Supplemental Notes: Line Integrals
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- Milton Atkinson
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1 Nottion: Supplementl Notes: Line Integrls Let be n oriented curve prmeterized by r(t) = x(t), y(t), z(t) where t b. denotes the curve with its orienttion reversed mens tke curve 1 nd curve 2 nd put them together. Let Let F = P, Q, R be vector field defined on (nd round). Define s(t) = t r (u) du. This is the rc length function. Note: s(b) is the rc length of. Recll tht T(t) = r (t) is the unit Tngent function. r (t) Also, N(t) = T (t) is the unit Norml nd B(t) = T(t) N(t) is the Binorml. T (t) Differentite the rc length function nd get ds dt = r (t) (by the fundmentl theorem of clculus). Treting ds nd dt like forml vribles, we get ds = r (t) dt. Following the sme convention we mke the following nottionl definitions: ds = r (t) dt = (x (t)) 2 + (y (t)) 2 + (z (t)) 2 dt x = x(t), y = y(t), nd z = z(t) so dx = x (t)dt, dy = y (t)dt nd dz = z (t)dt r(t) = x(t), y(t), z(t) nd so dr = r (t) dt = x (t), y (t), z (t) dt = dx, dy, dz F = F(r(t)) = P (x(t), y(t), z(t)), Q(x(t), y(t), z(t)), R(x(t), y(t), z(t)) So we cn define line integrls: b g(x, y, z) ds = g(r(t)) r (t) dt = F dr = = Note: b P dx + Q dy + R dz = b b = g(x(t), y(t), z(t)) (x (t)) 2 + (y (t)) 2 + (z (t)) 2 dt F(r(t)) r (t) dt P (x(t), y(t), z(t))x (t) + Q(x(t), y(t), z(t))y (t) + R(x(t), y(t), z(t))z (t) dt 1 ds = Arc Length of Remember tht both types of line integrls re independent of prmeteriztion (s long s the orienttion of is preserved) tht is using different prmeteriztion will not chnge the nswer. Also, if the line integrl is done with respect to rc length, then orienttion does not mtter. Wheres, with the other type of line integrl, reversing the orienttion negtes the result. g(x, y, z) ds = g(x, y, z) ds F dr = F dr 1
2 Work nd Flux Suppose F is force field (like the force due to grvity). We know tht work is force dotted with displcement vector (when force is constnt nywy). onsider prticle being moved long n oriented curve. If we focus on smll enough portion of our curve the force should be (pproximtely) constnt. So the work to move the prticle long this smll portion of the curve will be F T s where T is the unit tngent nd s is the length of this prt of the curve. Why T s? Well, since we re focusing on such smll prt of our curve, the curve should look (pproximtely) like stright line, so there is no difference between our curve nd it s tngent moving long the curve = moving long the tngent (pproximtely nywy). Now dding up the work moving long ll smll pieces of our curve we get ΣF T s which trnslted to the world of integrls is... Work = F T ds = F r r r dt = F dr Seeing the ppernce of T in the integrl bove might led you to sk, Wht hppens if we use N (the unit norml) insted? The following integrl is clled the flux of F cross (Note: we cn t get rid of N like we did T): Flux = F N ds Imgine tht F is the velocity of some fluid crossing the curve (pretend is filter or net). Then F N will grb the component of the velocity vector which is perpendiculr to. So it picks out the prt of the fluid crossing. Thus flux is mesuring how much fluid is crossing over the curve. Exmple: See the book for exmples involving work. Let s compute the flux of F(x, y) = 3y 2, 2x cross where is the qurter of the unit circle in the first qudrnt oriented counter-clockwise. is prmetrized by r(t) = cos(t), sin(t) with 0 t π/2 so tht r (t) = sin(t), cos(t) = T(t) since r (t) = 1. Next, T (t) = cos(t), sin(t) so N(t) = 1 T (t) T (t) = cos(t), sin(t). π/2 Flux = F N ds = 3(sin(t)) 2, 2 cos(t) cos(t), sin(t) 1 dt A plot of F (scled down) nd. = 0 π/2 0 3(sin(t)) 2 cos(t) + 2 sin(t) cos(t) dt = (sin(t)) 3 + (sin(t)) 2 π/2 = ( ) ( ) = 2 0 Note: Since the curve is oriented counter-clockwise, its norml vectors point inwrd nd thus we get positive flux. 2
3 enter of mss Suppose we hve wire bent in the shpe of the curve nd suppose this write hs density ρ(x, y, z) t ech point (x, y, z) long the curve. Then if we focus on little segment of the wire where the density is roughly constnt, the mss of the segment of wire will be pproximtely ρ(x 0, y 0, z 0 ) s where s is the length of this piece of the wire. So if we dd up Σρ s we should get the totl mss of the wire (pproximtely nywy). Trnslting to the world of integrls we hve... mss = m = ρ(x, y, z) ds Let x = 1 m xρ(x, y, z) ds, ȳ = 1 m yρ(x, y, z) ds, nd z = 1 m zρ(x, y, z) ds Where x, ȳ, nd z re weighted verges of x, y, nd z coordintes. We cll ( x, ȳ, z) the center of mss of the wire. If ρ(x, y, z) = c = constnt 0, then ( x, ȳ, z) is clled the centroid of. Exmple: Let be the helix prmeterized by r(t) = cos(t), sin(t), 3t for 0 t 2π. Then r (t) = sin(t), cos(t), 3 so r (t) = ( sin(t)) 2 + (cos(t)) 2 + ( 3) 2 = 4 = 2. Thus ds = 2 dt. The rc length of is... 2π 1 ds = 2 dt = 4π 0 Let s compute the centroid of. So let ρ be some non-zero constnt, sy ρ = 1 (the esiest constnt to work with). x = 1 x ds = 1 2π cos(t)2 dt = 0 m 4π 0 ȳ = 1 y ds = 1 2π sin(t)2 dt = 0 m 4π 0 z = 1 z ds = 1 2π 3 2π 3 2π 3 3t2 dt = 2t dt = m 4π 0 4π 0 4π t2 = 0 4π (2π)2 0 = π Notice the helix s z-coordintes rnge from z = 0 to z = 3 2π so z is exctly hlf wy between 0 nd 3 2π (looking t picture of this helix should convince you tht this is the right nswer). Answer: ( x, ȳ, z) = (0, 0, π 3) 3
4 onservtive Vector Fields Let F be vector field defined on R n nd let R be some (open) subset of R n. Here s few definitions: F is grdient vector field if there exists (sclr vlued) function f such tht F = f. f is clled potentil function. We sy n integrl F dr is pth independent if for ny other curve 2 with the sme end points nd orienttion s we hve tht F dr = F dr. 2 F is conservtive vector field on R if F dr is pth independent for ll curves which lie inside the region R. Let nd 2 be curves with the sme strt nd end points (nd the sme orienttion). The curve 1 2 is clled simple closed curve losed becuse it s strt nd end points mtch nd Simple becuse it does not cross itself. Then ssuming pth independence we hve... F dr = F dr F dr F dr = F dr + F dr = 0 F dr = So requiring pth independence is the sme s requiring tht ll line integrls round closed loops turn out to be 0. This explins the nme conservtive since if F is force field nd so these line integrls compute work, then when prticle strts nd ends in the sme plce no work hs been done becuse energy ws conserved. A non-simple closed curve. 4
5 Recll our nottion: is curve prmeterized by r(t) where t b. Let A = r() (the strting point) nd B = r(b) (the ending point). Theorem: (The Fundmentl Theorem of Line Integrls) f dr = f(b) f(a) proof: f dr = b f(r(t)) r (t) dt = b (f r) (t) dt = (f r)(t) b = f(r(b)) f(r()) = f(b) f(a) The second equlity is estblished using the chin rule nd the third equlity is the regulr fundmentl theorem of clculus. The fundmentl theorem of line integrls sttes tht line integrls involving grdient vector fields cn be computed using the endpoints of curves lone! This sys tht line integrls involving grdient vector fields re pth independent. Put nother wy... Theorem: Grdient vector fields re conservtive. The converse of this theorem lso holds, but it is bit hrder to show. Let R be connected open subset of R n. By connected we men tht ny two points in R re connected by curve in R. Open mens tht R hs fuzzy edges. Techniclly we men tht given ny point in R there is disk/bll surrounding tht point (possibly with n extremely smll rdius) which lies entirely inside R. Theorem: If F is conservtive on n (open connected) region R, then F is grdient vector field. proof: We will prove this for vector field F(x, y) = P (x, y), Q(x, y) defined on n open connected subset of R 2, but the theorem still holds in ny dimension (the sme proof works too but the nottion gets bit messy). First, fix point (, b) in R. Let (x, y) be some point in R nd let (x,y) denote some curve from (, b) to (x, y). Note tht such curve exists since R is connected. Let s define our potentil potentil function: f(x, y) = (x,y) F dr. We should mke the following very importnt note: since F is conservtive, the vlue of our integrl is independent of our choice of pth. Thus ny choice of (x,y) will yield the sme vlue f(x, y). In other words, our function f is well defined on ll of R. We wnt to show tht f = F. So we need to show f x = P nd f y = Q. Let s focus on f x = P. First, some prep work. Fix some point (x, y) in R nd let h be rel number smll enough so tht (x + h, y) is in R s well (such numbers exist becuse R is open). Next, let (x+h,y) = (x,y) + h where (x,y) is ny pth from (, b) to (x, y) nd h is the line segment prmeterized by r(t) = t, y where x t x + h. So (x+h,y) is pth from (, b) to (x, y) to (x + h, y). 5
6 f(x + h, y) = F dr = (x+h,y) F dr = (x,y) + h F dr + (x,y) F dr = f(x, y) + h F dr h We will compute the integrl over h by plugging in our prmeteriztion. Remember r(t) = t, y so r (t) = 1, 0 (since y is fixed nd t is our vrible). Thus x+h x+h f(x + h, y) f(x, y) = F dr = P (t, y), Q(t, y) 1, 0 dt = P (t, y) dt h x x Let g(x, y) be n ntiderivtive (integrting with respect to x) of P (x, y) (so g x = P ). Then Finlly, we find tht f(x + h), y) f(x, y) = x+h f x (x, y) = lim h 0 f(x + h, y) f(x, y) h x P (t, y) dt = g(t, y) x+h x = lim h 0 g(x + h, y) g(x, y) h = g(x + h, y) g(x, y) = g x (x, y) = P (x, y) A similr proof shows tht f y (x, y) = Q(x, y). Therefore, f(x, y) = f x (x, y), f y (x, y) = P (x, y), Q(x, y) = F(x, y) nd thus F is grdient vector field on R. Thus, on n (open connected) region R, every grdient vector field is conservtive nd every conservtive vector field is grdient vector field. This leves us with the question, Given vector field F, how cn we tell if F is grdient/conservtive? The following theorem helps us prtilly nswer the question: Theorem: Let F = P, Q nd suppose P nd Q hve continuous first prtils on some region R. If F is grdient on R, then P y = Q x. This lso tells us tht if P y Q x t some point in R then F cnnot be grdient on R. proof: Suppose tht F = f. Then P = f x nd Q = f y. Thus P y = f xy = f yx = Q x (since we ssumed tht these prtils re continuous lirut s theorem pplies). Next, we might sk, Wht bout vector fields in R 3? To discuss these we need to introduce the curl opertor. Definition: Given vector field F = P, Q, R, define curl(f) = F s follows: i j k curl(f) = F = x y z = y z P Q R Q R i x z P R j + x y P Q k = R y Q z, P z R x, Q x P y The nottion indictes tht you re (sort of) crossing the grdient opertor with your vector field. Quick Exmple: Let F(x, y, z) = (xyz, x 2 y, y 2 + z 2 ). Then curl(f) = F =... i j k x y z = y z xyz x 2 y y 2 + z 2 x 2 y y 2 + z 2 i x z xyz y 2 + z 2 j + x y k = 2y 0, xy 0, 2xy xz xyz x 2 y 6
7 An Importnt omputtion: If f hs continuous second prtils, then curl( f) = f = 0. i j k curl( f) = x y z = y z f x f y f z f y f z i x z f x f z j+ x y f x f y k = f zy f yz, f xz f zx, f yx f xy = 0, 0, 0 Since we ssumed tht f hs continuous second prtils, lirut s theorem pplies (mixed prtils re equl). This computtion shows us tht... Theorem: Let F = P, Q, R nd suppose tht P, Q, nd R hve continuous first prtils on some region R. If F is grdient vector field, then F = 0 on R. This lso tells us tht if F 0 t some point in R, then F cnnot be grdient vector field on R. Note: If F(x, y, z) = P (x, y), Q(x, y), 0 where P nd Q do not depend on z, then we hve: F = 0, 0, Q x P y = 0 if nd only if P y = Q x. So our 2-dimensionl theorem is just specil cse of the 3-dimensionl theorem. Our finl question is, Do the converses of this theorem nd the lst theorem hold? The nswer is YES on simply connected domins nd possibly NO on other domins. Definition: A region R of R n is simply connected if R is connected (every two points in R cn by joined by pth in R) nd every closed pth cn be contrcted to point. Without getting too technicl, the ide is: Given closed pth (one whose strt nd end points re the sme), we cn deform/stretch/contrct it until there is nothing left. Essentilly, we need R to be free of 1-dimensionl holes. R 1 is not connected, R 2 is connected but not simply, R 3 is simply connected On the left, the curve cn be contrcted to point. However, on the right, the curve gets stuck on the hole in R 2. 7
8 Theorem: Let R be simply connected open subset of R 3 nd suppose F = 0 on R. Then F is conservtive/grdient vector field on R. If R is simply connected open subset of R 2 nd F = P, Q where P y = Q x on R, then F is conservtive/grdient vector field on R. proof: The 2-dimensionl version follows from Green s theorem nd the 3-dimensionl version follows from Stoke s theorem. Accepting these theorems for the time being, here s the proof: Let be simple closed curve in R. If R is subset of R 2, then bounds some region D. If R is subset of R 3, since R is simply connected, we cn contrct to point, while doing this sweeps out surfce S in R. D is the region bounded by in R 2, S is the surfce swept out s is contrcted to point in R 3. In 2-dimensions use Green s theorem: F dr = Q x P y da = 0 da = 0 or In 3-dimensions use Stoke s theorem: So we hve (in either cse) tht F is conservtive. F dr = D S curl(f) ds = S D 0 ds = 0 Exmple: We showed tht for F(x, y, z) = xyz, x 2 y, y 2 + z 2, F = 2y, xy, x(2y z) 0. So this vector field is not conservtive. Likewise, if F(x, y) = x 2, xy then P y = 0 Q x = y so F is not conservtive. Exmple: Let F(x, y) = 2x + y 2, e y + 2xy. Then P y = 2y = Q x so F is conservtive (everywhere). Let s find potentil function for F. We need to find function f so tht f x = 2x + y 2 nd f y = e y + 2xy. Integrting both of these equtions we find tht f(x, y) = x 2 +xy 2 +g(y) nd f(x, y) = e y +xy 2 +h(x) (notice tht our rbitrry constnts need to be functions of the other vrible). Putting these nswers together we find tht f(x, y) = x 2 + xy 2 + e y + where is ny constnt (this time n honest to goodness constnt not function). In prticulr, F(x, y) = f(x, y) where f(x, y) = x 2 + xy 2 + e y (choosing = 0). Exmple: Let F(x, y, z) = 2xyz + z, x 2 z + 3y 2 + e z, x 2 y + ye z + x. Then i j k curl(f) = x y z = 2xyz + z x 2 z + 3y 2 + e z x 2 y + ye z + x y x 2 z + 3y 2 + e z z x 2 y + ye z + x i x 2xyz + z z x 2 y + ye z + x j + x 2xyz + z 8 y x 2 z + 3y 2 + e z k
9 = (x 2 + e z ) (x 2 + e z ), (2xy + 1) (2xy + 1), (2xz) (2xz) = 0, 0, 0 Thus F is conservtive. Agin, let s find potentil function. This time we need to find function f so tht f x = 2xyz + z, f y = x 2 z + 3y 2 + e z, nd f z = x 2 y + ye z + x. Integrting ll three of these equtions (gin remember tht the constnts re relly functions of the other vribles), we get f(x, y, z) = x 2 yz + xz + g 1 (y, z), f(x, y, z) = x 2 yz + y 3 + ye z + g 2 (x, z), nd f(x, y, z) = x 2 yz + ye z + xz + g 3 (x, y). So putting this ll together we hve tht f(x, y, z) = x 2 yz + xz + y 3 + ye z + where is ny constnt. In prticulr, F(x, y, z) = f(x, y, z) where f(x, y, z) = x 2 yz + xz + y 3 + ye z (choosing = 0). y Exmple: Let F(x, y) = x 2 + y, x. We should check to see if F is conservtive. After some 2 x 2 + y 2 work we find tht... P y = Q x = y2 x 2 (x 2 + y 2 ) 2 So F is conser...wait! Notice tht these prtils re not continuous t the origin (they re not even defined there). We cn only conclude tht F is conservtive on ny simply connected domin which does not contin the origin. Notice tht simply connected rules out the domin of F which is R 2 {(0, 0)} (the whole plne with the origin deleted). The prtils mtch on the entire domin of F, the domin is connected, but it isn t simply connected. In fct, to demonstrte wht cn hppen... Let be the unit circle oriented counter-clockwise. Prmeterize by r(t) = cos(t), sin(t) where 0 t 2π. 2π sin(t) F dr = cos 2 (t) + sin 2 (t), cos(t) cos 2 (t) + sin 2 sin(t), cos(t) dt (t) 0 = 2π 0 sin 2 (t) + cos 2 (t) dt = 2π 0 1 dt = 2π Since our integrl round the closed curve is not 0, F is not conservtive on R 2 {(0, 0)}. Note: Let be some closed curve. It cn be shown tht F dr = 2π(A B) where A is the number of times wrps itself round the origin in counter-clockwise fshion nd B is the number of times tht 1 wrps itself round the origin in clockwise fshion. F dr computes winding 2π numbers. Referring to the picture bove, wrps round (0, 0) twice in counterclockwise direction nd once in clockwise direction, so it nets 2 1 = 1 times round (counterclockwise): F dr = 2π(2 1) = 2π 9
10 Supplementl Problems: Line Integrls with respect to Arc Length: x 2 yz ds where is the curve prmeterized by r(t) = t, t 2, 23 t3 nd 0 t 1. xy 4 ds where is the right hlf of the circle centered t the origin of rdius 4. x 2 z ds where is the line segment from (0, 6, 1) to (4, 1, 5). x ds where is the line segment x = 1 + 2t, y = t where 0 t y2 e z ds where is the helix r(t) = 2 cos(t), 2 sin(t), t nd 0 t 2π. x 2 + y2 6. Find the centroid of the curve : the upper-hlf of the unit circle plus the x-xis from 1 to 1. Hint: Use geometry nd symmetry to compute 2 of the 3 line integrls. 7. Find the centroid of the helix prmeterized by r(t) = 2 sin(t), 2 cos(t), 3t where 0 t 2π. Pth Dependnce & Vector Fields: Determine if the following vector fields re conservtive. For those which re conservtive, find potentil function. 8. F(x, y) = x 2 + y 2, xy 9. F(x, y) = 2x + 3x 2 y 2 + 5, 2x 3 y 10. F(x, y) = e x, ye y2 11. F(x, y) = e xy, x 4 y 3 + y 12. F(x, y) = e x + y, rctn(x) + (1 + y)ey 1 + x2 13. F(x, y, z) = yz + y + 1, xz + x + z, xy + y F(x, y, z) = 2xyz + 3x 2, x 2 z + 6z, 6y 15. F(x, y, z) = yze xy + 2xz 2, xze xy + 7y 6 + 3y 2 z 2, e xy + 2x 2 z + (1 + 2z)e 2z + 2y 3 z 10
11 Let F(x, y) = P (x, y), Q(x, y) be vector field where P y = Q x except t the points P 1, P 2, nd P 3. Suppose tht F dr = 1, F dr = 2, F dr = 3, nd F dr = 4 where 1, 2,... 8 re pictured to the right. 4 ompute: 16. F dr [Answer: 6] F dr [Answer: 3] F dr [Answer: 0] F dr [Answer: 0] 8 Green s Theorem: 20. Let be the squre with vertices (0, 0), (1, 0), (1, 1), nd (0, 1) (oriented counter-clockwise). ompute the line integrl: y 2 dx + x 2 dy two wys. First, compute the integrl directly by prmeterizing ech side of the squre. Then, compute the nswer gin using Green s Theorem. 21. ompute F dr where F(x, y) = y ln(x 2 + y 2 ), 2 rctn(y/x) nd is the circle (x 2) 2 + (y 3) 2 = 1 oriented counter-clockwise. [Answer: π] 22. Let be the boundry of the prt of the disk x 2 + y 2 16 which lies in the first qudrnt (oriented counter-clockwise). ompute x 2 y dx y 2 x dy. [Answer: 32π] 23. Let be the line segments from (1, 0) to (1, 1), (1, 1) to (0, 1), nd (0, 1) to (0, 0). ompute 1 + y3 dx + (x 2 + e y2 ) dy. Hint: Use Green s Theorem to replce with the line segment from (1, 0) to (0, 0). 11
12 Surfce & Flux Integrls: 24. Find the centroid of the upper-hemisphere of x 2 + y 2 + z 2 = 9. [Answer: (0, 0, 3/2)] 25. Find the centroid of z = x 2 + y 2 where z Let r(u, v) = u cos(v), u sin(v), v where 0 u 1 nd 0 v 2π. Let be the surfce prmeterized by r. Grph in Mple. Then evlute x, y, z 2y ds if is given the orienttion n = r u r v r u r v. [Answer: π2 ] 27. Let be the prt of the prboloid z = 1 x 2 y 2 which lies bove the xy-plne (z 0) nd orient upwrd. Evlute x, y, z ds. [Answer: 3π/2] 28. Let be the upper-hemisphere of x 2 +y 2 +z 2 = 9 oriented upwrd. Evlute y, x, 1 ds. [Answer: 9π] 29. Let be the surfce of the cylinder bounded by x 2 + y 2 = 4, z = 0, nd z = 5 oriented outwrd (include the top nd the bottom of the cylinder). Evlute x 3, y 3, 0 ds. If you know the Divergence Theorem, use it to recompute your nswer. [Answer: 120π] 30. Let be the prt of the plne x + y + z = 1 which lies in the first octnt (x 0, y 0, z 0) nd oriented downwrd. Evlute (xze y i xze y j + zk) ds. [Answer: 1/6] 31. Let be the cube with vertices (±1, ±1, ±1) oriented outwrd. Evlute x, 2y, 3z ds. If you know the divergence theorem, reclculte this integrl using the theorem. [Answer: 48] Stokes Theorem: 32. Let S be the lower hlf of the sphere x 2 + y 2 + z 2 = 4 oriented downwrd with boundry. Also, let F(x, y, z) = 2y z, x + y 2 z, 4y 3x. Verify Stokes Theorem by computing both sides of F dr = ( F) ds. S 33. Verify Stoke s theorem for F(x, y, z) = y 2, x, z 2 nd the prt of the circulr prboloid z = x 2 + y 2 which lies below z = 1. Give the upwrd orienttion. 34. Let be the circle prmeterized by r(t) = cos(t), sin(t), 0 where 0 t 2π. Evlute z 2, 2x, y 3 dr. [Answer: 2π] 35. Let be the rectngulr boundry of the prt of the plne z = y which lies bove 0 x 1 nd 0 y 3. Give the counter-clockwise orienttion when viewed from bove. Evlute x 2, 4xy 3, y 2 x dr. [Answer: 90] 12
13 36. Let be the prt of the grph of z = e (x2 +y 2) which lies bove x 2 + y 2 1. Orient upwrd. Let F(x, y, z) = (e y+z 2y)i+(xe y+z +y)j+e x+y k. Evlute ( F) ds. Hint: Use Stokes theorem to replce with nother surfce which shres the sme boundry. [Answer: 2π] 37. be the four sides nd the top of the cube with vertices (±1, ±1, ±1) oriented outwrd. Let F(x, y, z) = xyzi+xyj+x 2 yzk. Evlute ( F) ds by using Stokes theorem to exchnge with n esier surfce whose boundry is the sme s. [Answer: 0] 38. ompute the work done by the force field F(x, y, z) = x x + z 2, y y + x 2, z z + y 2 when prticle moves under its influence round the edge of the prt of the sphere x 2 + y 2 + z 2 = 4 which lies in the 1 st octnt nd is oriented in counter-clockwise direction when viewed from bove. [Answer: 16] 39. Let be the curve of intersection of x + y + z = 1 nd x 2 + y 2 = 9. Orient counter-clockwise when viewed from bove. Evlute x 2 z, xy 2, z 2 dr. [Answer: 81π/2] The Divergence Theorem: 40. Let E be the solid bounded bove by z = 9 x 2 y 2 nd below by z = 0, let S be the boundry of E, nd let F(x, y, z) = x, y, z. Verify the Divergence Theorem by computing both sides of F ds = ( F) dv. S E 41. Let be the unit sphere x 2 + y 2 + z 2 = 1 oriented outwrd. Evlute z, y, x ds. [Answer: 4π/3] 42. Verify the divergence theorem when F(x, y, z) = xy, yz, zx nd E is the cylindricl region bounded by x 2 + y 2 = 1, z = 0, nd z = Let be the boundry of the upper-hlf of x 2 + y 2 + z 2 1 (the upper-hlf of the unit bll). Orient S outwrd. Evlute x 3, y 3, z 3 ds. [Answer: 6π/5] 44. Let be the surfce of the circulr prboloid bounded by z = x 2 + y 2 nd z = 1. Orient outwrd. Evlute x 2, y 2, z 2 ds. [Answer: 2π/3] 45. Let be the surfce of the cylinder bounded by y 2 + z 2 = 1, x = 1, nd x = 2 (including the front nd bck) oriented outwrd. Evlute 3xy 2, xe z, z 3 ds. [Answer: 9π/2] 46. Let be the upper-hemisphere of x 2 + y 2 + z 2 = 1 oriented upwrds (this is just the upper shell do not include the bottom). Also, let F(x, y, z) = z 2 x, y 3 /3 + tn(z), x 2 z + y 2. Evlute F ds. Hint: use the theorem to exchnge with n esier surfce. [Answer: 13π/20] 13
14 47. Let be the prt of the grph of z = (1 x 2 y 2 )e 1 x2 3y 2 which lies bove the xy-plne oriented upwrd. Also, let F(x, y, z) = e y cos(z), x sin(z), x 2 + y Evlute F ds by using the theorem to exchnge with n esier surfce. [Answer: 7π/2] 48. Prove Green s Formuls: Let f nd g be smooth functions nd let S be the surfce (oriented outwrd) of simple solid region E. f g + f ( g) dv = f g ds E (f ( g) g ( f)) dv = (f g g f) ds E S Hint: First, prove the following product rule: (f F) = f F + f F. S 14
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