F ds, where F and S are as given.

Transcription

1 Math 21a Integral Theorems Review pring, 29 1 For these problems, find F dr, where F and are as given. a) F x, y, z and is parameterized by rt) t, t, t t 1) b) F x, y, z and is parameterized by rt) t, t, 3 t t 1) c) F 2xy, x 2 + z, y + 2z and is parameterized by rt) t 2 t, sinπt), cos 2 πt) t 1) d) F y, x 2 +y 2 x and is the unit circle in the xy-plane, oriented counter-clockwise. x 2 +y 2 2 For these problems, find F d, where F and are as given. a) F curl 2y, z 2x, yz and is the hemisphere of radius 1, centered at the origin and above the xy-plane, oriented with the upward-pointing normal. b) F curl 2y, z 2x, yz and is the solid disk of radius 1 in the xy-plane, centered at the origin, oriented with the upward-pointing normal. c) F curl 2y, z 2x, yz and a) b) ) is the union of the two surfaces from parts a) and b), oriented with the outward-pointing normals. d) F x 2 y 3x, xy cosy)z, siny)z 2 and a) b) ) is the same as in part c). 3 For these problems, find div F dv, where F and are as given. a) F curl G where G is any appropriately smooth vector field) and is any simple solid b) F x, y 2, 2yz and is the solid ball of radius a centered at the origin c) F x 2, 2yz, x 2 z 2 and is the unit cube with corners at,, ) and 1, 1, 1)

2 Integral Theorems Review Answers and olutions 1 a) To compute this we just do it. Using rt) t, t, t, we get Frt)) t, t, t and dr 1, 1, 1 dt, so 1 1 F dr t, t, t 1, 1, 1 dt 3t dt 3 2 b) Here we could just do it again, but the parameterization is messier. It s simpler to notice that the vector field F x, y, z is conservative and thus independent of path. We see this by noticing that i j k curl F x y z x y z or simply that F f where f 1 2 x2 + y 2 + z 2 ). ince the path in part b) starts and ends at the same places as the path in part a), we can simply integrate over that path. This is what we ve done in part a), so we get the same answer: 3. 2 Another approach would be to use the fundamental theorem of line integrals: f dr fr1)) fr)) f1, 1, 1) f,, ) as before ) 3 2, c) Here the key is that the given vector field is conservative. We can see this by computing curl F or writing F f, where f x 2 y + yz + z 2. ince is a closed curve one that starts and ends at the point,, 1)), the integral must be zero. We could also see this via the fundamental theorem of line integrals: f dr fr1)) fr)) f,, 1) f,, 1). It s also possible to see this using tokes theorem where is some any!) oriented surface with as its boundary): F dr curl F d d. ither approach is fine. d) This is a cautionary tale. Let s think of our vector as F P, Q. Then it s straightforward to see that Q x P y y2 x 2 x 2 + y 2 ), 2 so we might think that we can proceed as in part c). That is, we could say that F is conservative and thus F dr. Or we might apply Green s theorem: let D be the unit disk with as it s boundary), so Q F dr x P ) da da. y D Both these conclusions are WRONG! and in fact the line integral has value 2π. D

3 What has gone wrong? We ve tried to blindly apply theorems without checking the hypotheses. In the first case we tried to apply theorem 6 on page 928) that says that F is conservative when P y Q x. But we must have this equality on a simply connected region D. In our case, both P and Q are undefined at the origin, ), so any region D containing the unit circle also contains a hole at the origin. imilarly, Green s theorem require that P and Q have continuous derivatives inside D, which again fails at the origin. We can compute the actual value using a simple parameterization of : rt) x, y cost), sint) and so drt) sint), cost) dt. Thus as claimed. F dr sint), cost) 1 1 sin 2 t) + cos 2 t) ) dt 1 dt 2π, sint), cost) dt 2 a) While we could compute this directly, it seems easier to use tokes theorem to compute a line integral instead. Here the boundary is simply the unit circle in the xy-plane, so we can parameterize as rt) x, y, z cost), sint),. Thus the vector we are integrating is 2y, z 2x, yz 2 sint), 2 cost), while dr sint), cost), dt. Thus curl 2y, z 2x, yz d 2y, z 2x, yz dr 2 sint), 2 cost), sint), cost), dt 2 sin 2 t) 2 cos 2 t) ) dt 2 dt 2t 2π 4π. b) As with part a), we could compute this integral directly. Now, however, we have even more of an incentive to use tokes theorem and compute the line integral we ve already computed this line integral! That is, our surface has the same boundary the unit circle in the xy-plane) as the surface from part a), so we can simply use our answer from there: curl 2y, z 2x, yz d 2y, z 2x, yz dr 4π.

4 curl 2y, z 2x, yz d + c) Our surface is a) b), the union of the surface from part a) with the same outwardpointing normal) and the surface from part b) with the downward-pointing normal, the opposite orientation from part b)). Thus curl 2y, z 2x, yz d curl 2y, z 2x, yz d a) b) curl 2y, z 2x, yz d curl 2y, z 2x, yz d a) 4π) 4π). b) Notice that it doesn t matter what the values of the surface integrals from part a) and part b) were, it only matters that they were the same. Thus the given surface integral over the closed surface a) b) will always be zero. This is because of two facts: i) Both a) and b) are oriented surfaces that have the curve as oriented) as their boundary, and ii) The flux integrals over a) and b) involve integrating a vector field that is a curl, and thus we can apply tokes theorem. Another approach is to apply the Divergence theorem. That is, since we re integrating over a closed surface that bounds a solid hemisphere we ll call this ), then curl 2y, z 2x, yz d div curl 2y, z 2x, yz ) dv. Now the key is the divcurl G) for any vector field G, so this integral is zero. d) This is very similar to part c). Perhaps the simplest approach from c) that we could use here is the last one, using the Divergence theorem. In this approach we again let be the solid unit hemisphere that has as its oriented) boundary. Thus F d div F) dv, where F x 2 y 3x, xy cosy)z, siny)z 2 so div F x 2 y 3x ) + xy cosy)z ) + ) siny)z 2 x y z 2xy + 2xy 2 siny)z) + 2 siny)z 3. Thus our flux integral is again!) zero. ) 2 F d 3) dv 3Vol) 3 3 π 2π, where we ve used the fact that the volume of a hemisphere is 2 3 πr3 and here r 1).

5 3 a) The key here is that div curl G) this is just an application of lairaut s theorem on order of partial derivatives). Thus div F dv div curl G) dv dv. b) Here Thus div F x x) + y y2 ) + 2yz) 1 + 2y 2y 1. z div F dv 1 dv Vol) 4 3 πa3, since is a ball of radius a. c) Here div F x x2 ) + y 2yz) + z x2 z 2 ) 2x + 2z 2z 2x. Thus, since {x, y, z) : x 1, y 1, z 1}, we get div F dv 2x dv x dx dy dz 1.