APPLIED MATHEMATICS Part 3: Vector Integral Calculus

Transcription

1 APPLIED MATHEMATICS Part 3: Vector Integral Calculus

2 Contents 9 Vector Integral Calculus and Integral Theorems Line Integrals Line Integrals Independent of Path Double Integrals (Review of Calculus) Green s Theorem in the Plane Surfaces Surface Integral Triple Integral (Volume Integral) and Divergence Theorem of Gauss Applications of Divergence Theorem Stokes Theorem

3 Chapter 9 Vector Integral Calculus and Integral Theorems 2

4 9.1 Line Integrals Line integral of F ( r) overacurvec : r(t) =x(t)î + y(t)ĵ + z(t)ˆk, c F ( r) d r = b a F ( r(t)) d r dt dt where r(a) is the initial point, r(b) is the terminal point of the integral. Line integral of F ( r) overaclosedcurvec, c F ( r) d r. F = F 1 î + F 2 ĵ + F 3ˆk = F ( r) d r = (F c c 1dx + F 2 dy + F 3 dz) = b (F dx a 1 dt + F dy 2 dt + F dz 3 dt )dt If F = force and r = displacement = F d r = work 3

5 Ex : F ( r) =zî + xĵ + yˆk (a) r =costî +sintĵ +3tˆk (t =0 2π) (b) r = î + tˆk (t =0 6π) (a) F c ( r) d r = 2π F ( r(t)) d r 0 dt dt = 2π (3tî +costĵ +sintˆk) ( sin tî +costĵ +3ˆk)dt 0 = 2π ( 3t sin t +cos 2 t +3sint)dt 0 = 7π (b) F c ( r) d r = 2π (tî + ĵ) (ˆk)dt =0 0 4

6 Properties of Line Integral c k F d r = k c F d r ( F + G) d r = F d r + G d r c c c F c d r = c1 F d r + c F d r 2 where C = C1 C 2 5

7 9.2 Line Integrals Independent of Path c1 F ( r) d r = c2 F ( r) d r = c 3 F ( r) d r The lime integrals may be independent of path of integration,which means as long as A and B are fixed, the integral is all the same no matter which path is chosen. Theorem 1 F ( r) d r with continuous F 1, F 2 and F 3 in D is independent of path in D if and only if F = f. Proof : r(t) =x(t)î + y(t)ĵ + z(t)ˆk, a t b F = f =( f )î +( f)ĵ +( f x y z )ˆk = F d r = B ( f f f dx + dy + c A x y z dz) = b a df dt dt = B df = f(b) f(a) A the result is independent of the path, depends only the initial and terminal points A and B. 6

8 Ex : F ( r) =3x 2 î +2yzĵ + y 2ˆk, A :(0, 1, 2), B :(1, 1, 7) If the potential f exists, f x =3x2 = f(x, y, z) =x 3 + g(y, z) f y =2yz = g y = g(y, z) =y 2 z + h(z) f z = y2 = g z = y2 + dh dz = h(z) =c (constant) the potential f(x, y, z) =x 2 + y 2 z + c exists. B A F d r = f(b) f(a) =(1+7+c) (2 + c) =6 7

9 Theorem 2 F ( r) d r is independent of path in D if and only if c F ( r) d r =0inD. Proof : c F ( r) d r =0 = c 1 F ( r) d r c 2 F ( r) d r =0 = c 1 F ( r) d r = c 2 F ( r) d r F d r is independent of path. Note: Since the work W done by a force F along a path C is W = c F ( r) r, if the integral is independent of path, then F ( r) r =0, and F = f. = F is called conservative if F is a force vector. F is called irrotational if F is a velocity vector. 8

10 Exact Differential c F ( r) d r = c (F 1dx + F 2 dy + F 3 dz) (F 1 dx + F 2 dy + F 3 dz) is called exact on an exact differential in D if F 1 dx + F 2 dy + F 3 dz = f f f dx + dy + dz = df x y z = F 1 = f x, i.e. F = f F 2 = f y, F 3 = f z Simply Connected Domain D If every closed curve in D can be continuously shrunk to any point in D without leaving D, then D is called a simply connected domain. e.g. interior of a sphere, a cube, a sphere with finitely many points removed. Note: yes: no: domain between two concentric spheres interior of a torus or 9

11 Theorem 3 c F ( r) r = c (F 1dx + F 2 dy + F 3 dz) ( ) where F 1,F 2 and F 3 and their first derivatives are continuous in D then (1) if (*) is independent of path, i.e. F = f or F 1 dx + F 2 dy + F 3 dz is exact = F =0inD, (2) if F =0inD and D is simply connected = (*) is independent of path in D. Ex : I = c [(2xyz2 ) dx +(x 2 z 2 + z cos yz) dy +(2x 2 yz + y cos yz) dz] }{{}}{{}}{{} = F 1 = F 2 = F 3 F =[(F 3 ) y (F 2 ) z ] î +[(F 1 ) z (F 3 ) x ] ĵ +[(F 2 ) x (F 1 ) y ] ˆk (F 3 ) y =2x 2 z +cosyz yz sin yz =(F 2 ) z (F 1 ) z =4xyz =(F 3 ) x (F 2 ) x =2xz 2 =(F 1 ) y F =0 10

12 F = f f x =2xyz2 = f(x, y, z) =x 2 yz 2 + g(y, z) f y = x2 z 2 + z cos yz = x 2 z 2 + g y = x2 + z cos yz = g(y, z) =sinyz + h(z) f z =2x2 yz + y cos yz = 2x 2 yz + y cos yz + dh dz =2x2 yz + y cos yz = h(z) =c = f(x, y, z) =x 2 yz 2 +sinyz + c If C is any curve from A :(0, 0, 1) to B :(1, π 4, 2) F c d r = B df = f(b) f(a) =(π +sinπ A 2 + c) (c) =π +1 11

13 Ex : Consider a two-domensional domain D and a vector function F = y x 2 + y 2 î + x x 2 + y 2 ĵ F = 0 except at (x, y) =(0, 0) If D excludes (0, 0), then c F d r is independent of path. However, if D includes (0, 0), then F = 0 only in D, where D {(0, 0)} = D. D is not simply connected. c F d r is dependent of path. 12

14 9.3 Double Integrals (Review of Calculus) definite integral indefinite integral b a f(x)dx c f( r) d r R f(x, y)dxdy V f(x, y, z)dxdydz... S f nda 13

15 Definition: Double integral over region R: R f(x, y)dxdy or R f(x, y)da = lim N N k=1 f(x k,y k ) A k Properties of Double Integral: R kfda = k R fda k = constant R (f + g)da = R fda+ R gda R fda = R 1 fda+ R 2 fda, R = R 1 R 2 Mean Value Theorem for Double Integral: R f(x, y)da = f(x 0,y 0 )A, where A = area of R and (x 0,y 0 ) R. 14

16 Evaluation of double integral: (1) If R can be described by a x b g(x) y h(x) = R f(x, y)dxdy = b (2) If R can be described by a [ h(x) g(x) f(x, y)dy ] dx c y d p(y) x q(y) = R f(x, y)dxdy = d c [ q(y) p(y) f(x, y)dx ] dy (3) If R cannot be represented by those inequalities in (1) and (2), but can be subdivided into portions which can be represented by those inequalities,then 15

17 Applications of double integral: A = R dxdy = area of region R V = R plane f(x, y)dxdy = volume between the surface z = f(x, y) andxy Change of Variables in Double Integrals: Recall that b f(x)dx = β=u(b) f(x(u))dx a α=u(a) du du. For double integral, if x = x(u, v) andy = y(u, v) R f(x, y)dxdy = R f (x(u, v),y(u, v)) (x, y) (u, v) }{{} J dudv where R is region in uv plane, and Jacobian J = (x, y) (u, v) = x u y u x v y v 16

18 Ex : R y2 dxdy R :0 y 1 x 2, 0 x 1 (1) R y2 dxdy = x 2 0 y 2 dy dx (2) = x 2 3 y3 y=0 dx = [1 0 x2 ] 3/2 dx = π 16 R y2 dxdy = y 2 0 = 1 1 y 2 0 y2 [x] x=0 = 1 0 y2 = 1 y 2 dy y 2 dx dy dy y2 (1 y 2 ) 3/2 4 = 1 8 π 2 = π 16 + y(1 y2 ) 1/ sin 1 y

19 (3) x = r cos θ y = r sin θ J = x r y r x θ y θ = cos θ r sin θ sin θ rcos θ = r cos 2 θ + r sin 2 θ = r R f(x, y)dxdy = R r 2 sin 2 θ r drdθ = π/2 0 = π/2 0 = π/2 0 [ 1 0 r3 sin 2 ] θdr dθ r4 sin 2 θdθ r=0 1 4 sin2 θdθ = π 16 18

20 9.4 Green s Theorem in the Plane Consider a closed region R with boundary C consists of finitely many smooth curves. If F 1 (x, y), F 2 (x, y) and their first derivatives are continuous = R ( F 2 x F 1 y )dxdy = C (F 1dx + F 2 dy) (1) The direction of integration along C is such that R is always on the left hand side. Equation (1) can be written in vectorial form as ( F ) ˆk dxdy= F d r, R C where F = F 1 î + F 2 ĵ. 19

21 ProofofGreen stheorem: R ( F 2 x F 1 y )dxdy = C (F 1dx + F 2 dy) If R can be represented in the forms: a x b c y d u(x) y v(x) p(y) x q(y) For those regions which cannot be represented by the above forms, the domain is divided into subregions each of them can be represented by the required forms. 20

22 R F 1 y dxdy = b a v(x) u(x) F 1 y dy dx = b [F a 1(x, v(x)) F 1 (x, u(x))] dx = b F a 1(x, u(x))dx a F b 1(x, v(x))dx = F c 1(x, y)dx ( ) R F 2 x dxdy = d c q(y) p(y) F 2 x dx dy = d [F c 2(q(y),y)dy F 2 (p(y),y)] dy = d F c 2(q(y),y)dy + c F d 2(p(y),y)dy = F c 2(x, y)dy ( ) (*) and (**) = R F 2 x F 1 = y (F c 1dx + F 2 dy) 21

23 Ex : F 1 = y 2 7y, F 2 =2xy +2x R = R ( F 2 x F 1 y )dxdy [(2y +2) (2y 7)]dxdy = R 9dxdy =9π (F C 1dx + F 2 dy) = F d r C = C F r dt = 2π 0 = 2π 0 = 2π 0 [ (y 2 7y)î +(2xy +2x)ĵ ] [ sin tî +costĵ ] dt [ (sin 2 t 7sint)( sin t)+(2cost sin t +2cost)(cos t) ] dt [ sin 3 t +7sin 2 t +2cos 2 t sin t +2cos 2 t) ] dt = 7π +2π =9π 22

24 Ex : Area of a plane region (a) If F 1 0, F 2 x R ( F 2 x F 1 )dxdy = (1 0)dxdy y = C (F 1dx + F 2 dy) = C xdy R A = R dxdy = c xdy (b) If F 1 = y, F 2 =0 R ( F 2 x F 1 )dxdy = (0 + 1)dxdy y = C (F 1dx + F 2 dy) = C ( y)dx R A = R dxdy = C ydx Combine (a) and (b), A = 1 2 C (xdy ydx) 23

25 For example, for an ellipse x2 a + y2 =1,orx = a cos t, y = b sin t, 2 b2 A = c xdy = 2π 0 x du dt dt = 2π (a cos t)(b cos t)dt 0 = ab 2π cos 2 tdt 0 = πab 24

26 Ex : Area of a plane region in polar coordinates x = r cos θ, y = r sin θ dx = x r dy = y r x dv + dθ = cosθdr r sin θdθ θ y dv + dθ = sinθdr + r cos θdθ θ A = 1 2 = 1 2 = 1 2 = 1 2 C C (xdy ydx) [(r cos θ)(sin θdr + r cos θdθ) (r sin θ)(cos θdr r sin θdθ)] C [(r cos θ sin θ r sin θ cos θ)dr +(r2 cos 2 θ + r 2 sin 2 θ)dθ] 2 0 πr2 dθ For example, a cardioid is represented by r = a(1 cos θ), 0 θ 2π. A = a π(1 cos θ)2 dθ = 3π 2 a2 25

27 Ex : Transformation of a double integral of the Laplacian of a function into a line integral of its normal derivative. 2 wdxdy = w R C n ds, where w = w(x, y), 2 = 2 x x 2. Proof : Let F 1 = w y, R F 2 = w x ( F 2 x F 1 )dxdy = y R ( 2 w x w y 2 )dxdy (F C 1dx + F 2 dy) = (F dx C 1 ds + F dy 2 ds )ds = C ( w y dx ds + w x dy ds )ds Since w = w xî + w y ĵ ˆt = d r ds = x sî + dy dsĵ dy = ˆn = dsî dx dsĵ w ˆn = w n = w dy x ds w dx y ds R ( 2 w x w y 2 )dxdy = C w n ds 26

28 9.5 Surfaces Representation of surface: S : z = f(x, y) S : g(x, y, z) =0 S : r(u, v) =x(u, v)î + y(u, v)ĵ + z(u, v)ˆk Notes: curve C : r(t) =x(t)î + y(t)ĵ + z(t)ˆk 27

29 Ex : (1) Circular cylinder: S : x 2 + y 2 = a 2, 1 z 1 S : r(u, v) =a cos uî + a sin uĵ + vˆk, 0 u 2π, 1 v 1 (2) Sphere: S : x 2 + y 2 + z 2 = a 2 S : r(u, v) =a cos u cos vî + a sin u cos vĵ + a sin vˆk, π 0 u 2π, 2 v π 2 S : r(u, v) =a cos u sin vî + a sin u sin vĵ + a cos vˆk, 0 u 2π, 0 v π 28

30 (3) Cone: S : z = x 2 + y 2, 0 z h S : r(u, v) =u cos vî + u sin vĵ + uˆk 0 u h 0 v 2π 29

31 Tangent plane and surface normal vector: A tangent plane at p contains all the tangent vectors of curves on S passing through p. A normal vector of S at p is a vector normal to the tangent plane at p. 30

32 To find the normal vector N of S : r = r(u, v), consider two curves on S, C 1 : r(u) = r(u, v 0 ),v 0 = constant C 2 : r(v) = r(u 0,v), u 0 = constant normal vector N = r u r v N unit normal vector ˆn = N = r u r v r u r v Note :IfSis represented by S : g(x, y, z) =0 = N = g, ˆn = g g 31

33 Ex : Surface of a sphere S : g(x, y, z) =x 2 + y 2 + z 2 a 2 =0 g = g xî + g yĵ + g zˆk =2xî +2yĵ +2zˆk g = 4x 2 +4y 2 +4z 2 =2 x 2 + y 2 + z 2 =2a ˆn = 2xî +2yĵ +2zˆk 2a = x aî + y aĵ + z aˆk Or S : r(u, v) =a cos u cos vî + a sin u cos vĵ + a sin vˆk r u r v = a sin u cos vî + a cos u cos vĵ = a cos u sin vî a sin u sin vĵ + a cos vˆk r u r v = î ĵ ˆk a sin u cos v acos u cos v 0 a cos u sin v a sin u sin v acos v = î(a 2 cos u cos 2 v)+ĵ(a 2 sin u cos 2 v) +ˆk(a 2 sin 2 u sin v cos v + a 2 cos 2 u sin v cos v) = î(a 2 cos u cos 2 v)+ĵ(a 2 sin u cos 2 v)+ˆk(a 2 sin v cos v) r u r v = a 2 cos 2 u cos 4 v +sin 2 u cos 4 v +sin 2 v cos 2 v = a cos 2 4 v +sin 2 v cos 2 v = a 2 cos v cos 2 v +sin 2 v = a 2 cos v 32

34 ˆn = î a 2 cos u cos 2 v a 2 + ĵ cos v = î (cos u cos v) }{{} +ĵ (sin u cos v) }{{} a 2 sin u cos 2 v a 2 + cos v ˆk +ˆk (sin v) }{{} a 2 sin v cos v a 2 cos v = x a = y a = z a 33

35 9.6 Surface Integral Consider a surface, S : r(u, v) =x(u, v)î + y(u, v)ĵ + z(u, v)ˆk, with the normal vector: N N = ru r v, ˆn = N. Surface integral of a vector function F ( r) overs is defined as: S F ( r) ˆn da, where F ˆn = component of F normal to S. e.g. : If F = ρ vda = F ˆn = ρ v ˆndA represents the mass flux across S. S S 34

36 Evaluation of surface integral: S F ( r) ˆn da r(u + u, v) r(u, v) = r u +... u r(u, v + v) r(u, v) = r v +... v The area of A is A = r r u u v v da = lim A = u 0 v 0 = r u r v r u r v u v du dv 35

37 S F ( r) ˆndA = R F ( r) ( r u r v ) r u r v r u r v dudv = R F ( r) ( r u r v )dudv = R F ( r) Ndudv i.e. ˆndA = Ndudv Now represent F, N and ˆn by: F F 1 î + F 2 ĵ + F 3ˆk N N 1 î + N 2 ĵ + N 3ˆk ˆn cos αî +cosβĵ +cosγˆk where ˆn î =cosα, ˆn ĵ =cosβ, ˆn ˆk =cosγ, α is the angle between ˆn and x axis β is the angle between ˆn and y axis γ is the angle between ˆn and z axis = S F ( r) ˆndA = (F 1 cos α + F 2 cos β + F 3 cos γ)da S = R (F 1 N 1 + F 2 N 2 + F 3 N 3 )dudv 36

38 da cos γ = dxdy Similarly, da cos α = dydz da cos β = dzdx = S F nda = (F 1 dydz + F 2 dzdx + F 3 dxdy) S If S : z = h(x, y) S F 3 (x, y, z)cosγda = ± F 3 [x, y, h(x, y)]dxdy R xy where R xy is the projection of A on the xy plane. S F 3 (x, y, z)cosγda = cos γ F 3 [x, y, h(x, y)]dxdy cos γ R xy 37

39 Similarly, if S : y = g(x, z) S F 2 (x, y, z)cosβda = cos β F 2 [x, g(x, z),z]dzdx cos β R zx If S : x = f(y, z) S F 1 (x, y, z)cosαda = cos α F 1 [f(y, z),y,z]dydz cos α R yz 38

40 Ex : Compute the flux of water through the parabolic cylinder S : y = x 2, 0 x 2, 0 z 3 and the velocity vector is V = yî +2ĵ + xzˆk. Flux = S V ˆndA = R V Ndudv S : r = uî + u 2 ĵ + vˆk, 0 u 2, 0 v 3 r u = î +2uĵ r v = ˆk N = r u r v = î ĵ ˆk 1 2u =2uî ĵ V = u 2 î +2ĵ + uvˆk = Flux = R V Ndudv = 3 0 = (u2 î +2ĵ + uvˆk) (2uî ĵ)dudv 2 0 (2u3 2)dudv =12 39

41 Or S : f(x, y, z) =x 2 y =0 0 x 2, (i.e. 0 y 4) 0 z 3 ˆn = f f = 2xî ĵ 4x2 +1 cos α = 2x 4x2 +1 > 0 cos β = 1 4x2 +1 < 0 the flux is s = + = 3 0 = 3 y2 2 = 12 V ˆndA R yz v 1 dydz 4 ydydz R zx v 2 dzdx 3 0 2dzdx 2 2 3=

42 Integral over oriented surface: S F ˆndA We have chosen one of the two possible unit normal vectors (ˆn or ˆn). The integral is over an oriented surface. Smooth surface: S is smooth if we can choose a normal vector at any point P of S. The following are not smooth surfaces, but are piecewise smooth. Orientable surface: Smooth surface S is orientable if the positive (or negative) normal direction at any point is always in the same direction. Möbius strip is not oriented surface. 41

43 Integral over non-oriented surface: s G( r)da Recall that da = N dudv = r u r v dudv s G( r)da = R G[ r(u, v)] N(u, v) dudv If G( r) = 1 S da = R r u r v dudv = area of S A(S) 42

44 Ex : Area of a torus surface (doughnut) r(u, v) =(a + b cos v)cosuî +(a + b cos v)sinuĵ + b sin vˆk 0 u 2π 0 v 2π r u = (a + b cos v)sinuî +(a + b cos v)cosuĵ r v = b sin v cos uî b sin v sin uĵ + b cos vˆk r u r v = î ĵ ˆk (a + b cos v)sinu (a + b cos v)cosu 0 b sin v cos u b sin v sin u bcos v = î(a + b cos v)cosu cos v + ĵ(a + b cos v)sinu cos v +ˆk[ b sin v(a + b cos v)sin 2 u b sin v(a + b cos v)cos 2 u] = b(a + b cos v)[cos u cos vî +sinu cos vĵ +sinvˆk] r u r v = b(a + b cos v) 43

45 2π A(S) = 2π b(a + b cos v)dudv 0 0 = 2π (abu + b 2 sin v) 2π 0 u=0dv = 2π 0 2πabdv =4π 2 ab 44

46 9.7 Triple Integral (Volume Integral) and Divergence Theorem of Gauss Triple integral (volume integral) of f(x, y, z) over a region T is defined as: T f(x, y, z)dxdydz or T f(x, y, z)dv Theorem: Divergence Theorem (or Gauss s Theorem) (Transformation between volume integral and surface integral) Let T be a closed bounded region in space with boundary S which is a piecewise smooth orientable surface. Vector function F ( r) and its first derivatives are continuous. = T FdV = S F ˆndA, where ˆn is outward unit normal vector of S 45

47 Proof : T FdV = S F ˆndA F = F 1 î + F 2 ĵ + F 3ˆk ˆn = n 1 î + n 2 ĵ + n 3ˆk =cosαî +cosβĵ +cosγˆk i.e. we want to prove = S T ( F 1 x + F 2 y + F 3 z )dxdydz (F 1 cos α + F 2 cos β + F 3 cos γ)da. We first prove T F 3 dxdydz = z s F 3 cos γda T F 3 dxdydz = [ h(x,y) F 3 z g(x,y) R z dz]dxdy = F 3 (x, y, h(x, y))dxdy R R F 3 (x, y, h(x, y))dxdy 46

48 Recall that cos γda = dxdy = S F 3 dxdy = S F 3 cos γda = S F 3 n 3 da T F 3 dxdydz = z S F 3 n 3 da (1) Similarly T F 1 dxdydz = x S F 1 n 1 da (2) T F 2 dxdydz = y S F 2 n 2 da (3) (1), (2) and (3) = T FdV = s F ˆndA 47

49 Physical interpretation of divergence theorem T FdV = S F ˆnds Divide T into many small boxes: Flux of F out of the small box = (F 1 + F 1 x x) y z F 1 y z +(F 2 + F 2 y y) z x F 2 z x +(F 3 + F 3 z z) x y F 3 x y = ( F 1 x + F 2 y + F 3 z ) x y z = ( F ) V T FdV = lim 0 V ( F ) V 48

50 But, only the surfaces on S contribute to the flux in lim V 0 V ( F ) V = S F ˆnds =Fluxof F out of S 49

51 9.8 Applications of Divergence Theorem The theory of solutions of Laplace s equation 2 f = 2 f x f y f z 2 =0 is call potential theory, andf is called harmonic function. Application of divergence theorem in potential theory T FdV = S F ˆndA (1) F f F ˆn = f ˆn = f n = T 2 fd = S f n da Theorem : If f is a harmonic function i.e. 2 f =0,then S f da =0 n 50

52 (2) F f g F = (f g) = f g + f 2 g F ˆn = f g n = f g n = T (f 2 g + f g)dv = S f g n da ( ) This is called the Green s first formula (identity). Interchange f and g i.e. F g f = T (g 2 f + f g)dv = S g f n da ( ) ( ) ( ) = T (f 2 g g 2 f)dv = S (f g n g f n )da This is called Green s second formula (identity). 51

53 (3) If f satisfies 2 f =0inT and f =0onS, ( ) = T = T (f 2 f + f f)dv = S f 2 dv =0 f f n da = f 2 =0 = ( f x )2 +( f y )2 +( f z )2 =0 f x =0, f y =0, f z =0 i.e. f = constant in T. But since f is a continuous function,if f =0onS, thenf =0inT too Theorem If f is a harmonic function in T,andf =0onS, thenf =0inT. If f 1 = f 2 on S, i.e. (f 1 f 2 )=0onS,andf 1 and f 2 are both harmonic functions in T,then 2 f 1 =0and 2 f 2 =0 Since (f 1 f 2 )=0onS = 2 (f 1 f 2 )=0inT = (f 1 f 2 )=0inT i.e. f 1 = f 2 in T. Theorem If f is a harmonic function in T,thenf is uniquely determined in T by its value on S. 52

54 9.9 Stokes Theorem Transformation between Surface and Line Integrals If S is a piecewise smooth oriented surface and C is a piecewise smooth simple closed curve which is the boundary of S, F (x, y, z) and its first derivatives are continuous in S, then S ( F ) nda = C F r ds, where r = d r ds = ˆt. If S lies on the xy plane (R), ( F ) ˆn =( F 2 x F 1 y )ˆk ˆk = F 2 x F 1 y F d r ds ds = F d r = F 1 dx + F 2 dy R ( F 2 x F 1 y )dxdy = (F 1 dx + F 2 dy) Green s theorem in 9.4 C 53

55 Ex : F = yî + zĵ + xˆk S : z = f(x, y) =1 (x 2 + y 2 ), z 0 x = u cos v y = u sin v z =1 (u 2 cos 2 v + u 2 sin 2 v)=1 u 2 S : r(u, v) =u cos vî + u sin vĵ +(1 u 2 )ˆk 0 u 2π, 0 v 1 N = r u r v = î ĵ ˆk cos v sin v 2u u sin v ucos v 0 = î(+2u 2 cos v)+ĵ(+2u 2 sin v)+ˆk(u) F = î ĵ ˆk x y z y z x = î ĵ ˆk S ( F ) ˆndA = 2π ( 2u2 cos v 2u 2 sin v u)dudv 54

56 = 2π ( cos v 2 3 sin v 1 )dv = π 2 55

57 C : r(s) =cossî +sinsĵ, 0 s 2π r (s) = sin sî +cossĵ F (s) =sinsî +cossˆk F C r ds = 2π ( sin 2 s)ds = π 0 56

58 ProofofStoke stheorem: S ( F ) nda = C F r ds Recall that nda = Ndudv, N =[N1,N 2,N 3 ]= r u r v We want to prove = C R ( F 3 y F 2 z )N 1 +( F 1 z F 3 x )N 2 +( F 2 x F 1 y )N 3 dudv (F 1 dx + F 2 dy + F 3 dz) where R is the region with boundary C in the uv-plane. Proof of ( F 1 R z N 2 F 1 y N 3)dudv = C F 1 dx Consider S which can be represented simultaneously by z = f(x, y), y = g(x, z), x = h(y, z) Let u = x, v = y, then r(u, v) = r(x, y) =xî + yĵ + f(x, y)ˆk N = r u r v = r x r y = (î + f xˆk) (ĵ + fyˆk) = f x î f y ĵ + ˆk 57

59 R = = = ( F 1 z N 2 F 1 y N 3)dudv ( F 1 S z ( f y) F 1 y )dxdy F 1 (x, y, f(x, y)) dxdy S y C F 1 dx = C F 1 dx (a) (by Green s Theorem) Similarly, R ( F 2 z N 1 + F 2 x N 3)dudv = C F 2 dy (b) R ( F 3 y N 1 F 3 x N 2)dudv = C F 3 dz (c) From (a), (b) and(c) = C R ( F 3 y F 2 z )N 1 +( F 1 z F 3 x )N 2 +( F 2 x F 1 y )N 3 dudv (F 1 dx + F 2 dy + F 3 dz) 58

60 Physical meaning of Stoke s theorem If F v (velocity), v(x, y, z) =uî + vĵ + wˆk, s ( v) ˆndA = v d r }{{}} c {{} ω Γ ω is called vorticity Γ is called circulation Stoke s theorem applied to independent of path Recall that in 9.2, Theorem 3: If a line integral F ( r) d r is independent of path, C = F =0inD. If v =0inD, by Stoke s theorem: C 1 + C 2 F d r = S F =0 = C 1 F d r = C 2 F d r = C 2 F d r ( C 2 = C 2 ) = C F ( r) d r is independent of path. 59