University of. d Class. 3 st Lecture. 2 nd

Transcription

1 University of Technology Electromechnicl Deprtment Energy Brnch Advnced Mthemtics Line Integrl nd d lss st Lecture

2 nd Advnce Mthemtic Line Integrl lss Electromechnicl Engineer y Dr.Eng.Muhmmd.A.R.Yss Dr.Eng Dr.Eng Muhmmd.A.R.yss Pge of Line Integrl's <<->>

3 Multiple Integrl Green's Theorm Stoke's Theorm Dr.Eng Dr.Eng Muhmmd.A.R.yss Pge of Line Integrl's <<->>

4 Line Integrl (with Respect To rc Length) The line integrl of f (, ) xy long is denoted y, f (, ) xy ds where dx dy ds = + dt dt dt dx dy f ( xy, ) ds = f ( ht (), g() t ) + dt dt dt If we use the vector form of the prmeteriztion we cn simplify the nottion up somewht y noticin tht where r ( t) Exmple Evlute counter clockwise direction. is the mgnitude or norm of ( ) forright hlf of the circl nd dx dy + = r () t dt dt r t. Using this nottion the line integrl ecomes, f xy ds f ht g t r t dt (, ) = ( (), ()) () xy ds where is the right hlf of the circle, so r = s we know tht x=rcost x= cost y = sin t e nd y=r sint so π π t dx dy = sint = cost dt dt + y = 6 rotted in the Dr.Eng Muhmmd.A.R.yss ds= 6sin t+ 6cos tdt = dt. x -. The line integrl is then, ( ) ( ) π xy ds= cost sint π dt π = 96 cost sin = π 96 sin 5 5 t π π 89 = Pge of 5 Line Integrl's <<->> tdt

5 Exmple Evlute x ds where is the curve shown elow. So, first we need to prmeterize ech of the curves. : x= t, y =, t Now let s do the line integrl over ech of these curves. x t y t t : =, =, : x=, y = t, t () ( ) x ds = t + dt = t dt = t = 6 x ds = t + t dt () ( ) = t + 9t dt Dr.Eng Muhmmd.A.R.yss = ( + 9t ) = = () ( ) () x ds = + dt = dt = 8 Finlly, the line integrl tht we were sked to compute is, x ds= x ds+ x ds+ x ds = = 5.7 Pge 5 of Line Integrl's <<->>

6 Exmple Evlute x ds were is the line segment from (, ) to ( ),. From the prmeteriztion formuls t the strt of this section we know tht the line segment strt t (, ) nd ending t (, ) is given y, r t = t, + t, ( ) ( ) = +, t + t for t. This mens tht the individul prmetric equtions re, x= + t y = + t Using this pth the line integrl is, Exmple Evlute ( ) = x ds t dt ( )( t ) = + 5 = = 5 =. x ds were is the line segment from ( ), to (, ). This one isn t much different, work wise, from the previous exmple. Here is the prmeteriztion of the curve. r t = t, + t, ( ) ( ) =, t t for t. Rememer tht we re switch the direction of the curve nd this will lso chnge the prmeteriztion so we cn mke sure tht we strt/end t the proper point. Here is the line integrl. ( ) = 9+ 9 x ds t dt Dr.Eng Muhmmd.A.R.yss ( )( t) = + 5 = = 5 = We then hve the following fct out line integrls with respect to rc length. Pge 6 of Line Integrl's <<->>

7 Fct (, ) = (, ) f xy ds f xy ds So, for line integrl with respect to rc length we cn chnge the direction of the curve nd not chnge the vlue of the integrl. This is useful fct to rememer s some line integrls will e esier in one direction thn the other. Exmple 5 Evlute xds for ech of the following curves. () () (c) : = y x x : The line segment from (,) to ( ) : The line segment from (, ) to (,),.. () : y = x, x Here is prmeteriztion for this curve. : x= t, y = t, t Here is the line integrl. Dr.Eng Muhmmd.A.R.yss ( ) xds = t + t dt = + t = () : The line segment from (,) to ( ) ( ) ( ),. : r t = t, + t, = t, for t.., : x= t, y =, t Pge 7 of Line Integrl's <<->>

8 This will e much esier prmeteriztion to use so we will use this. Here is the line integrl for this curve. xds = t + dt = t = (c) : The line segment from (, ) to (,). since we know tht =. The fct tells us tht this line integrl should e the sme s the second prt (i.e. zero). However, let s verify tht, Here is the prmeteriztion for this curve. : r t = t, + t, for t. Here is the line integrl for this curve. ( ) ( ) =, t ( ) ( ) xds = t + dt = t t = Sure enough we got the sme nswer s the second prt. Let s suppose tht the three-dimensionl curve is given y the prmeteriztion, x= xt, y = y t z = z t t then the line integrl is given y, ( ) ( ) ( ) dx dy dz f ( xyz,, ) ds = f ( xt (), y() t, z() t ) + + dt dt dt dt Note tht often when deling with three-dimensionl spce the prmeteriztion will e given s vector function. r t = xt, y t, z t ( ) ( ) ( ) ( ) Notice tht we chnged up the nottion for the prmeteriztion little. Since we rrely use the function nmes we simply kept the x, y, nd z nd dded on the ( t ) prt to denote tht they my e functions of the prmeter. Dr.Eng Muhmmd.A.R.yss Also notice tht, s with two-dimensionl curves, we hve, dx dy dz + + = r () t dt dt dt nd the line integrl cn gin e written s, f xyz ds f xt y t z t r t dt (,, ) = ( (), (), ()) () So, outside of the ddition of third prmetric eqution line integrls in three-dimensionl spce work the sme s those in two-dimensionl spce. Let s work quick exmple. Pge 8 of Line Integrl's <<->>

9 Exmple 6 Evlute t π. xyzds where is the helix given y, r( t) = cos ( t),sin( t),t, Note tht we first sw the vector eqution for helix ck in the Vector Functions section. Her is quick sketch of the helix. () () π cos sin sin cos 9 xyzds = t t t t+ t+ dt π = t sin( t) + 9dt = π ( ) tsin t dt t = sin cos = π ( t) ( t) Dr.Eng Muhmmd.A.R.yss π Pge 9 of Line Integrl's <<->>

10 Line Integrl-Prt II (with respect to x nd/or y) As with the lst section we will strt with two-dimensionl curve with prmeteriztion, x= xt y = y t t ( ) ( ) The line integrl of f with respect to x is, (, ) = ( (), ()) () The line integrl of f with respect to y is, Also f xy dx f x t y t x t dt (, ) = ( (), ()) () f xy dy f xt y t y t dt (, ) (, ) Pdx+ Qdy = P xy dx+ Q xy dy Exmple Evlute sin ( π y) dy+ yx dx where is the line segment from (, ) to ( ) Here is the prmeteriztion of the curve. r t = t, + t, = t,+ t t ( ) ( ) The line integrl is, sin( π ) + = sin ( π ) + y dy yx dx y dy yx dx = cos( π + πt) + t + t π 7 = 6 ( π ( ))( ) ( )( ) () = sin + t dt+ + t t dt Dr.Eng Muhmmd.A.R.yss,. Exmple Evlute sin ( π y) dy+ yx dx where is the line segment from (, ) to ( ) So, we simply chnged the direction of the curve. Here is the new prmeteriztion. r t = t, + t, = t, t t ( ) ( ) The line integrl in this cse is, sin( π ) + = sin ( π ) + y dy yx dx y dy yx dx ( π ( ))( ) ( )( ) ( ) cos 5,. = sin t dt+ t t dt 8 = ( π πt) t + t t + t π Pge of 7 Line Integrl's <<->> = 6

11 Fct If is ny curve then, f xy, dx = f xy, dx nd f xy, dy = f xy, dy ( ) ( ) ( ) ( ) With the comined form of these two integrls we get, Pdx+ Qdy = Pdx+ Qdy We cn lso do these integrls over three-dimensionl curves s well. In this cse we will pick up third integrl (with respect to z) nd the three integrls will e. (,, ) = ( (), (), ()) () f xyz dx f x t y t z t x t dt (,, ) = ( (), (), ()) () f xyz dy f xt y t z t y t dt (,, ) = ( (), (), ()) () f xyz dz f xt y t z t z t dt where the curve is prmeterized y x= xt y = y t z = z t t ( ) ( ) ( ) As with the two-dimensionl version these three will often occur together so the shorthnd we ll e using here is, (,, ) (,, ) (,, ) Pdx+ Qdy+ Rdz = P xyz dx+ Q xyz dy+ R xyz dz Exmple Evlute ydx+ xdy+ zdz where is given y x= cost, y = sint, z = t, t π. So, we lredy hve the curve prmeterized so there relly isn t much to do other thn evlute the integrl. ydx+ xdy+ zdz = ydx+ xdy+ zdz ( ) ( ) ( ) π π π = sint sint dt+ cost cost dt+ t t dt π π π = sin tdt+ cos tdt+ t dt = ( ( )) ( ( )) π π π cos t dt cos t dt t dt Dr.Eng Muhmmd.A.R.yss = t sin( t) + t+ sin( t) + t = 8π π Pge of Line Integrl's <<->>

12 Line Integrl of Vector Fields In the previous two sections we looked t line integrls of functions. In this section we re going to evlute line integrls of vector fields. We ll strt with the vector field, F xyz,, = P xyz,, i + Q xyz,, j + R xyz,, k ( ) ( ) ( ) ( ) nd the three-dimensionl, smooth curve given y r t = xt i + y t j + z t k t ( ) ( ) ( ) ( ) The line integrl of F long is. Also, F( r( t) ) Fdr i = F r t i r t dt ( ()) () is shorthnd for, F r t = F xt, y t, z t ( ( )) ( ( ) ( ) ( )) We cn lso write line integrls of vector fields s line integrl with respect to rc length s follows, Fdr i = FTds i where T ( t) is the unit tngent vector nd is given y, r ( t) T() t = r () t If we use our knowledge on how to compute line integrls with respect to rc length we cn see tht this second form is equivlent to the first form given ove. Fdr i = FTds i r () t = F( r() t ) i r () t dt r () t = F( r() t ) i r () t dt Exmple Evlute Fdr i where F( xyz,, ) = 8xyzi + 5zj xyk nd is the curve given y r( t) = ti + t j + t k, t. Dr.Eng Muhmmd.A.R.yss Oky, we first need the vector field evluted long the curve. 7 F r t = 8t t t i + 5t j t t k = 8t i + 5t j t k ( ( )) ( )( ) ( ) Next we need the derivtive of the prmeteriztion. r t = i + tj + t k ( ) Finlly, let s get the dot product tken cre of. F r t i r t = 8t + t t The line integrl is then, 7 5 ( ( )) ( ) 7 5 Fdr i = 8t + t t dt ( t t t ) = + = Pge of Line Integrl's <<->>

13 Exmple Evlute (,,) nd ( ) Fdr,,. i where (,, ) F xyz = xzi yzk nd is the line segment from Here is the prmeteriztion for the line. r t = t,, + t,, ( ) ( ) = t,, tt, t So, let s get the vector field evluted long the curve. F( r( t) ) = ( t )( t) i ( t)( t) k = t t i t t k Now we need the derivtive of the prmeteriztion. r t = The dot product is then, The line integrl ecomes, ( ) ( ) ( ),, = = ( ( )) i ( ) ( ) ( ) F r t r t t t t t 8t 6t Fdr i = t tdt 8 6 ( 6t t ) = = Given the vector field F( xyz,, ) = Pi + Qj + Rk nd the curve prmeterized y r t = xt i + y t j+ z t k, t the line integrl is, ( ) ( ) ( ) ( ) Fdr i = Pi + Qj + Rk xi + y j + zk dt ( ) i( ) = Px + Qy + Rz dt = Px dt+ Qy dt+ Rz dt = Pdx+ Qdy+ Rdz Dr.Eng Muhmmd.A.R.yss = Pdx+ Qdy+ Rdz So, we see tht, Fct Fdr i = Pdx+ Qdy+ Rdz Fdr i = Fdr i Pge of Line Integrl's <<->>

14 Fundementl Theorem for Line Integrls Green's Theorm Green s Theorem Let e positively oriented, piecewise smooth, simple, closed curve nd let D e the region enclosed y the curve. If P nd Q hve continuous first order prtil derivtives on D then, Q P Pdx+ Qdy = da x y Green s Theorem we will often denote the line integrl s, Exmple Use Green s Theorem to evlute D Pdx+ Qdy or Pdx+ Qdy vertices (, ), (, ), (, ) with positive orienttion. xydx+ xy dy where is the tringle with Green s Theorem nd we cn see tht the following inequlities will define the region enclosed. x y x We cn identify P nd Q from the line integrl. Here they re. P= xy Q= xy So, using Green s Theorem the line integrl ecomes, xydx+ xy dy = xy xda D x = xy xdydx x xy xy dx = Dr.Eng Muhmmd.A.R.yss 5 = 8x x dx 6 = x x = Pge of Line Integrl's <<->>

15 Exmple Evlute centered t the origin. y dx x dy where is the positively oriented circle of rdius Let s first identify P nd Q from the line integrl. P= y Q= x Now, using Green s theorem on the line integrl gives, y dx x dy = x y da Exmple Evlute D ( ) y dx x dy = x + y da = = π π r = dθ = π D π r drd dθ y dx x dy where re the two circles of rdius nd rdius centered t the origin with positive orienttion. Fct θ ( ) = + y dx x dy x y da = = = D π r drd π π 5π = r 5 dθ dθ Dr.Eng Muhmmd.A.R.yss A= xdy = ydx= xdy ydx θ o Exmple Use Green s Theorem to find the re of disk of rdius. We cn use either of the integrls ove, ut the third one is proly the esiest. So, A= xdy ydx Pge 5 of Line Integrl's <<->>

16 where is the circle of rdius. So, to do this we ll need prmeteriztion of. This is, x= cost y = sint t π The re is then, A= xdy ydx ( ( ) ( ) ) π π cost cost dt sint sin t dt = π = cos sin t+ tdt π = dt = π Dr.Eng Muhmmd.A.R.yss Pge 6 of Line Integrl's <<->>

17 Stokes' Theorm The following surfce with the indicted orienttion. Stokes Theorem Let S e n oriented smooth surfce tht is ounded y simple, closed, smooth oundry curve with positive orienttion. Also let F e vector field then, Fidr = curl Fi ds Exmple Use Stokes Theorem to evlute nd S is the prt of S = ove the plne z 5 x y Let s strt this off with sketch of the surfce. S curl F i ds where F = z i xy j + xy k z =. Assume tht S is oriented upwrds. Dr.Eng Muhmmd.A.R.yss In this cse the oundry curve will e where the surfce intersects the plne z = nd so will e the curve = 5 x y x y z + = t = So, the oundry curve will e the circle of rdius tht is in the plne z =. The prmeteriztion of this curve is, r t = ti + tj + k t π ( ) cos sin, Pge 7 of Line Integrl's <<->>

18 The first two components give the circle nd the third component mkes sure tht it is in the plne z =. Using Stokes Theorem we cn write the surfce integrl s the following line integrl. π curl FidS = Fidr = F r t i r t dt S ( ()) () So, it looks like we need couple of quntities efore we do this integrl. Let s first get the vector field evluted on the curve. Rememer tht this is simply plugging the components of the prmeteriztion into the vector field. F( r( t) ) = ( ) i cos ( t)( sint) j + ( cost) ( sin t) k = i costsintj + 6cos tsin tk Next, we need the derivtive of the prmeteriztion nd the dot product of this nd the vector field. r ( t) = sinti + costj F r t ir t = sint sintcos t ( ()) () We cn now do the integrl. curl i sin sin cos S π F ds = t t tdt ( cost 8cos t) = + = Exmple Use Stokes Theorem to evlute π F i dr where the tringle with vertices (,, ), (,, ) nd ( ) = + + nd is F z i y j xk,, with counter-clockwise rottion. We re going to need the curl of the vector field eventully so let s get tht out of the wy first. i j k curlf = = zj j = ( z ) j x y z z y x Dr.Eng Muhmmd.A.R.yss Pge 8 of Line Integrl's <<->>

19 Since the plne is oriented upwrds this induces the positive direction on s shown. The eqution of this plne is, x+ y+ z= z = g xy, = x y ( ) Now, let s use Stokes Theorem nd get the surfce integrl set up. Fidr = curl FidS S S D ( z ) = jds i f = ( z ) j i f da f Oky, we now need to find couple of quntities. First let s get the grdient. Recll tht this comes from the function of the surfce. f ( xyz,, ) = z g( xy, ) = z + x+ y f = i + j + k Note s well tht this lso points upwrds nd so we hve the correct direction. Now, D is the region in the xy-plne shown elow, We get the eqution of the line y plugging in z = into the eqution of the plne. So sed on this the rnges tht define D re, x y x+ The integrl is then, ( ) i( ) Dr.Eng Muhmmd.A.R.yss Fidr = z j i + j + k da D x+ ( ) = x y dydx Don t forget to plug in for z since we re doing the surfce integrl on the plne. Finishing this out gives, x+ Fidr = x ydydx ( ) = x = x x Pge 9 of Line Integrl's <<->> = 6 x+ = y xy y dx xdx

20 Home Work Dr.Eng Muhmmd.A.R.yss Pge of Line Integrl's <<->>

21 Dr.Eng Muhmmd.A.R.yss Pge of Line Integrl's <<->>