ES.182A Topic 40 Notes Jeremy Orloff

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1 ES.182A opic 4 Notes Jeremy Orloff 4 Flux: ormal form of Gree s theorem Gree s theorem i flux form is formally equivalet to our previous versio where the lie itegral was iterpreted as work. Here we will use a lie itegral for a differet physical quatity called flux. he formal equivalece follows because both lie itegrals are ultimately writte i terms of M, N, dx, dy, which do t eed ay physical iterpretatio. At the ed of these otes, for your ejoymet, we will give physical iterpretatios of curlf ad (a ew gadget) divf. 4.1 he lie itegral for flux Here we ll give the lie itegral for flux without ay motivatio. Below we ll give a physical iterpretatio that shows it arises aturally. Suppose is a curve i the plae traversed i a specified directio. Suppose F = P, Q is a vector field. he flux of F across is defied as flux = P dy Q dx (1) Itrisic formula for flux ecall the itrisic formula for work: F ds, where is the uit taget vector to ad ds is the differetial of arclegth. As with the work itegral, we have a itrisic formula for flux. First we eed to defie : at each poit of, is the uit ormal vector, foud by rotatig clockwise by 9. Usig, the itrisic formula for flux is he uit ormal is rotated 9 clockwise. flux = F ds. (2) Proof. We eed to show that the two formulas for flux give the same value. We kow that, ds = dr = dx, dy. Sice is rotated 9 clockwise, ds is ds rotated 9 clockwise. hat is ds = dy, dx. 1

2 4 FLUX: NOMAL FOM OF GEEN S HEOEM 2 hus F ds = P, Q dy, dx = P dy Q dx. his shows that the itegrad i Equatios 1 ad 2 are the same. QED We will give a physical motivatio for flux below. Example 4.1. ompute the flux of F = x 2 i + y j across the lie segmet from (, ) to (1, 2). y (1,2) x Parametrize the curve as x = x, y = 2x, x from to 1. Writig F = P, Q, we have i terms of x: So, flux = P dy Q dx = dx = dx, dy = 2 dx, P = x 2, Q = y Gree s theorem i flux form x 2 2dx 2x dx = = 1 3. heorem. Gree s theorem i flux or ormal form. Suppose that is a simple closed curve surroudig a regio. We assume is positively orieted. ( is to the left as you traverse the curve.) Also, suppose that F = P, Q is a vector field, which is differetiable o all of. I flux form, Gree s theorem says P dy Q dx = P x + Q y dx dy. (3) We ca also write this i vector form. I two dimesios we defie the divergece by divf = P x + Q y. Note, this takes a vector field ad produces a regular fuctio. Usig the itrisic formula for flux Equatio 3 ca be writte as F ds = divf dx dy. (4) Proof. Gree s theorem writte with differetials does t eed ay physical iterpretratio. So, the proof of the ormal form is just traslatig the origial work form of Gree s theorem to the ew otatio. he origial form says M dx + N dy = M x N y dx dy If we let N = P ad M = Q the this becomes Q dx + P dy = M dx + N dy = N x M y dx dy = P x + Q y dx dy. Before workig more examples, let s take the time to gai a physical ituitio for flux. QED

3 4 FLUX: NOMAL FOM OF GEEN S HEOEM Flux Flux meas flow, as i the Eglish word iflux. Newto s origial ame for the derivative was fluxio. Sice our ituitio is three dimesioal we ll start by describig flux i 3D Flux i 3D he figure below shows rai fallig straight dow. here are two widows i the rai. he orage widow o the left is orieted horizotally ad the blue widow o the right is orieted vertically. z y x wo widows i the rai he flux through each widow is the rate i volume/time that water pours through the frame of the widow. It should be clear that the flux through the left had widow is ot zero. It should also be clear that the flux through the right had widow is zero, that is o water passes through the frame of the widow. I geeral, the flux will deped o the orietatio ad area of the widow ad the velocity of the water Flux i 2D he idea is the same as i 3D. he figure below shows rai fallig straight dow i a two dimesioal world. I such a world a widow is a lie. he figure shows 3 widows with differet orietatios. y hree 2D widows i the rai x

4 4 FLUX: NOMAL FOM OF GEEN S HEOEM 4 I this case the flux is rate i area/time that water pours through the frame of each widow. Lookig at the above figure, it should be clear that the flux is greatest for the horizotal widow, is for the vertical widow, ad is somewhere i betwee for the agled widow. I geeral, the flux will deped o the orietiatio ad legth of the widow ad the velocity of water Basic formula for flux i 2D As usual, we start by fidig a basic formula for a costat velocity vector field ad a curve which is a straight lie. We will also be careful with orietatio, so that we ca say whether the flux is positive or egative. he figure below shows water flowig straight dow at a costat velocity v ad a segmet of legth s. I time t the water that flows across the segmet fills up the parallelogram. he uit ormal is perpedicular to the segmet. s v θ θ Altitude = v t cos(θ) = v t Area = v t cos(θ) s = v t s v t Flux = area/time I time t, the water flows a distace v t. his is the legth of each vertical segmet. ake the segmet of legth s as the base of the parallelogram. he altitude is the parallel to the ormal. learly it has legth v t cos(θ) = v t. (Do t forget is a uit vector.) hus, the area of the parallelogram is base height = v t s. he flux is area per uit time, i.e. flux = his is our basic formula for flux. v t s t = v s. Eve if the vector field is ot a velocity field we ca use this formula to defie the flux of the field across the segmet Negative flux he followig picture shows water flowig straight dow, but this time the segmet is orieted so that whe we rotate its taget vector 9 clockwise the uit ormal poits upward. his meas that the water us flowig agaist the ormal ad the flux should be egative. his is reflected i the fact that v <.

5 4 FLUX: NOMAL FOM OF GEEN S HEOEM 5 s v v t Flow agaist the ormal has egative flux I geeral, if you face i the directio the curve is traversed the the ormal will be o your right ad positive flux will be for a field that flows from your left to your right across the curve. 4.4 Itegral formula for flux i 2D Oce we have the basic formula, it is a small step to get the lie itegral for flux. s 5 s 4 F 5 5 F 3 F 4 F 1 F 2 s 4 3 s s he figure shows a curve divided ito small segmets. We ca use the basic formula to approximate the total flux of the field F across : otal flux 5 F i i s i. i=1 As usual, we shrik the segmets to ifiitesimal legth ad the approximate sum becomes a exact itegral: otal flux = F ds. his is exactly the itrisic formula i Equatio Examples Example 4.2. ompute the flux of F = y i + x j (field of a rotatig rigid fluid) across a circle of radius a ad ceter.

6 4 FLUX: NOMAL FOM OF GEEN S HEOEM 6 y a x Write F = P, Q = y, x. Method 1: Direct computatio, x = a cos(t), y = a si(t), t 2π. Flux = = = F ds = P dy Q dx = y dy x dx a si(t)(a cos(t)) dt a cos(t)( a si(t)) dt a 2 ( si(t) cos(t) + cos t si(t)) dt =. Method 2: Use Gree s theorem. divf = P x + Q y = +. So, flux = F ds = divf dx dy =. Method 3: Use the itrisic formula. F is always taget to, so F =. hus, flux = F ds =. ( x 2 + y 2 ) Method 4: Exact differetials. F ds = P dy Q dx = x dx y dy = d. 2 Sice this is a exact differetial its itegral over ay closed curve is. So the flux =. Example 4.3. Let F = x 2, 2xy. Use Gree s theorem to compute the flux of F across the uit circle. aswer: Sice F is differetiable o the etire uit disk, Gree s theorem tells us that flux = divf dx dy. If F = P, Q the divf = P x + Q y. I this case, divf = 2x 2x =. herefore, the flux is. Example 4.4. Let F = r m x, r m y, with m a positive iteger. Use Gree s theorem to compute the flux of F across the uit circle. aswer: Sice F is differetiable o the etire uit disk, Gree s theorem tells us that flux = divf dx dy. ememberig that r x = x r divf = rm x y x + rm y we have m 1 x2 = mr r +rm m 1 y2 +mr r +rm = 2r m + mrm 1 (x 2 + y 2 ) = (2+m)r m. r

7 4 FLUX: NOMAL FOM OF GEEN S HEOEM 7 By Gree s theorem flux = Example 4.5. Let F = flux = (2 + m)r m dx dy. Usig polar coordiates 2π 1 y r, x r flux of F across the uit circle? (2 + m)r m r dr dθ = 2π.. Why ca t we use Gree s theorem to compute the aswer: Because F is ot differetiable at the origi, i.e. it is ot differetiable o the etire regio iside the uit circle. Example 4.6. Let F = y r, x. Use the itrisic formula for flux to compute the flux r of F across the uit circle. aswer: he itrisic formula is flux = F ds. Sice F is a tagetial vector field ad is ormal to the uit circle, their dot product is. hus, the flux =. 4.6 Positive ad egative flux If is a simple closed curve surroudig a regio. If is positively orieted the the ormal poits outward Normals poit outward from a positively orieted closed curve. I this case, if the flux of a field F across is positive, it represets a et flow out of. If the flux is egative it represets a et flow ito. Example 4.7. Water added at origi. Imagie water beig poured o a table at the origi. (Or a sprig bubblig up at the origi.) he water will spread radially outward. Sice water is icompressible the flux across every circle cetered o the origi will be the same. x y osider F = x r 2 i + y r 2 j. (a) Show divf =.

8 4 FLUX: NOMAL FOM OF GEEN S HEOEM 8 (b) Why ca t we use Gree s theorem to show the flux across circles is? (c) ompute the flux of F across the circle of radius a cetered at. (d) Note that F is a possible velocity field for this example. aswer: (a) Agai use r r = x/r, x y = y/r. P = xr 2, so P x = r 2 2xr 3 x r = r2 2x 2 r 4 = x2 + y 2 r 4. Q = yr 2, so Q y = r 2 2yr 3 y r = r2 2y 2 r 4 = x2 y 2 r 4. So divf = P x + Q y =. (b) Sice F has a sigularity at the origi Gree s theorem does ot apply o circles aroud the origi. (c) Sice F is radial, o a circle of radius a, F = F = 1. hus, usig the itrisic a formula 1 2πa flux = F ds = ds = a a = 2π. (d) he field is radial ad the flux across every circle is the same. So, this could represet a velocity field for water pourig oto the origi. 4.7 Physical iterpretatios of curl ad div We start by usig Gree s theorem to give geometric defiitios of curl ad div. First curl: Gree s theorem says F dr = curlf da. If F is a force, we kow that the left-had side is the work doe by F as it pushes a body aroud. Now if, the regio is very small we ca approximate the right-had side as curlf (area of ). hus, curlf work of F aroud. area of As the regio shriks to a poit the approximatio becomes exact, i.e. curlf = lim shriks to a poit he argumet for div usig Gree s theorem is the same. divf = lim shriks to a poit work of F aroud. area of flux of F across. area of Our ext goal is to show that curlf measures the ifiitesimal tedecy of the field to cause rotatio. Suppose F is the velocity field of flowig water. Suppose also that we put a small paddle wheel of radius a i the water at the poit (x, y ).

9 4 FLUX: NOMAL FOM OF GEEN S HEOEM 9 (x, y ) a (x, y ) Paddle wheel i a velocity field op view of paddle wheel i the xy-plae he flowig water will push o the paddles geerally causig the wheel to tur. (If the field is costat the force o diametrically opposed paddles will cacel ad the wheel wo t spi.) Let be a circle of radius a aroud (x, y ). At each poit o the circle the water has a give velocity F. he compoet of the velocity taget to the circle is F. Now, for circular motio we kow that velocity = radius agular velocity. hus, o the circle we ca write v = aω, where ω is the agular velocity of the water. If the velocity field is ot costat, ad if the wheel has a small umber of arms, the its agular velocity will deped o the positio of the paddles i the movig water. I this case, the agular velocity of the paddle will chage as the wheel spis. If there are may (ifiite) arms the agular velocity of the wheel will be the average of the agular velocities of the water over the etire circle. laim. curlf(x, y ) represets twice the average agular velocity of the water aroud a ifiitesimal circle cetered at (x, y ) o led credece to this claim let s look at water flowig i a circle with costat agular velocity ω. For this flow the velocity F(x, y) = ωy, ωx. omputig the curl is easy (remember ω is costat) curlf = N x M y = 2ω. his is exactly as claimed. Now let s look at a arbitrary velocity field F. We ca parametrize the circle as (x, y) = x + a cos(θ), y + a si(θ)). So, ds = a dθ. Let be the regio iside the circle ad suppose a is small. hus 2π 2π F ds = aω(θ) a dθ = a 2 ω(θ) dθ curlf da curlf(x, y ) area of = curlf(x, y ) πa 2. Gree s theorem tells us that the left-had sides of both equatios above are equal, so Dividig both sides by 2πa 2 gives 2π ω(θ) dθ 2π 2π a 2 ω(θ) dθ πa 2 curlf(x, y ). curlf(x, y ). 2

10 4 FLUX: NOMAL FOM OF GEEN S HEOEM 1 he left-had side is exactly the average value of ω(θ) over the circle. If we let a shrik to, the approximatio becomes exact ad the claim is proved.

Math 5C Discussion Problems 2 Selected Solutions

Math 5C Discussion Problems 2 Selected Solutions Math 5 iscussio Problems 2 elected olutios Path Idepedece. Let be the striaght-lie path i 2 from the origi to (3, ). efie f(x, y) = xye xy. (a) Evaluate f dr. olutio. (b) Evaluate olutio. (c) Evaluate