MA2331 Tutorial Sheet 5, Solutions December 2014 (Due 12 December 2014 in class) F = xyi+ 1 2 x2 j+k = φ (1)

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1 MA2331 Tutorial Sheet 5, Solutions. 1 4 Deember 214 (Due 12 Deember 214 in lass) Questions 1. ompute the line integrals: (a) (dx xy dy x2 + dz) where is the line segment joining the origin and the point (1,1,2). (b) (dx yz +dy xz +dz yx2 ) where is the same line as in the previous part A quik way here is to note that F is onservative. F = xyi+ 1 2 x2 j+k = φ (1) where φ = 1 2 x2 y +z. Hene F dl = φ(1,1,2) φ(,,) = 5 2. (2) For the next part, use the parametrization x(u) = u, y(u) = u, z(u) = 2u ( u 1). so dr du = i+j+2k, du = 2u2 +2u 2 +2u 3 = 4u 2 +2u 3 (3) F dl = du ( 4u 2 +2u 3) = = (4) 2. For eah of the following vetor fields ompute the line integral F dl where is the unit irle in the xy-plane taken anti-lokwise. 1 StefanSint,sint@maths.td.ie,seealsohttp:// 1

2 (a) F = xi+yj (b) F = yi x 2 yj. In the first part F = 1 2 (x2 + y 2 ) so that F is onservative giving F dl. In the seond part parametrize urve: where u 2π or r(u) = osui+sinuj. Now and Thus dr(u) du x(u) = osu y(u) = sinu z(u) = (5) = sinui+osuj. (6) du = ysinu x2 yosu = sin 2 u os 3 usinu. (7) F dl = 2π du ( sin 2 u os 3 usinu ) = π, (8) sine the average value of sin 2 u is 1 2 and 2π du os 3 usinu = by symmetry. 3. Evaluate the line integrals F dl for (a) F = (x 2 y,4,) with given by r(t) = (exp(t),exp( t),) with t going from zero to one; (b) F = (z,x,y) with given by r(t) = (sint,3sint,sin 2 t) with t going from zero to π/2. For the first one so dr dt r = (exp(t),exp( t),) (9) = (exp(t), exp( t),) (1) 2

3 and, on the urve, and hene F dl = dt dt = dt = e2t 4e t (11) (e 2t 4e t )dt = 1 2 e2 +4e (12) For the seond one so and, on the urve, dr dt r = (sint,3sint,sin 2 t) (13) = (ost,3ost,2sintost) (14) and hene dt = (7sin2 t+3sint)ost (15) F dl = = = π/2 π/2 dt dt (7sin 2 t+3sint)ostdt (7u 2 +3u)du = 23 6 where we have used a substitution u = sint. (16) 4. For eah of these fields determine if F is onservative, if it is, by integration or otherwise, find a potential: φ suh that F = φ. (a) F = xi+yj (b) F = 3y 2 i+6xyj () F = e x osyi e x sinyj (d) F = (osy +yosx)i+(sinx xsiny)j 3

4 So, in the first ase, it is easy to see the url is zero, having done that we want F = φ, hene F 1 = x φ or x φ = x (17) and hene φ = x 2 /2+(y,z), where (y,z) is an arbitrary funtion of y and z, substitute that bak in to get y = y (18) giving φ = x 2 /2+y 2 /2+(z) where (z) = a onstant follows from F 3 =. For the next one the url is again zero so there is a potential, x φ = 3y 2 (19) so φ = 3y 2 x+(y,z). Substituting into the y equation gives y φ = 6xy + y (y,z) = 6xy (2) and hene y (y,z) = so (y,z) = (z), further substituting this into z φ = shows (z) = a onstant and φ = 3y 2 x+. For the next one the url is again zero so there is a potential, x φ = e x osy (21) so φ = e x osy+(y,z). Substituting into the y equation and z equation show that (y,z) = a onstant and φ = e x osy +. Finally the last one also has zero url and x φ = osy +yosx (22) giving φ = xosy + ysinx + (y,z) and, again, substituting in to the y equation and z equation show that (y,z) = a onstant and φ = xosy + ysinx +. It won t always work out like this with arbitrary funtion turning out to be an arbitrary onstant, it is just an aident that I ask you three examples like this! 4

5 5. onsider the point vortex vetor field F = y x x 2 +y 2i x 2 +y 2j. Show that url F = away from the z-axis. Establish that F is not onservative in the (non simply-onneted) domain x 2 +y 2 1. Is F 2 onservative in the domain defined by x 2 +y 2 1, y? If so obtain 2 a salar potential for F. F = 1 [ ( ) x 2 k x x 2 +y 2 = 1 [ 2 k 1 ( )] y + y x 2 +y 2 ] x 2 +y + 2x 2 2 (x 2 +y 2 ) 1 2 x 2 +y + 2y 2 2 (x 2 +y 2 ) 2 =. (23) To show that F is not onservative onsider F dl where is the unit irle. Using the obvious parametrization 2π F dl = du ( sin 2 u os 2 u ) therefore F is not onservative. = 2π, (24) Thedomainx 2 +y 2 1 2, y issimplyonnetedandfisirrotational and smooth is the domain. Thus F is onservative. Write F = φ. Seek a φ(x,y) suh that φ x = y x 2 +y 2, φ y = x x 2 +y2. (25) Integrate first equation by treating y as a onstant dx φ(x,y) = y x 2 +y = x 2 tan 1 +(y). (26) y Assume that x and y are non-negative, then tan 1 x y +tan 1 y x = π 2, 5

6 so that φ(x,y) = tan 1 y + a possibly y-dependent onstant. However it is easy to hek that φ = tan 1 y φ x satisfies = x. x y x 2 +y 2 learly, tan 1 y is the usual polar angle θ, that is φ = θ. x an try to extend this bak to the original domain x 2 +y 2 1, but φ 2 will suffer a branh ut disontinuity at, say θ = 3π Find the flux of F = e y i yj + xsinzk aross the portion of the paraboloid r(u,v) = 2osvi+sinvj+uk (27) with u 5 and v 2π, oriented to give a positive answer. NOTE: this is not really a paraboloid, there was a mix-up with a different question! Anyhow, using the given parameterization, we have giving On the surfae and the flux is Now, 2π u = k = 2sinvi+osvj (28) v u v = osv 2sinv (29) F(r(u,v)) = e sinv i sinvj+2osvsinuk (3) φ = 5 dvosve sinv = 2π du dv ( osve sinv +2sin 2 v ) (31) 2π dv d dv e sinv = e sinv 2π = 1 1 = (32) This leaves the other bit of the integral, whih we do using the usual 2sin 2 x = 1 os2x (33) 6

7 giving φ = 1π (34) whih is a positive answer, so the orientation was hosen orretly. 7. Use Green s Theorem to evaluate (y 2 dx+x 2 dy) (35) where is the square with vertie (,), (1,), (1,1) and (,1) and oriented anti-lokwise. By Green s theorem (y 2 dx+x 2 dy) = dx dy(2x 2y) = dx(2x 1) = (36) 8. alulate diretly and using Stokes Theorem F ds (37) S where F = (z y)i + (z + x)j (x + y)k and S is the paraboloid z = 9 x 2 y 2 oriented upwards with z >. So, to alulate diretly, hoose some parameterization works. Now r = ρosφi+ρsinφj+(9 ρ 2 )k (38) φ = ρsinφi+ρosφj = osφi+sinφj 2ρk (39) ρ and, hoosing the other order to make the normal point upwards, ρ 2ρ 2 osφ φ = 2ρ 2 sinφ (4) ρ 7

8 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 Stokes S urla ds = A dl (41) hene, to apply Stokes, 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 (42) 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. Sine this formula is ompliated it would ertainly be a good idea to hek F = F. Now, to apply Stoke s theorem: urla ds = A dl (43) S where is the irle of radius three around the origin in the xy-plane: x 2 +y 2 = 9 and z =. We parameterize with r = 3osti+3sintj (44) so that dr dt = 3sinti+3ostj (45) 8

9 Restriting A to the urve and doing the dot produt gives A dl = 9 2π ( s 2 s s)dt = (46) where = ost and s = sint and we are using the usual anti-symmetry argumentthatoddpowersofsineandosineintegratetozeroovertheir entire period. 9