t i [t i 1, t i ] nd use δ(x i, yi ), where x i = x(t i ), yi = y(t i ) s our estimte for the density. We hve previously estimted the rc length ( ) 2 ( ) 2 dy s i + t i, where the derivtives re evluted t t i. We could then estimte the mss m i of the i th piece of the curve by ( ) 2 ( ) 2 dy δ(x i, yi ) + t i, so s n estimte for the entire mss m we would get ( m = n i=1 m i ) 2 n i=1 δ(x i, yi ) + The Line Integrl 1 ( ) 2 dy t i.

2 ( ) 2 ( ) 2 n dy i=1 δ(x i, yi ) + t i is Riemnn Sum, which, if everything is continuous, will pproch the ordinry integrl ( ) 2 ( ) 2 b dy δ(x(t), y(t)) +. This integrl is clled line integrl nd is denoted by δ(x, y) ds. efinition of Line Integrl In generl, given curve with prmetriztion x = x(t), y = y(t), t b nd function f(x, y) defined on the curve, we define ( f(x, y) ds = ) 2 ( ) 2 b dy f(x(t), y(t)) +. In R 3, we would hve, with the nlogous sitution, ( f(x, y, z) ds = ) 2 b f(x(t), y(t), z(t)) + ( ) 2 dy + 2 ( ) 2 dz. Note: This generlizes in the obvious wy to higher dimensions nd there re other types of line integrls. Work Suppose force F(x) is exerted long curve : x(t), t b. Along short intervl of length s, the force won t chnge much nd the mount of work performed will pproximte the tngentil component of the force multiplied by the length of tht short piece of curve. The tngentil component of the force will equl F T, where T is the unit tngent. We thus conclude the totl mount of work my be clculted by the line integrl W = F T ds. Another Look t the Line Integrl Since T = nd ds =, when we evlute the line integrl we get

3 W = b F = b F, which we my wnt to write s F. Another Type of Line Integrl If we write F =< F x, F y > nd b (F x + F dy y ), or =<, dy >, reclling dy = nd = dy, we might write the integrl s F x + F y dy. This suggests nother type of line integrl, which we might write s P (x, y) + Q(x, y) dy. P (x, y) + Q(x, y) dy If we prtition curve, in ech piece of the curve from (x i 1, y i 1 ) to (x i, y i ) choose point (x i, yi ) nd clculte n i=1 P (x i, yi ) x i + Q(x i, yi ) y i, s limit we will get the line integrl P (x, y) + Q(x, y) dy. To clculte such line integrl, we tke prmetriztion nd clculte b P (x, y) + Q(x, y)dy. Sometimes, it s convenient to split the integrl into the sum P (x, y) + Q(x, y) dy nd use different prmetriztions for ech. Integrls Along Horizontl Line Segments onsider P (x, y) + Q(x, y) dy where is horizontl line segment y = k, x b. In this cse, we cn clculte P (x, y) +Q(x, y) dy = b P (x, k). To see this, use the prmetriztion x = t, y = k, t b, so = 1, dy = 0 nd P (x, y) +Q(x, y) dy = b dy P (x, y) +Q(x, y) = b P (t, k) 1 + Q(t, k) 0 = b P (t, k) = b P (x, k). Integrls Along Verticl Line Segments 3

4 onsider P (x, y) + Q(x, y) dy where is verticl line segment x = k, y b. In this cse, we cn clculte P (x, y) +Q(x, y) dy = b P (k, y) dy. We cn see this using the prmetriztion x = k, y = t, t b. Integrls Along urves y = f(x) onsider P (x, y) + Q(x, y) dy where is curve y = f(x), x b. In this cse, we my clculte P (x, y) + Q(x, y) dy = b P (x, y) + Q(x, y) dy, where we replce y by f(x). This is shown immeditely using the cnonicl prmetriztion x = t, y = f(t), t b, getting P (x, y) + Q(x, y) dy = b P (x, y) + Q(x, y)dy = b P (x, y) 1 + Q(x, y) dy = b P (t, f(t)) + Q(t, f(t))f (t) = b P (x, f(x)) + Q(x, f(x))f (x). Independence of Pth If, in some domin, 1 P (x, y) + Q(x, y) dy = 2 P (x, y) + Q(x, y) dy whenever 1 nd 2 re piecewise smooth curves with the sme initil nd terminl points, we sy P (x, y) + Q(x, y) dy is independent of pth in. Independence of Pth nd losed urves A closed curve is curve for which the initil nd terminl points coincide. A circle is the simplest exmple of closed curve. It is esily seen tht n integrl P + Q dy is independent of pth in some domin if nd only if P + Q dy = 0 for every closed curve in the domin. The only if prt is obvious. To prove the if prt, suppose 1 nd 2 re two curves in the domin with the sme initl nd terminl points nd let be the curve obtined by first trveling long 1 nd then trveling bckwrds long 2. Then P + Q dy = 0, but lso P + Q dy = 1 P + Q dy 2 P +Q dy, so 1 P +Q dy 2 P +Q dy = 0 nd 1 P + Q dy = 2 P + Q dy. Independence of Pth nd Potentil Functions 4

5 It cn esily be shown tht P (x, y) + Q(x, y) dy is independent of pth if there exists some potentil function φ(x, y) such tht φ =< P, Q >, i.e. such tht φ x = P, φ y = Q. Suppose such function exists nd goes from (x 1, y 1 ) to (x 2, y 2 ). Let x = f(t), y = g(t), t b be prmetriztion of. We then clculte: P (x, y) + Q(x, y) dy = b φ φ + x y dy = b dφ = [φ(f(t), g(t))] b = φ(f(b), g(b)) φ(f(), g()) = φ(x 2, y 2 ) φ(x 1, y 1 ). This obviously doesn t depend on. A Necessry ondition If φ φ = P nd x y derivtives we get 2 φ y x = P y nd 2 φ x y = Q x. = Q, when we clculte the second prtil If these second prtils re continuous, then they must be equl, so P y = Q x. This is obviously necessry condition for the existence of potentil function φ. It cn be shown this is lso sufficient condition for the integrl P + Q dy to be independent of pth. This will be consequence of Green s Theorem. Green s Theorem Theorem 1 (Green s Theorem). Under suitble hypothesis, given plne region with boundry oriented positively, P +Q dy = Q x P y da. If we hve closed curve bounding region nd Q x = P y, it follows tht P + Q dy = the integrl is independent of pth. Proof of Green s Theorem Q x P y da = 0 da = 0, so 5

6 We will verify Green s Theorem for some specil curves, but will not give generl proof. We will first ssume region cn be described s {(x, y) f(x) y g(x), x b. We cn then prmetrize the boundry by the union = , with 1 is the curve y = f(x), with x going from to b, 2 is the verticl line segment x = b, with y going from f(b) to g(b), 3 is the curve y = g(x), with x going from b to nd 4 is the verticl line segment x =, with y going from g() to f(). Hlf the Integrl Now consider P. We cn split the integrl into four prts. P = 1 P + 2 P + 3 P + 4 P. The portions long the verticl line segments re obviously 0, so we re left with P = 1 P + 3 P = b P (x, f(x)) + P (x, g(x)) = b b P (x, f(x)) b P (x, g(x)) = b P (x, f(x)) P (x, g(x)). But P (x, f(x)) P (x, g(x)) = f(x) P dy, so g(x) y P = b f(x) P g(x) y dy = b g(x) P f(x) y dy = P y da. More on Green s Theorem Essentilly, we hve shown tht if region cn be divided up into verticl strips going between two curves, then P (x, y) = P y da. A similr clcultion shows tht if the sme region cn be divided up into horizontl strips going between two curves, then Q(x, y) dy = Q x da. If the region cn be divided up both wys, we cn dd these integrls together to get P (x, y) + Q(x, y) dy = Q x P y da. Simple losed urves nd Independence of Pth A curve for which the initil nd terminl points coincide is clled closed curve. If those re the only points which coincide, the curve is clled simple closed curve. We will ssume ll our curves re piecewise smooth. 6

7 lim 1. In domin, P (x, y) + Q(x, y) dy is independent of pth if nd only if P (x, y) + Q(x, y) dy = 0 over ll simple closed curves. lerly, if, P (x, y) +Q(x, y) dy is independent of pth, then for ny closed curve P (x, y) + Q(x, y) dy = 0 since the vlue of the integrl would be the sme if we used degenerte curve. On the other hnd, suppose P (x, y) + Q(x, y) dy = 0 over every simple closed curve nd consider curves 1, 2 with the sme ini- til nd terminl points. Let = 1 ( 2 ). By 2 we men the curve trcing over the sme points s 2 but going in the opposite direction. is closed curve, so P (x, y) + Q(x, y) dy = 0. But P (x, y) + Q(x, y) dy = 1 2 P (x, y) + Q(x, y) dy = 1 P (x, y) + Q(x, y) dy 2 P (x, y) + Q(x, y) dy. It follows tht 1 P (x, y) + Q(x, y) dy = 2 P (x, y) + Q(x, y) dy. Green s Theorem nd Independence of Pth If Q x = P, then over ny simple closed curve, which we my ssume trverses counterclockwise round the boundry of plne region y, P (x, y) + Q(x, y) dy = Q x = P y da = 0 da = 0. It follows immeditely tht if Q x = P y in, then P (x, y) + Q(x, y) dy is independent of pth in. The Other irection Theorem 2. If P (x, y) + Q(x, y) dy is independent of pth in, then Q x = P in. y We cn prove this by finding potentil function φ(x, y) such tht φ =< P, Q >, i.e. such tht φ x = P, φ y = Q. To define φ(x, y), tke n rbitrry point (x 0, y 0 ) in nd let φ(x, y) = P (x, y) +Q(x, y) dy, where is ny smooth curve in going from (x 0, y 0 ) to (x, y). This is n unmbiguous definition, since the integrl is independent of pth. If we tke curve where the lst prt is horizontl line segment, we x+h φ(x, y) φ(x + h, y) φ(x, y) P (t, y) x get = lim h 0 = lim h 0 x h h 7

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