Integrals over Paths and Surfaces
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- Godwin Hoover
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2 Î 7 á Integrls over Pths nd urfes Î 1 Å Pth Integrl Pth integrl R 2 or R 3. A prmeterized urve n be written s (t) = (x(t),y(t),z(t)). If x(t), y(t), z(t) re ontinuous then we sy is ontinuous, nd if x(t),y(t),z(t) re differentible, then is differentible. If x (t), y (t), z (t)re ontinuous then we sy us C 1 -urve. Let be defined on [,b] nd let P : = t 0 < t 1 < < t k = b be the prtition of [,b]. t i is point between t i 1 nd t i. Then the Riemnn sum is k f((t i )) (t i ) (t i 1 ) i=1 As P pprohes 0 the sum pprohes k f((t i )) (t i) (t i t i 1 ) i=1 efinition 1.1. If is defined over I = [,b] hving vlues in R 3 - C 1 -urve, f is defined over region ontining the imge of. Then f is rel vlued funtion defined on I. We define the pth integrl of s: b f((t)) (t) dt 3
3 4 7 ì INTEGRAL OVER PATH AN URFACE z (t k ) = t 0 t 1 t i 1 t i t k = b x O (t 0) (t i 1) (t i) f((u i)) (t i) (t i 1) y (t 1) 7.1: Riemnn sum over pth We denote it by f ds or f(x,y,z)ds. If f = 1 then 1ds is the length of. Exmple 1.2. Find pth integrl of f(x,y,z) = x 2 + y 2 + z 2 over. (t) = (os t,sin t,t), t [0,2π] sol. ine (t) = ( sin t,os t,1), the line integrl is f ds = 2π 0 2π f((t)) (t) dt = (1 + t 2 ) 2 dt 0 = ( ) 2 2π + 8π 3 /3 Pth integrl over plnr Curves If f(x, y) is ontinuous funtion defined over region ontining the imge of, then the pth integrl of f long is given by f(x,y)ds = b If f = 1, it is nothing but the r -length. f(x(t),y(t)) x (t) 2 + y (t) 2 dt Exmple 1.3 (Tom wyer s fene). Find the re of fene long prmeterized urve in R 2 nd height is given by f(x,y) = 1 + 3y.
4 2 LINE INTEGRAL 5 sol. (t) = (30os 3 t,30sin 3 t) for t [0,π/2]. The re of one side is C f(x,y)ds where ds = (t) dt = 90sin t os tdt. o C f(x,y)ds = = 90 π/2 0 π/2 0 ( ) sin 3 t 90sin t os tdt (sin t + 10sin 4 t)os t dt = 225. This is hlf of the fene. Totl re of fene(both sides) is 900 squre ft. If he n get.05 dollr per squre feet, he n mke = Î 2 Å Line integrls First onsider work by fore fields. uppose prtile move long urve while ted upon by fore F. If is stright line segment given by the vetor d nd F is onstnt fore, then the work is, by definition F d = mgnitude of fore displement in the diretion of fore If the pth is urved, we brek the urve into smll piees nd dd the work t eh piee then tke the limit. o the work is defined by n 1 lim n i=0 F((t i )) [(t + t) (t)] = b Here (t)dt (t + t) (t) represent line segment. The work done by the fore is the following sum. k F((t i )) ((t i ) (t i 1 )) i=1 F((t)) (t)dt As P 0 the sum is k F((t i )) (t i ) t i i=1 the limit is b F((t)) (t)dt.
5 6 7 ì INTEGRAL OVER PATH AN URFACE We use the nottion: In the limit this beomes ds. s = (t + t) (t) (t) t efinition 2.1. Let F be vetor field on R 3 tht is ontinuous on the C 1 - pth : [,b] R 3. efine the line integrl F ds = b F((t)) (t)dt A nie interprettion in terms of slr integrl is s follows: For (t) 0, we see, if T(t) = (t)/ (t) is the unit tngent vetor, then F ds = = = = b b b C F((t)) (t)dt [ F((t)) ] (t) (t) dt (t) [F((t)) T(t)] (t) dt F((t)) T(t)ds. o the line integrl is like the pth integrl of the tngentil omponent: F((t)) T(t) long. Another Nottion for Line integrl Let us write (t) = (x(t),y(t),z(t)) nd F = (F 1,F 2,F 3 ). Then ds = (dx,dy,dz) = ( dx dt, dy dt, dy dt )dt. o the integrl Exmple 2.2. F ds = = = b (F 1,F 2,F 3 ) (dx,dy,dz) F 1 dx + F 2 dy + F 3 dz ( dx F 1 dt + F dy 2 dt + F 3 x 2 dx + xydy + dz = where (t) = (t,t 2,1) = (x,y,z) on [0,1]. ) dy dt dt
6 2 LINE INTEGRAL 7 Exmple 2.3. os zdx + e x dy + e y dz where (t) = (1,t,e t ) on [0,2]. Exmple 2.4. zdx + os zdy (xy) (sin 1/3 dz) = 1 2 where x = os 3 θ, y = sin 3 θ,z = θ on [0,7π/2]. rw the urve in R 3. Exmple 2.5. uppose F(x,y,z) = x 3 i+yj+zk nd prmeterize the irle x = 0, y = os θ, z = sin θ, 0 θ 2π (t) = (0, sin θ,os θ) ine F((θ)) (θ) = 0, the work must be zero. You n verify by finding the vlue. Reprmetriztion The line integrl depends not only on F but lso depends on the pth. If 1, 2 re two different prmetriztion of the sme urve, we shll see F ds = ± F ds 1 2 efinition 2.6. Let h : I I 1 be rel vlued C 1 urve tht is one-to-one. Let : I 1 R 3 be C 1 urve. Then the omposition p = h : I R 3 is lled reprmetriztion of. Theorem 2.7. Let : [ 1,b 1 ] R 3 be urve nd p: [,b] R 3 is given by p = h where h: [,b] [ 1,b 1 ] stisfies p() = 1, h(b) = b 1 (Orienttion preserving)
7 8 7 ì INTEGRAL OVER PATH AN URFACE C (t) p(t) h(t) b 1 b 1 7.2: Reprmetriztion of urve or h() = b 1, h(b) = 1 (Orienttion reversing) Then we hve p F ds = ± F ds Here we hve + sign, if p is orienttion preserving, nd sign, if p is orienttion reversing. Proof. If h is orienttion preserving then h() = 1, h(b) = b 1. In this se, p F((s)) ds = b F((h(t))) (h(t))h (t)dt Let s = h(t). Then b = F((h(t))) (h(t))h (t)dt h(b) = F((s)) (s)ds h() b1 = F((s)) (s)ds 1 = F((s))ds
8 2 LINE INTEGRAL 9 If p is orienttion reversing, then the integrl beomes 1 = b 1 = F((s)) (s)ds F((s))ds Exmple 2.8. (1) Given : [,b] R n. As typil exmple, onsider op : [,b] R n defined by op = ( + b t): [,b] R n. op is lled opposite pth. This is orienttion reversing. We see b F ds = F ( op ) (t)dt op b = F (b + t)( 1)dt = F (u)du b b = F (u)du = F ds (2) The pth p;[0,1] R 3 given by p(t) = ( + (b )t) is n orienttion preserving reprmetriztion. Exmple 2.9. Find the line integrl of f(x,y,z) = x 2 y 2 z 2 on C. C = {(x,y,z): x 2 + y 2 + z 2 = 2, x 2 + y 2 = z 2, z > 0}
9 10 7 ì INTEGRAL OVER PATH AN URFACE sol. ubtrting we see z 2 = 1. o z = 1. A prmetriztion of C is (t) = (os t,sin t,1), t [0,2π] (t) = ( sin t,os t,0) Hene the integrl is f ds = = = 1 8 = π 4 2π 0 2π 0 2π 0 f((t)) (t) dt os 2 t sin 2 t dt (1 os 4t)dt The line integrl is n oriented integrl, in the sense tht hnge of sign ours if the orienttion is reversed. The pth integrl does not hve this property. Theorem 2.10 (Pth integrl is independent of prmetriztion). If nd p re two prmetriztion of pieewise C 1 -urve C, nd f is ny rel vlued ontinuous funtion, then f(x,y,z)ds = f(x,y,z)ds p Let (t) : [,b] R n. As n exmple, let p defined by p(t) = ( + b t). Then p Here u = b + t. f ds = b b = = = = b b f( (t)) ( ) (t) dt f((b + t)) (b + t)( 1) dt f((b + t)) (b + t) dt f((u)) (u) ( 1)du f ds
10 2 LINE INTEGRAL 11 Exmple Find pth integrl of f(x,y,z) = x 2 + y 2 + z 2 over C. C = {(os t,sin t,t): t [0,2π]} {(1,0,t): t [0,2π]} sol. C is the union of C 1 nd C 2. C 1 = {(os t,sin t,t): t [0,2π]}, C 2 = {(1,0,t): t [0,2π]} We prmeterize C 1 nd C 2 s follows: 1 = (os t,sin t,t) t [0,1], 2 = (1,0,t) t [0,2π] Then C f ds = f ds + f ds C 1 C 2 = f ds + f ds 1 2 2π 2π = (1 + t 2 ) 2dt + (1 + t 2 )dt 0 0 = (1 + ( ) 2) 2π + 8π 3 /3 Line integrls of Grdient Fields A vetor field F is lled grdient vetor field if F = f for some rel vlued funtion f. Thus F = f x i + f y j + f z k. Theorem uppose f : R 3 R is lss C 1 nd : [,b] R 3 is smooth. Then f ds = f((b)) f(()). Proof. Apply hin rule to f((t)) (f ) (t) = f((t)) (t)
11 12 7 ì INTEGRAL OVER PATH AN URFACE o f ds = b f((t)) (t)dt = b f (u)du = f((b)) f(()). o the line integrl is independent of prmetriztion. Exmple Let (t) = (t 4,sin 3 (tπ/2),0), t [0,1]. Evlute ydx + xdy whih mens ydx + xdy + 0dz. sol. We reognize (y,x,0) s grdient of f(x,y,z) = xy. Hene the vlue is f((1)) f((1)) = 1 0 efinition We sy urve is simple if it is the 1-1 imge of pieewise C 1 mp : I R 3. A simple urve is one whih does not interset itself. If I = [,b], then (), (b) re lled end points of the urve. Eh simple urve hs two orienttions. (From P to Q) It is oriented or direted urve. efinition If is 1-1 exept t end points nd, () = (b), it is simple losed urve if it is () = (b), but not 1-1, then it is losed urve. Line integrls over oriented simple urves uppose is ny orienttion preserving prmetriztion of C, then the line integrl is independent of prmetriztion: Hene we n define C F ds = F ds If C is the sme urve s C but with the opposite orienttion. Then C F ds = F ds C Line integrls over urves with severl omponents Let C be n oriented urve whih is mde up of severl oriented urves C i,i = 1,2, ine eh C i n be prmeterized seprtely, we n show tht the
12 2 LINE INTEGRAL 13 integrl stisfies F ds = C F ds + C 1 F ds + + C 2 F ds C k Thus the following sum of oriented urves mkes sense. C = C 1 + C C k C 3 C 2 C 1 7.3: um of severl urves Exmple Find the line integrl of F(x, y, z) = xi + yj + zk over. (t) = (os t,sin t,t), t [0,2π] sol. We hve (t) = ( sin t,os t,1) nd F((t)) = os ti+sintj+tk. Hene F((t)) (t) = os t( sin t) + sin t os t + t = t F ds = 2π 0 t dt = 2π 2 More exmples in the book. The nottion dr for line integrls ometimes we use the nottion for line integrl: F dr C
13 14 7 ì INTEGRAL OVER PATH AN URFACE Here r denotes the position vetor r = xi + yj + zk. C F dr = b F(r(t)) dr dt dt Î 3 Å Prmeterized urfes Grphs re too restritive. ee the exmple of surfe in the book. Or simply sphere or torus. Those re importnt exmples of figures tht rise often in rel life. But those figures nnot be represented s the grphs of funtions. 7.4: A surfe tht is not the grph of funtion efinition 3.1. A prmeteriztion of urfe is funtion Φ: R 2 R 3 where is domin in R 2. The surfe orresponding to the funtion Φ is the imge = Φ(). Φ(u,v) = (x(u,v),y(u,v),z(u,v)) If Φ is differentible or C 1, then we sy is differentible or C 1 -surfe. The grph of funtion is speil se.
14 3 PARAMETERIZE URFACE 15 Φ(u, v) v y x u 7.5: A prmetriztion Tngent Vetors nd Tngent Plne to urfe Consider the mpping Φ: R 3 through (x,y,z) = Φ(u,v). First look t the se the surfe is the grph of f : R we n write Φ: R 3 s Φ(x,y) = (x,y,f(x,y)) First fix x = x 0 nd then y = y 0. Then tngent vetor long y-xis nd x-xis t Φ(x 0,y 0 ) = (x 0,y 0,f(x 0,y 0 )) is Φ x (x 0,y 0 ) = i + f x (x 0,y 0 )k, Φ y (x 0,y 0 ) = j + f y (x 0,y 0 )k Hene tngent plne is perpendiulr to the norml vetor given by the ross produt Φ x (x 0,y 0 ) Φ y (x 0,y 0 ) = (i + f x (x 0,y 0 )k) (j + f y (x 0,y 0 )k) i j k = 1 0 f x (x 0,y 0 ) 0 1 f y (x 0,y 0 ) = f x (x 0,y 0 )i f y (x 0,y 0 )j + k In generl, we see two tngent vetors re T u = Φ u = x u i + y u j + z u k (u0,v 0 )
15 16 7 ì INTEGRAL OVER PATH AN URFACE This is obtined by fixing u 0 imilrly, If the norml vetor T v = Φ v = x v i + y v j + z v k n = Φ u Φ v = Φ u Φ v (u0,v 0 ) is nonzero, then we see the surfe is smooth. o ll it regulr. efinition 3.2. ine n is norml to tngent plne the eqution of tngent plne t Φ(u 0,v 0 ) = (x 0,y 0,z 0 )is given by n (x x 0,y y 0,z z 0 ) = 0 Or if n = (n 1,n 2,n 3 ) then the tngent plne is n 1 (x x 0 ) + n 2 (y y 0 ) + n 3 (z z 0 ) = 0 Exmple 3.3 (Prboloid). Consider Find tngent plne t Φ(1,0). x = uos v, y = usin v, z = u 2 + v 2 sol. Φ(u,v) = (uos v,usin v,u 2 + v 2 ). o T u = (os v,sin v,2u), T v = ( usin v,uos v,2v) o T u T v = ( 2u 2 os v+2v sinv, 2u 2 sin v 2v os v,u). At Φ(1,0) = (1,0,1), n = T u T v (1,0) = ( 2,0,1) Exmple 3.4 (Cone). Consider Is ti regulr, differentible? Φ(u,v) = (uos v,usin v,u), u 0 sol. T u = (os v,sin v,1), o T u T v = 0 t (0,0,0). T v = ( usin v,uos v,0)
16 3 PARAMETERIZE URFACE 17 Exmple 3.5. Find prmetriztion of the following hyperboloid of one sheet. x 2 + y 2 z 2 = 1 sol. ine the grph is symmetri in x nd y it is nturl to use polr oordinte x = os θ, y = sinθ Then we hve r 2 z 2 = 1 Thus we use r = osh u, z = sinhu to get x = osh uos θ, y = osh usin θ, z = sinhu so Φ(u,θ) = (x(u,θ),y(u,θ),z(u,θ)) = (osh uos θ,osh usin θ,sinhu) Here 0 θ 2π, < u < Φ(u,θ) belongs to the surfe sine os 2 θ + sin 2 θ = 1, osh 2 θ sinh 2 θ = 1 Exmple 3.6. Find eqution of tngent plne to Φ t (1,0,1). Φ(u,v) = (uos v,usin v,u 2 ), 0 v 2π, < u <
17 18 7 ì INTEGRAL OVER PATH AN URFACE sol. Φ(1,0) = (1,0,1). Find norml vetor n t (1,0,1). n = Φ u (1,0) Φ v (1,0) = (os vi + sinvj + 2uk) ( usin vi + uos vj) (u,v)=(1,0) = (i + 2k) j = 2i + k Then eqution is 2(x 1) + (z 1) = 0 Î 4 Å Are of Prmeterized urfe Find the re of U = Φ() where Φ: R 3 is surfe prmetriztion. ivide into smll retngles. Consider smll retngle R = [u,u + u] [v, u + v] The re of smll retngle under Φ is pproximted by prllelogrm by Φ(u,v), Φ(u + u,v), Φ(u,v + v), Φ(u + u,v + v). Two sides re given by Φ u (u,v) u nd Φ v (u,v) v. (Fig 7.6) Hene the re of Φ(R) is T u = Φ u = x u i + y u j + z u k T v = Φ v = x v i + y v j + z v k T u T v u v Hene Are of surfe is the limit of the following Tu T v u v efinition 4.1. We define the surfe re A() of prmeterized surfe by A() = Lter we shll lso denote it by T d. T u T v dudv
18 4 AREA OF PARAMETERIZE URFACE 19 v z Φ() (u, v+ v) (u+ u, v+ v) Φ u u R Φ Φ(u, v) Φ v v (u, v) (u+ u, v) O u x O y 7.6: Are of surfe Let Φ(u,v) = (x(u,v),y(u,v),z(u,v)). We define d by d = Φ u Φ v dudv, nd ll the (re element). Rell the nottion (x, y) (u,v) = x u y u Then we see (Not trivil. Need justifition). 1 Φ d = = Φ u Φ v dudv [ (y,z) ] 2 + (u, v) x v y v [ ] (z,x) 2 + (u, v) Are of surfe is independent of prmetriztion. [ ] (x,y) 2 dudv (u, v) Exmple 4.2 (Cone). Let be given by x = r os θ, y = r sinθ, z = r. sol. Either use formul bove or T r T θ drdθ. T r T θ = r 2 1 Φ is ssumed to be 1-1.
19 20 7 ì INTEGRAL OVER PATH AN URFACE Exmple 4.3 (Helioid). Let the helioid given by x = r os θ, y = r sin θ, z = θ. nd let be the region where 0 θ 2π nd 0 r 1. Find the re. T r T θ = r Need tble to see π[ 2 + log(1 + 2)]. urfe Are of Grph When surfe U is given by the grph of funtion z = f(x,y) on, we see U is prmeterized by Φ(x,y) = (x,y,f(x,y)). Find Φ x Φ y by Φ x = i + f x k, Φ y = j + f y k ine Are is Φ x Φ y = (i + f f k) (j + k) = f x y x i f y j + k, Φ d = The unit norml vetor n(x,y,z) on U is ( ( f/ x) 2 + ( f/ y) 2 + 1) 1/2 dxdy n = n(x,y,z) = f x i f y j + k We n find the formul using the ngle between n nd k. Let ϕ be the ngle between n nd k. Then os ϕ stisfies os ϕ = n k n = 1 ( f/ x) 2 + ( f/ y) Hene nd we get d = ( f/ x) 2 + ( f/ y) 2 + 1dxdy = dxdy os ϕ, Φ d = dxdy os ϕ Exmple 4.4. Find the surfe re of unit bll.
20 4 AREA OF PARAMETERIZE URFACE 21 sol. From x 2 + y 2 + z 2 = 1, we let z = f(x,y) = 1 x 2 y 2. f x = x 1 x 2 y 2, f y = y 1 x 2 y 2 Are of hlf sphere is Φ d = = 2π 1 0 = 2π 1 1 x 2 y 2dxdy 0 r 1 r 2 drdθ urfe of revolution The lterl surfe re generted by revolving the grph y = f(x) 0 is b A = 2π y 1 + (f (x)) 2 dx Exmple 4.5. Use prmetriztion to express the re generted by revolving the grph y = f(x). We n hoose the prmetriztion Φ(u,v) = (x,y,z) = (u,f(u)os v,f(u)sin v) over the region u b, 0 v 2π.
21 22 7 ì INTEGRAL OVER PATH AN URFACE sol. Find the derivtives (y, z) (u,v) = f(u)f (u), Hene the re is (z, x) (u,v) = f(u)os v, (x,y) (u,v) = f(u)sin v Φ d = = = [ (y,z) ] 2 f(u) + (u, v) f(u) 1 + [f (u)] 2 dudv b 2π = 2π 0 b f(u) 1 + [f (u)] 2 dvdu f(u) 1 + [f (u)] 2 du [ ] (z,x) 2 + (u, v) [ ] (x,y) 2 dudv (u, v) This formul oinide with erlier formul. Î 5 Å Integrls of lr funtions over urfe Integrls of lr funtions over urfe Let Φ: R 3 be prmeterized surfe = Φ() nd let f : R be rel vlued funtion defined on Φ. First, if f = 1, it represents the re. d = Φ u Φ v dudv, In generl, the integrl of f on f d is defined by is efinition 5.1. f d = f(φ(u,v)) Φ u Φ v dudv If the surfe is prmeterized by Φ(u,v) = (x(u,v),y(u,v),z(u,v)) then the integrl beomes f d = f(φ(u,v)) T u T v dudv [ (y,z) ] 2 f(x(u,v),y(u,v),z(u,v)) + (u, v) [ ] (z,x) 2 + (u, v) [ ] (x,y) 2 dudv (u, v)
22 5 INTEGRAL OF CALAR FUNCTION OVER URFACE 23 If Φ = the sum of Φ 1,Φ 2,...,Φ m, f d = Φ f d + Φ 1 f d + + Φ 2 f d Φ m urfe integrls over grphs uppose is the grph of C 1 funtion z = g(x,y). Then we prmeterize it by x = u, y = v, z = g(u,v) nd o T u T v = f(x,y,z)d = 1 + ( g u )2 + ( g v )2 f(x,y,g(x,y)) 1 + ( g x )2 + ( g y )2 dxdy Exmple 5.2. Let be defined by z = x 2 + y, where is 0 x 1, 1 y 1. Find xd sol. xd = x 1 + ( g x )2 + ( g 1 1 y )2 dxdy = x 1 + 4x 2 + 1dxdy 1 0 = 1 1 [ 1 ] (2 + 4x 2 ) 1/2 (8xdx) dy = [(2 + 4x 2 ) 3/2] dy 1 0 = Exmple 5.3. Evlute z2 d when is the unit sphere.
23 24 7 ì INTEGRAL OVER PATH AN URFACE sol. Use spheril oordinte for x 2 +y 2 +z 2 = 1. Then ρ = 1 nd z 2 = os 2 φ. z 2 d = os 2 φ T θ T φ dθdφ ine T θ T φ = sinφ 2π π f(x,y,z)d = os 2 φsin φdφdθ 0 0 = 4π 3 Integrl over Grphs We show f(x,y,z)d = f(φ(u,v)) T u T v dudv = f(x,y,g(x,y)) dxdy os θ where θ is the ngle between norml vetor nd k vetor. = A/os θ. N k θ 7.7: Rtio between two surfe os θ = N k N = 1 ( g/ x) 2 + ( g/ y) Exmple 5.4. Compute xd where is tringle with verties (1,0,0), (0,1,0) nd (0,0,1).
24 6 URFACE INTEGRAL OF VECTOR FIEL 25 k n θ sol. os θ = n k = 1/ 3. 3 xdxdy = x 0 0 xdydx = 3 6 Exmple 5.5. Let Φ = (r os θ,r sinθ,θ) be the helioid where 0 r 1. uppose hs density m equl to twie the distne to the entrl xis, m = 2 x 2 + y 2 = 2r. Totl mss of the surfe. sol. M = But T r T θ = 1 + r 2. Hene 2rd = 2 r T r T θ drdθ M = 2π r 1 + r 2 drdθ = 4 3 π(23/2 1). Î 6 Å urfes Integrls of vetor Fields This setion we develop the notion of integrl of vetor field over surfe. Rell the line integrl of vetor field hs physil interprettion: Work. imilrly, the notion of integrl of vetor field over surfe is Flux. uppose F represent the veloity of fluid.(like river) Then you ple net into the wter nd imgine the mount of wter tht psses through your net per unit time(= rte t whih wter pss through the net) Let F: V R 3 defined over the surfe = Φ(), Φ: R 3 we define Φ F d
25 26 7 ì INTEGRAL OVER PATH AN URFACE efinition 6.1. F d = Φ Note we used the nottion Φ insted of. F(Φ(u,v)) (Φ u Φ v )dudv If we let n = Φ u Φ v / Φ u Φ v be the unit norml vetor to the surfe, then F d = Φ (F n) (Φ u Φ v )dudv = F nd. Hene if F hs sme diretion s n = Φ u Φ v / Φ u Φ v, i.e, ( F(x,y,z) = f(x, y, z)n for some slr funtion f) then f(x,y,z)d = F(x,y,z) d Φ Φ Exmple 6.2. Exmple 6.3 (pheril oordinte). Let be the unit sphere prmeterized by : 0 θ 2π, 0 φ π Φ: R 3, Φ(θ,φ) = (os θ sin φ,sin θ sin φ,os φ) Compute r d where r = xi + yi + zk denotes the position vetor.
26 6 URFACE INTEGRAL OF VECTOR FIEL 27 sol. Φ θ = sin θ sin φi + os θ sinφj Φ φ = os θ os φi + sin θ os φj sin φk Now ompute Φ θ Φ φ : Φ θ Φ φ = sinφ(os θ sin φi + sin θ sin φj + os φk) = ( sin φ)r(φ(θ,φ)) Hene r d = ( sin φ)r rdθdφ Φ = ( sin φ)dθdφ = 4π Orienttion As with line integrl, the surfe integrl lso hs the notion of diretion. First we need to define the orienttion of surfe. It depends on the prtiulr prmetriztion. efinition 6.4 (Oriented urfe). Oriented urfe is two sided surfe with one side speified s outside(or positive side) At eh point there re two unit norml vetors n 1 nd n 2, where n 1 = n 2. Eh of these normls n be ssoited with one side of the surfe. (orienttion) orientble surfe. For orientble surfe, there re two possible norml vetors. There re nonorientble surfe.(mobius strip) Let Φ: R 3 represent oriented surfe. If n(φ) is the unit norml to, then n(φ) = ± Φ u Φ v Φ u Φ v A prmetriztion is lled Orienttion-preserving if T u T v T u T v = +n(φ)
27 28 7 ì INTEGRAL OVER PATH AN URFACE Otherwise, it is Orienttion-reversing. Exmple 6.5. The prmetriztion of sphere by spheril oordinte is orienttionreversing. By hnging the order of θ nd φ, we n get orienttion-preserving prmetriztion. U n 1 n 2 ounter-lokwise lokwise n 1 7.8: lokwise, ounter-lokwise (n 1 nd n 2 re normls the orienttion points) : Möbius strip Orienttion of grph Exmple 6.6. Let be the grph of funtion z = g(x,y). Then the unit norml
28 6 URFACE INTEGRAL OF VECTOR FIEL 29 The norml is given by n = g x i g y j + k ( ) 1 + g 2 ( ) dxdy x + g 2 y We n give orienttion of suh surfe by tking the positive side to be the side wy from whih n points. (Fig 7.7) Independene of prmetriztion Theorem 6.7. Let be n oriented surfe nd let Φ 1, Φ 2 be two regulr orienttion preserving prmetriztion, then for ontinuous F defined on, then F d = Φ 1 F d Φ 2 If one of them is orienttion reversing, then F d = Φ 1 F d Φ 2 For slr f, we hve for ny Φ 1,Φ 2 fd = Φ 1 fd Φ 2 If Φ onsists of Φ 1, Φ 2,..., Φ m F d = Φ F d + Φ 1 F d + + Φ 2 F d Φ m Hene we n define the sum of surfes s Φ = Φ 1 + Φ Φ m n = f/ xi f/ y j + k ( f/ x) 2 + ( f/ y) urfe U is given by Φ(x,y) = (x,y,f(x,y)). If F = F 1 i + F 2 j + F 3 k is
29 30 7 ì INTEGRAL OVER PATH AN URFACE vetor field then F d is expressed s Φ F d = = Reltion with slr integrls Rell efinition 6.8. If we write Φ F d = F (Φ x Φ y )dxdy ( [F 1 f ) ( + F 2 x f y ) + F 3 ] dxdy F(Φ(u,v)) (Φ u Φ v )dudv n = (Φ u Φ v )/ Φ u Φ v, d = (Φ u Φ v )dudv, d = nd then it n be written s F d = F nd Φ Φ Physil Interprettion of urfe Integrls Consider the prllelogrm determined by three vetors F, T u u nd T v v. The volume is F (T u u T v v) = F (T u T v ) u v If F is veloity of fluid, the volume is the mount of fluid to flow ourwrd the surfe per unit time. Hene F d is the net quntity of fluid to flow ross the surfe per unit time, i.e, the rte of fluid flow. It is lso lled flux of F ross. Exmple 6.9 (Het flow). Let T denote the temperture t point. Then T = T x i + T y j + T z k
30 6 URFACE INTEGRAL OF VECTOR FIEL 31 F v Φ(u,v) d U =Φ() v v d 7.10: Are of shdow region nd flux ross represent the temperture grdient nd het flows with the vetor field k T. Exmple uppose temperture is T = x 2 + y 2 + z 2 on where : x 2 + y 2 + z 2 = 1. Find the flux ross if k = 1 sol. We hve F = 2r. Then r n = 2. o F d = 2 d = 8π Exmple 6.11 (Guss Lw). The sum of the flux of n eletri field E over losed surfe is the net hrge Q ontined in the surfe. Nmely, E d = Q
31 32 7 ì INTEGRAL OVER PATH AN URFACE 7.11: Wter through pipe nd surfe uppose E = En(onstnt multiple of norml vetor) then E d = Ed = Q = EA() o If is sphere of rdius R then E = Q A() E = urfe Integrl over Grphs uppose is the grph of z = g(x,y). We show F d = F (T x T y )dxdy = Q 4πR 2 (7.1) [ F 1 ( g x ) + F 1( g ] y ) + F 3 dxdy We prmeterize the surfe by Φ(x, y) = (x, y, g(x, y)) nd ompute T x = i + g x k, T y = j + g y k Hene nd we proved the formul. Exmple The eqution T x T y = ( g )i ( g x y )j + k z = 12, x 2 + y 2 25
32 6 URFACE INTEGRAL OF VECTOR FIEL 33 desribes disk of rdius 5 lying on the plne z = 12. Compute r d when r = xi + yj + zk. sol. T x T y = i j = k o r (T x T y ) = z nd r d = zdxdy = 12A() = 300π. ummry (1) Prmeterized urfe Φ(u, v) () Integrl of slr f: fd = f(φ(u,v)) T u T v dudv Φ (b) lr surfe element: d = T u T v dudv () Integrl of vetor field: F d = F(Φ(u,v)) (T u T v )dudv Φ (d) Vetor surfe element: d = (T u T v )dudv = nd (2) Grph z = g(x, y) () Integrl of slr f: fd = f(x,y,g(x,y)) dxdy os θ
33 34 7 ì INTEGRAL OVER PATH AN URFACE (b) lr surfe element: d = dxdy os θ = () Integrl of vetor field: F d = (d) Vetor surfe element: (3) phere x 2 + y 2 + z 2 = R 2 ( g ) 2 + x d = nd = ( ) g dxdy y ( ) g F 1 x F g 2 y + F 3 dxdy ( g x i g y j + k ) dxdy () lr surfe element: d = R 2 sin φdφdθ (b) Vetor surfe element: d = (xi + yj + zk)r sin φdφdθ = rr sin φdφdθ = nr 2 sin φdφdθ Î 7 Å Applition to ifferentil Geometry Curvture Let Φ: R 3 be surfe prmetriztion of. Then T u = Φ u, T v = Φ v re tngent vetors. Assume the normls re well defined so tht T u T v 0. Let E = T u 2, F = T u T v, G = T v 2 We n show tht T u T v 2 = EG F 2
34 7 APPLICATION TO IFFERENTIAL GEOMETRY 35 If we let N = T u T v T u T v = T u T v EG F 2 denote the unit norml vetor to the surfe t point p = Φ(u, v). We define Guss urvture K(p) nd men urvture H(p). To define these first define three new funtions l,m,n on : l(p) = N(u,v) 2 Φ u = N(u,v)Φ 2 uu m(p) = N(u,v) 2 Φ u v = N(u,v)Φ uv (7.2) n(p) = N(u,v) 2 Φ v = N(u,v)Φ 2 vv The Guss urvture K(p) is given by K(p) = nd men urvture H(p) is given by ln m2 EG F 2 (7.3) H(p) = Gl + En 2Fm 2(EG F 2 ) (7.4) 7.1 Guss-Bonnet Theorem The Guss urvture to the sphere of rdius R is 1/R 2. The Guss urvture is in generl funtion. We onsider For the sphere, we hve 1 R 2 K da da = 4π Guss-Bonnet found out tht this quntity is preserved for ny sphere like objet. This is topologil invrint. Now onsider torus like objet. (This hs one hndle) In generl onsider n objet with g-hndles. Then K da = 4π(1 g)
35 36 7 ì INTEGRAL OVER PATH AN URFACE
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