We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.


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1 Mth Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn be nturlly pplied to function of n vribles. Assume f(x, y) is defined on smooth curve of finite length, prmeterized in terms of rclength s r(s) = x(s), y(s) for t b. For the purposes of introducing this concept, we ssume f(x, y). We will eventully mention wht the difference is if this ssumption is tken wy. We prtition into n smll rcs by forming prtition of [, b] by picking s i s follows: = s < s < < s n = b. Let s i for i =, 2,..., n denote the width of [s i, s i ]. Pick smple points s i from [s i, s i ] for i =, 2,..., n. Then we will pproximte the re of the curtin under the curve x(s), y(s), f(x(s), y(s)) (over the curve the xyplne see the picture t the top of the next pge) by the Riemnn sum n A f(x(s k), y(s k)) s k. k= Geometriclly ech summnd is the re of curvy pnel (to visulize them, tke rectngulr piece of pper nd bend it). Let (n) be the mximum vlue mong the s i for i =, 2,..., n. If the limit of the Riemnn sums s n exists for ny sequence of prtitions chosen such tht (n) s n then this limit is clled the pth integrl of f over (k sclr line integrl), nd it converges to the re of the curtin. See the picture t the top of the next pge. Nottion: When the limit exists, we sy f is integrble on the smooth curve of finite b length nd write f ds = f(r(s)) ds (where r(s) is n rclength prmetriztion of ). IMPORTANT: If we remove the ssumption tht f be nonnegtive, the pth integrl gives the net re (re bove the xyplne minus re below the xyplne). This construction nd these definitions nturlly extend to sclrvlued function f on subset S of R n contining smooth curve of finite length.
2 We will compute the re of the curtin bove, but for now we strt with more simplistic exmple: A thin wire in plne represented by smooth curve with density function ρ(x, y) (units of mss per length) hs Mss M = ρ ds. Suppose thin wire occupies the curve y = x 2 nd hs density ρ(x, y) = y g/cm (ssume x, y crry units of cm). Let s find the mss of this wire! An rclength prmetriztion of the upper hlf of the unit circle is r(s) = cos (s), sin (s) where s π. So the pth integrl is π M = ρ ds = ρ(cos (s), sin (s)) ds = π ρ(cos (s), sin (s)) ds = π sin (s) ds = 2 g.
3 Wht do we do if we hve n rbitrry prmetriztion of smooth curve of finite length? We could try to find n rclength prmetriztion, but often tht cretes n extremely ugly prmetriztion. Insted we will develop formul tht is independent of the prmetriztion! Let smooth curve of finite length nd the vectorvlued function r(t) for t b be n rbitrry differentible prmetriztion (moving long the curve in one direction tht is, not bck nd forth). Recll the rclength function below computes the rclength long from r() to r(t): s(t) = t r (u) du. Differentiting (nd using the Fundmentl Theorem of lculus) gives s (t) = r (t). Then we hve tht ds = s (t)dt = r (t) dt. Then if f is integrble, f ds = b f(r(t)) r (t) dt. So we cn compute the pth integrl with ny prmetriztion! Nturlly we cll r (t) the speed fctor. Let s find the re of the curtin t the top of the lst pge. Let f(x, y) = 3 2 cos (x). Let be given by r(t) = cos (t/3), sin (t/3) for t 2π. So we find f ds: f ds = 2π 3 2 cos (cos (t/3)) sin (t/3), cos (t/3) 3 dt = 2 2π t 3 dt = 2 (2π)2 = π2 3. Note: This mtches the simple solution which is to relize tht this curtin is just bent tringle of bse length 2π 3 nd height π.
4 Everything we hve done here generlizes for function of n vribles: Let f be n integrble sclrvlued function on some subset S of R n contining smooth curve of finite length prmeterized by the differentible vectorvlued function r(t) = x (t), x 2 (t),..., x n (t) (moving long the curve in one direction) for t b. Then f ds = b f(r(t)) r (t) dt = b f(x (t), x 2 (t),..., x n (t)) (x (t)) 2 + (x 2(t)) (x n(t)) 2 dt. Let s clculte the pth integrl for t π 2. z 2 ds = π 2 z 2 ds for the helicl pth given by r(t) = 4 cos (t), 4 sin (t), 3t (3t) 2 π ( ( 4 sin (t)) 2 + (4 cos (t)) 2 + (3) 2 2 π ) 3 dt = 45 t 2 5π 3 dt = 5 = 2 8. Let f be function on subset of R n contining smooth curve of finite length. The verge vlue of f on is given by f ds where l() is the length of. l() Find the verge vlue of f(w, x, y, z) = w 2 + x + 2y + 3z on the line segment joining (,,, 2) to (, 2, 2, ). Let s find prmetriztion of the line segment: r(t) = +t, +t, +t, 2 t for t. The length of this curve is l = = 2. This is lso the speed fctor! f ds = ( t 2 + ( + t) + 2( + t) + 3(2 t) ) (2) dt = 2 So the verge vlue of f on the line segment is ( ) t dt = = 56 3.
5 Now we introduce the concept of line integrl of vector field. First we strt with some bckground. An oriented curve is prmeterized curve for which direction is specified. The positive orienttion is the direction the curve is generted s the prmeter increses; the negtive orienttion is direction s the prmeter decreses. For instnce, the positive orienttion of r(t) = cos (t), sin (t) is counterclockwise. Let F be continuous vector field on some subset of R n tht contin smooth curve of finite length with n rclength prmetriztion r(s). Let T be the unit tngent vector t ech point of (with the positive orienttion given by the prmetriztion). The line integrl of F over curve (with given orienttion) is F T ds. IMPORTANT: F T is the component of F in the direction of T. Hence, geometriclly, the integrl is dding up the components of the vector field tht re tngentil to the curve. As we will see this hs mny pplictions. We immeditely seek wy to evlute line integrl tht does not require n rclength prmetriztion: Let F be vector field on some subset of R n tht contins smooth curve prmeterized by differentible vectorvlued function r(t) for t b such tht r (t) for ll t [, b]. Then T = r r. Then mking use of the fct tht ds = r dt we get b r b F T ds = F r r dt = F r dt. There re other wys to write this: F dr where dr = r dt. f dx + f 2 dx 2 + f n dx n where F = f, f 2,..., f n nd r(t) = x (t), x 2 (t),..., x n (t). Here is the connection: dr = x (t), x 2(t),..., x n(t) dt = dx, dx 2,..., dx n
6 Let s work through some exmples: Let s evlute the line integrl of F = x y, x on the prbol y = x 2 from (, ) to (, ): We cn use the grph prmetriztion r(t) = t, t 2 (which hs the right orienttion). lling the prbol we get F dr = t t 2, t, 2t dt = t + t 2 dt = 5 6. This yields mesure of the strength of the field long the curve in the direction of the curve. Here is picture:
7 Work: If F is vector field in region D of R 3 representing the force (s vector) t points inside D then the work done in moving n object long curve in some direction is the line integrl of F over (using suitbly oriented prmetriztion). An electric chrge t (,, ) produces force field on other point chrges given by k F = x, y, z where k is constnt. Let s clculte the work done in moving (x 2 + y 2 + z 2 ) 3/2 point chrge from (,, ) to (,, ): Using r(t) = t, t, t for t, we get W = k t, t, t,, dt = (3t 2 ) 3/2 k(3t) 3 3t 3 dt = k 3 t dt = k ( ) 2 3 ircultion: Let be closed (ends where it begins) smooth oriented curve in R n prmeterized once round by r(t) for t b (consistent with the orienttion) in region D where continuous vector field F is defined. The circultion of F on is the line integrl F dr. Let s clculte the circultion of F = using counterclockwise orienttion: x2 + y 2 x, y on the circle (x 2)2 + (y 2) 2 = 4 Let s lbel the circle by. First we prmeterize the once counterclockwise with r(t) = cos (t), sin (t) for t 2π. Then we integrte: F dr = = 2 = 2 2π 2π 2π cos (t), sin (t) 2 sin (t), 2 cos (t) dt (2 + 2 cos (t))2 + (2 + 2 sin (t)) 2 cos (t) sin (t) cos (t) + 2 sin (t) dt cos (t) sin (t) cos (t) + 2 sin (t) dt = cos (t) + 2 sin (t) =. 2π The circultion is zero becuse s much of the vector field points long the pth s ginst the pth when trversing. See the picture on the next pge!
8 Think of it like the vector field is the wind nd you wlk round. In this cse, s much is t your bck s t your fce... Let F = x 2, 2xy + x, z nd let be the circle x 2 + y 2 = oriented counterclockwise in the plne z =. Let s find the circultion of F on : Let s use the prmetriztion r(t) = cos (t), sin (t),. Then F dr = = = = 2π 2π 2π 2π cos 2 (t), 2 cos (t) sin (t) + cos (t), sin (t), cos (t), dt sin (t) cos 2 (t) + 2 cos 2 (t) sin (t) + cos 2 (t) dt cos 2 (t) sin (t) + cos 2 (t) dt cos 2 (t) sin (t) + ( + cos (2t)) dt 2 (t + 2 ) 2π sin (2t) = 3 cos3 (t) + 2 = π. Note: ircultion my be computed on closed curve mde up of piecewise smooth curves by just computing the line integrl on ech curve nd dding up the results.
9 Flux: The line integrl of vector field F dds up the components of F long smooth curve of finite length. Wht if we were to do the sme thing, but with the norml components? Tht is clled the flux of F cross. We will only focus on how this is developed for vector field in plne: To come up with the flux integrl in plne, we first hve to determine wy to find the unit norml vector to smooth curve of finite length. Let r(t) = x(t), y(t) for t b be differentible prmetriztion of (with some fixed orienttion) such tht r for ll t b. Recll: T = r r is the unit tngent vector to in the direction of the prmetriztion. Second we embed the xyplne into R 3, where the vector k =,, is orthogonl to the xyplne. Then unit vector orthogonl to in the xyplne would be orthogonl to both T nd k. So we set N = T k (the cross product of two orthogonl unit vectors gives unit vector orthogonl to both inputs) nd drop its third component, which is, so tht it ultimtely lives in R 2. So the integrl for the flux of F cross is F N ds where s is rclength. IMPORTANT: F N is the norml component of F to the curve. In the cse where is smooth closed curve using counterclockwise orienttion, N points outwrd to, nd thus, wherever F points outwrd cross there is positive contribution to the flux integrl, while wherever F point inwrd cross there is negtive contribution to the flux integrl. Our first gol is to find nicer formul for flux: Embedded in R 3, T = (x (t)) 2 +(y (t)) 2 x (t), y (t),. Then N = T k = (x (t)) 2 +(y (t)) 2 y (t), x (t),. After dropping the third component, N = (x (t)) 2 + (y (t)) 2 y (t), x (t). Then with F = f, g y (t), x (t) F N ds = F (x (t)) 2 + (y (t)) 2 dt = (x (t)) 2 + (y (t)) 2 fdy gdx.
10 Let s find nd simplify the flux integrl of F = x, y on the circle x2 + y2 (x 2) 2 + (y 2) 2 = 4 using counterclockwise orienttion: Let s lbel the circle by. First we prmeterize the once counterclockwise with r(t) = cos (t), sin (t) for t 2π. Then we integrte: F N ds = = 2 2π 2π 4.8. (2 + 2 cos (t))(2 cos (t)) (2 + 2 sin (t))( 2 sin (t)) (2 + 2 cos (t))2 + (2 + 2 sin (t)) 2 + cos (t) + sin (t) cos (t) + 2 sin (t) dt Note: The integrl ws pproximted numericlly with Mthemtic. Anywy, the positive result mens tht more of the vector field flows out of the region thn into the region (just tke look t the picture couple pges bck). dt Let s clculte the upwrd flux of the rottionl vector field F = y, x cross the curve given by y = x for x 4. While we could use the grph prmetriztion, it turns out to be nicer to use r(t) = t 2, t for t 2. Then r, 2t, 2t (t) = 2t,, which mkes for norml N = ± = ± r (t) 4t2 +. For n upwrd norml we choose the minus sign. Then F N ds = 2 t, t 2, 2t dt = 2 t + 2t 3 dt =. See the picture below!
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