231 Outline Solutions Tutorial Sheet 4, 5 and November 2007

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1 31 Outline Solutions Tutorial Sheet 4, 5 and 6. 1 Problem Sheet 4 November 7 1. heck that the Jacobian for the transformation from cartesian to spherical polar coordinates is J = r sin θ. onsider the hemisphere defined by x + y + z 1, z. Using spherical polar coordinates compute its volume and centroid. Solution:Spherical polar coordinates are defined by x = r sin θ cos φ, y = r sin θ sin φ, z = r cosθ. (1) The Jacobian is x x x (x, y, z) r θ φ = y y y (r, θ, φ) r θ φ z z z = sin θ cos φ r cosθ cosφ r sin θ sin φ sin θ sin φ r cosθ sin φ r sin θ cos φ cosθ r sin θ. r θ φ = r sin θ [ cos θ cos φ + cos θ sin φ + sin θ cos φ + sin θ sin φ ] = r sin θ. () Volume = dv. entroid x = ȳ = by symmetry and z = dv z/ dv. Now D D D π π/ π/ dv = dφ dθ r sin θ = π dθ sin θ 1 3. D as expected. The other integral is π dv z = D = π = π/ 3 π cosθ = π/3 (3) π and therefore z = 3/8. π dφ dθ dθ 1 sin θ = π 4 r π sin θ r cosθ = π dθ sin θ cos θ onor Houghton, houghton@maths.tcd.ie, see al Including material from hris Ford, to whom many thanks. (4) 1

2 . Show divr = 3 and grad r = r/ r. Solution:Well and and and by the chain rule. 3. Find (1/ r ). r = xi + yj + zk (5) r = x x + y y + z z = 3 (6) r = r x i + r y j + r z k (7) x r = x x + y + z = x r Solution:So this is similar to the previous one (8) 1 r = 1 x r i + 1 y r j + 1 z r k (9) and hence 1 x r = 1 r x r = x (1) r 3 1 r = r r 3 (11) and this, or course, is consistient with curlr/r 3 =. 4. Show grad f(r) = f (r)ˆr where r = r. If F(r) = f(r)r find divf(r). Find div grad f(r). Solution:So, now, we use the chain rule to show x f(r) = f (r) r x = xf (r) r (1) and, since the gradient has three terms of this form, it is easy to see grad f(r) = f (r)ˆr. As for the divergence, F 1 = xf(r) and x xf(r) = f(r) + x f r and, adding three similar terms together, we get (13) F = 3f + rf (14)

3 Finally,, we know the gradf(r) and,, using f /r for f in the divergence formula we get f(r) = 3f ( ) f r + r = f r r + f (15) which gives us a formula for the laplacian of a spherically symmetric field in polar coördinates, later we will see how to convert partial differential equations from one coördinate system to another. 5. Show that away from the origin the vector field F = ˆr r = r r 3 (16) is irrotational (here r = xi + yj + zk and r = r = x + y + z ). Solution:Note that F = grad ( 1/r) and curl F =. This can al be done by direct calculation. 3

4 Problem Sheet 5 1. alculate curlr/r and divr/r away from the origin. What is r? Solution:So it is easy enough to check these by hand, for example, the bfi component of curlr/r is given by ( r ) = z r 1 y r y z r = (17) and, with the other two components similar curlr/r =. As for the divergence x x r = 1 r x (18) r 3 and hence divr/r = /r. Finally, gra = bfr/r r = /r.. Prove the identity ( F) = (19) Solution:So this is easy by direct calculation, ( F) = x ( y F 3 z F ) + y ( z F 1 x F 3 ) + z ( x F y F 1 ) () and expanding out, all the terms cancel, assuming the partial derivative commute. 3. Prove the identity ( F) = ( F) F. (1) Solution:Lets do the first component: [ ( F)] 1 = y (F,x F 1,y ) z (F 3,x F 1,z ) () where F = F 1 i + F j + F 3 k and I am using a comma notation for differenciation for example Now, taking away me brackets oming from the other side F,x = F x (3) [ ( F)] 1 = F,xy F 1,yy F 3,xz F 1,zz (4) [ ( F)] 1 = x (F 1,x + F,y + F 3,z ) = F 1,xx + F,yx + F 3,zx (5) [ ( F)] 1 [ ( F)] 1 = F 1,xx + F 1,yy + F 1,zz = [ F] 1 (6) and similarily for the other components. 4

5 4. ompute the line integrals: (a) (dx xy + 1 dy x + dz) where is the line segment joining the origin and the point (1, 1, ). (b) (dx yz + dy xz + dz yx ) where is the same line as in the previous part. Solution:A quick way here is to note that F is conservative. F = xyi + 1 x j + k = φ (7) where φ = 1 x y + z. Hence φ(1, 1, ) φ(,, ) = 5. (8) For the next part, use the parametrization x(u) = u, y(u) = u, z(u) = u ( u 1). du = i + j + k, du = u + u + u 3 = 4u + u 3 (9) du ( 4u + u 3) = = (3) 5. For each of the following vector fields compute the line integral F dl where is the unit circle in the xy-plane taken anti-clockwise. (a) F = xi + yj (b) F = yi x yj. Solution:In the first part F = 1 (x + y ) that F is conservative giving F dl. In the second part parametrize curve: where u π or r(u) = cos ui + sin uj. Now and Thus (u) du x(u) = cos u y(u) = sin u z(u) = (31) = sin ui + cosuj. (3) du = y sin u x y cosu = sin u cos 3 u sin u. (33) π du ( sin u cos 3 u sin u ) = π, (34) since the average value of sin u is 1 and π du cos 3 u sinu = by symmetry. 5

6 Problem Sheet 6 1. For each of the following vector fields compute the line integral F dl where is the semi-circle of radius two around the origin in the xy-plane with y positive, taken anti-clockwise. (a) F = x i + xyj (b) F = yi + xj. Solution:So, once again we parameterize the curve: r = costi + sin tj (35) = sin ti + costj (36) dt and,, for the first part, on the curve, x = cost and y = sin t dt = 8 cos t sin t + 8 cos t sin t = (37) and hence the integral is zero. For the second part dt = 4 cos t + 4 sin t = 4 (38) π. Evaluate the line integrals F dl for π dt dt = 4 dt = 4π (39) (a) F = (x 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) = (sin t, 3 sin t, sin t) with tgoing from zero to π/. (c) F = λ(x, y) with λ = (x + y ) 3/ and with given by r(t) = (e t sin t, e t cost) with t going from zero to one. Solution:For the first one and, on the curve, dt r = (exp (t), exp ( t), ) (4) = (exp (t), exp ( t), ) (41) dt = et 4e t (4) 6

7 and hence dt dt = (e t 4e t )dt = 1 e + 4e 1 9 (43) For the next one and, on the curve, and hence = π/ π/ dt r = (sin t, 3 sint, sin t) (44) = (cos t, 3 cost, sin t cost) (45) dt = (7 sin t + 3 sint) cost (46) dt dt (7 sin t + 3 sin t) costdt = where we have used a substitution u = sin t. Finally which is an exponential spiral with (7u + 3u)du = 3 6 (47) r = ( e t sin t, e t cost ) (48) dt = ( e t (sin t + cost), e t (cost sin t) ) (49) and, thanksfully, this is simplifies on the curve to and hence dt = e t (5) dt dt = e t dt = e 1 1 (51) 3. For each of these fields determine if F is conservative, if it is, by integration or otherwise, find a potential: φ such that F = φ. (a) F = xi + yj (b) F = 3y i + 6xyj (c) F = e x cosyi e x sin yj (d) F = (cosy + y cos x)i + (sin x x sin y)j 7

8 Solution:So, in the first case, it is easy to see the curl is zero, having done that we want F = φ, hence F 1 = φ,x or x φ = x (5) and hence φ = x / + (y, z), where (y, z) is an arbitrary function of y and z, substitute that back in to get y = y (53) giving φ = x / + y / + (z) where (z) = a constant follows from F 3 =. For the next one the curl is again zero there is a potential, x φ = 3y (54) φ = 3y x + (y, z). Substituting into the y equation gives y φ = 6xy + y (y, z) = 6xy (55) and hence y (y, z) = (y, z) = (z), further substituting this into z φ = shows (z) = a constant and φ = 3y x +. For the next one the curl is again zero there is a potential, x φ = e x cosy (56) φ = e x cosy + (y, z). Substituting into the y equation and z equation show that (y, z) = a constant and φ = e x cosy +. Finally the last one al has zero curl and x φ = cosy + y cos x (57) giving φ = x cosy + y sin x + (y, z) and, again, substituting in to the y equation and z equation show that (y, z) = a constant and φ = x cosy + y sin x +. It won t always work out like this with arbitrary function turning out to be an arbitrary constant, it is just an accident that I ask you three examples like this! 4. onsider the point vortex vector field F = y x + y i x x + y j. Show that curl F = away from the z-axis. Establish that F is not conservative in the (non simply-connected) domain x + y 1. Is F conservative in the domain defined by x + y 1, y? If obtain a scalar potential for F. Solution: F = 1 [ ( ) ( )] x y k x + x + y y x + y 8

9 = 1 [ k 1 ] x + y + x (x + y ) 1 x + y + y (x + y ) =. (58) To show that F is not conservative consider Using the obvious parametrization therefore F is not conservative. π F dl where is the unit circle. du ( sin u cos u ) = π, (59) The domain x +y 1, y is simply connected and F is irrotational and smooth is the domain. Thus F is conservative. Write F = φ. Seek a φ(x, y) such that φ x = y x + y, φ y = x x + y. (6) Integrate first equation by treating y as a constant dx φ(x, y) = y x + y = x tan 1 + (y). (61) y Assume that x and y are non-negative, then tan 1 x y + tan 1 y x = π, that φ(x, y) = tan 1 y + a possibly y-dependent constant. However it is easy to x check that φ = tan 1 y φ satisfies = x. learly, tan 1 y is the usual polar x y x +y x angle θ, that is φ = θ. an try to extend this back to the original domain x + y 1, but φ will suffer a branch cut discontinuity at, say θ = 3π. 9

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