231 Outline Solutions Tutorial Sheet 7, 8 and January Which of the following vector fields are conservative?
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1 231 Outline olutions Tutorial heet 7, 8 and Problem heet 7 18 January Whih of the following vetor fields are onservative? (a) F = yz sin x i + z osx j + y os x k. (b) F = 1 2 y i 1 2 x j. () F = 1 (B r) where B is a onstant vetor. 2 olution: (a) F = yz osx so F is onservative. (b) url F = k so F is not onservative. () A short alulation gives url F = B so F is not onservative. Remark: 1 2 (B r) is a vetor potential for the onstant vetor field B. 2. Find a potential for olution:o, as before implies F 1 = x φ and hene so integrating F = (e yz + zye xy )i + (xze yz + xze xy )j + (xye yz + e xy )k (1) F = φ (2) x φ = e yz + zye xy (3) φ = xe yz + ze xy + C(y, z) (4) where C(y, z) is a funtion of y and z to be determined, in fat, it turns out to be onstant, it is easy to hek with C a onstant, is a potential for the field. φ = xe yz + ze xy + C (5) 1 Conor Houghton, houghton@maths.td.ie, see also 2 Inluding material from Chris Ford, to whom many thanks. 1
2 3. Compute the flux of the vetor field F = (x + x 2 )i + yj out of the ylinder defined by x 2 + y 2 = 1 and z 1. olution:as we have seen before, r(u, v) = os ui + sin uj + vk giving and u v = osui + sin uj. (6) F u v = (x + x2 ) osu + y sin u = os 2 + os 3 u + sin 2 u = 1 + os 3 u. (7) and so, noting that the F is perpendiular to the normal at both ends and so we need only inlude the urved surfae F da = sine the os 3 u integral is zero by symmetry. dv du (1 + os 3 u) = 2π, (8) Note: This problem an be solved by noting that x 2 i makes no ontribution (by symmetry) and F = xi + yj has flux 2π sine F n = Find the flux of F = z 3 k upwards through the part of the sphere x 2 + y 2 + z 2 = a 2 above the z = plane. olution:to parametrize the sphere hoose (u, v) = (θ, φ), that is, spherial polar angles. ine the radius r = a this gives x(u, v) = a sin u osv, y(u, v) = a sin u sinv, z(u, v) = a os u with θ π/2 and φ < 2π. ine F only has a non-zero z-omponent we just need ( u ) v 3 = x y u v y x u v = a 2 (osuosv sin u osv + osusinv sin u sinv) = a 2 osusinu, (9) whih is positive. o the orientation is upwards. Now F z = z 3 = a 3 os 3 u, and so F da = a 5 2 π = 2πa 5 du 2 π = 2πa 5 os5 u 5 dv os 4 u sin u du os 4 u sin u ] 1 2 π = 2πa5 5. (1) 2
3 Problem heet 8 1. Find the flux of F = xi + yj + k aross the portion of the paraboloid r(u, v) = u osvi + u sin vj + (1 u 2 )k (11) with 1 u 2 and v 2π, oriented to give a positive answer. olution:o, giving Now, on the parabola u v = osvi + sin vj 2uk = u sin vi + u osvj (12) u v = 2u 2 osv 2u 2 sin v u (13) F = u osvi + u sinvj + k (14) so, taking the dot produt, the flux, φ, is 2 φ = du dv ( 2u 3 os 2 v + 2u 3 sin 2 v + u 2) 1 = = 1 (15) whih is a positive answer, so the orientation was orret. 2. Find the flux of F = e y i yj + x sin zk aross the portion of the paraboloid r(u, v) = 2 osvi + sin vj + uk (16) with u 5 and v 2π, oriented to give a positive answer. olution:and again with the paraboloid: giving u = k = 2 sin vi + osvj (17) v u v = osv 2 sin v (18) 3
4 On the parabola and the flux is Now, φ = F = e sinv i sin vj + 2 osv sin uk (19) 5 du dv ( osve sinv + 2 sin 2 v ) (2) dv osve sinv = (21) there are lots of ways to see this, one way is to note that the integrand is odd about the point v = π/2 and a hange of variable and the periodiity ould be used to make the integral symmetri about this point dv osve sinv = 3π/2 π/2 dv osve sinv (22) and then let w = v π/2. This leaves the other bit of the integral, whih we do using the usual 2 sin 2 x = 1 os 2x (23) giving φ = 1φ (24) 3. Use Green s Theorem to evaluate (y 2 dx + x 2 dy) (25) where C is the square with vertie (, ), (1, ), (1, 1) and (, 1) and oriented antilokwise. olution:by Green s theorem (y 2 dx + x 2 dy) = dx dy(2x 2y) = dx(2x 1) = (26) 4. Calulate diretly and using toke s Theorem F d (27) where F = (z y)i + (z + x)j (x + y)k and is the paraboloid z = 9 x 2 y 2 oriented upwards with z >. olution:o, to alulate diretly, hoose some parameterization r = ρ osφi + ρ sin φj + (9 ρ 2 )k (28) 4
5 works. Now φ φ = ρ sin φi + ρ osφj = osφi + sin φj 2ρk (29) and, hoosing the other order to make the normal upward points v 2ρ 2 osφ u = 2ρ 2 sin φ (3) ρ Now, writing this as (2ρx, 2ρy, ρ) and doing the dot produt with F we are left with only terms whih are linear in x or y and sine the φ integral goes all the way around, we see the answer is zero. Next, using tokes urla d = A dl (31) hene, to apply tokes, we have to write F as urla, in other words, find a vetor potential for F. It is easy to hek that divf = so this should be possible. We will use the formula that was used to prove the existene of vetor potential for divergeneless fields on star-shapped domains. Hene A = F(tr) tr = 1 i j k 3 z y z + x (x + y) x y z = 1 z 2 + xz + xy + y 2 x 2 xy z 2 + yz (32) 3 zy y 2 xz x 2 where I got the third by noting that the overall fator of t 2 ame out of the determinant, and then integrating it. ine this formula is ompliated it would ertainly be a good idea to hek F = F. Now, to apply toke s theorem: urla d = A dl (33) where C is the irle of radius three around the origin in the xy-plane: x 2 + y 2 = 9 and z =. We parameterize with r = osti + sin tj (34) 5
6 so that dr = sin ti + os tj (35) dt Restriting A to the urve and doing the dot produt gives A dl = ( s 2 s s)dt = (36) where = ost and s = sin t and we are using the usual anti-symmetry arguement that odd powers of sine and osine integrate to zero over their entire period. 6
7 Problem heet 9 1. Let be the losed surfae onsisting of the portion of the paraboloid z = x 2 + y 2 for whih z 1 and apped by the disk x 2 + y 2 1 in the plane z = 1. Find the flux of the vetor field F(x, y, z) = zj yk in the outrward diretion aross. olution:use Gauss s theorem and the fat that divf = to get zero. 2. For any ondution loop C the eletri field E and magneti indution B are related by B E dr = d (37) t where C = δ, use toke s theorem to find a differential equation relating E and B. olution:by toke s theorem or B E dr = d (38) t ( E B ) d (39) t and sine this is true for any simple it must hold, by ontinuity, that E B t = (4) 3. Using Gauss theorem or otherwise ompute the flux of the vetor field F = x 3 i + y 3 j + z 3 k through the hemisphere x 2 + y 2 + z 2 = 1, z with the orientation taken upwards. What is the flux out of the whole sphere? olution:let be the losed surfae omprising the hemisphere x 2 + y 2 + z 2 = 1, z and the disk (needed to lose the surfae) z =, x 2 + y 2 1. Using Gauss theorem the flux of F out of is F da = div F dv = 3 (x 2 + y 2 + z 2 ) dv, D where D is the region enlosed by. This integral an be worked out through spherial polar oordinates or just by splitting D into small spherial half-shells of volume 2πr 2 δr: F da = 2π dr r 2 3r 2 = 6π 5. Now the flux out of the disk is zero sine here F is perpendiular to the outward normal n = k. Thus the flux through the hemisphere is 6π/5. The flux out of the whole sphere is 12π/5. D 7
8 4. Consider, again, the vetor field F = r r 3, r = xi + yj + zk. (a) Compute the flux of F out of a sphere of radius a entred at the origin. (b) Compute the flux of F out of the box 1 x 2, y 1, z 1. () Compute the flux of F out of the box 1 x 1, 1 y 1, 1 z 1. olution: (a) Flux integral trivial sine F n is onstant over the sphere (n is the outward normal). Here F n = 1/a 2. Therefore F da = 4πa2 = 4π. Note that C a 2 this is independent of the radius of the sphere. (b) We know divf =. Therefore Gauss theorem implies that the flux out of the box is zero. Note that this result is very diffiult to obtain via a diret surfae integral omputation. () Naively, Gauss theorem also gives a zero flux in this ase. However, this is not orret. Indeed, an indisriminate use of Gauss theorem would also give a zero flux in part i). The point is that in parts i) and iii) (but not ii) ) the inside region ontains the origin where F is singular. As in part i) the orret answer to this question is 4π. This an be obtained by ombining Gauss theorem with the result of the diret alulation of part a). To do this remove a sphere (entred at the origin) of radius less than one from the box. In this region F is smooth and so Gauss theorem implies that the flux out of this region is zero. However this flux omprises two parts; the flux out of the box (whih we wish to ompute) and the flux into the ut sphere (whih from part i) is 4π). Therefore the flux out of the box must be 4π. 5. Obtain a vetor potential for the solenoidal vetor field: F = xi + yj 2zk olution:use the formula A(r) = dt F(tr) rt. Now F = xi+yj 2zk = r 3zk so that A(r) = dt 3ztk rt = 3 dt t 2 (zxj zyi) = zyi zxj. 8
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