Green s theorem is true, in large part, because. lim

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1 Recall the statement of Green s Theorem: If is the smooth positively oriented boundary of a simply connected region D in the xy-plane, and if F is a vector field in the xy-plane with continuous first partial derivatives, then F dr = Q x P y da. Green s theorem is true, in large part, because r + F dr = Q x (x, y) P y (x, y) r D where r is the circle of radius r centered at (x, y) with positive orientation (see lecture 2). Notice that Green s theorem is a statement about regions in the xy-plane. On the other hand, there s nothing special about the xy-plane other than the fact that it s planar. o Green s theorem should hold for surfaces in R 3 that are contained within planes. To be precise, we should have the following: let be a positively oriented smooth boundary of a simply connected surface that is contained within some plane, say P. Let F be a vector field in R 3 tangent to P, i.e., F(x, y, z) is a vector in P for all (x, y, z) P. We call F planar in this case. uppose also that F has continuous first partial derivatives. Then F dr = f(x, y, z) d, where f(x, y, z) = r + F dr, r where r is the oriented circle of radius r in the plane P centered at (x, y, z). When P is the xy-plane, f(x, y, z) = Q x (x, y, z) P y (x, y, z). What does f(x, y, z) look like for a general plane P? laim. uppose P has unit normal vector n = a, b, c. Then f(x, y, z) = curl F n. The normal vector n is chosen to agree with the orientation of r. explained in the proof below. This is Proof. We consider the following special case: P is the plane ax + by + cz = and (x, y, z) = (,, ). The general case follows in an identical way but with more It would be a good exercise to verify this.

2 2 notation. The circle r is parameterized by x(t) = y(t) = r [ac cos t b sin t] a2 + b2 r [bc cos t + a sin t] a2 + b2 z(t) = r a 2 + b 2 cos t. Here, n = a, b, c is chosen so that r has the proper orientation. If n is used instead, then r will be traversed in the other direction. This is what is meant by n agreeing the the orientation of r. The derivation of this parameterization is somewhat long and tedious, but the reader can quickly verify that r(t) = (x(t), y(t), z(t)) satisfies both x 2 + y 2 + z 2 = r 2 and ax + by + cz =. Now, πr 2 r F dr = A + B + where A = B = = where, as usual, F = P, Q, R. onsider A. We use integration by parts with P(r(t))x (t)dt Q(r(t))y (t)dt R(r(t))z (t)dt, u = P(r(t)) dv = x (t) dt. The chain rule gives us that du = P(r(t)) r (t), while v = x(t). o A = = P(r(t)) r (t)x(t)dt P x (r(t))x (t)x(t) + P y (r(t))y (t)x(t) + P z (r(t))z (t)x(t) dt. Let X(t) = x(t)/r, and similarly for y(t) and z(t). Then we could rewrite the above as A = π P x (r(t))x (t)x(t) + P y (r(t))y (t)x(t) + P z (r(t))z (t)x(t) dt. Let r +. Then P x (r(t)) P x (,, ), and similarly for P y (r(t)) and P z (r(t)). It s left as an exercise to the reader to verify that X (t)x(t) dt = Y (t)x(t) dt = cπ Z (t)x(t) dt = bπ.

3 3 Putting it all together we get A = r + π [P x(,, ) + P y (,, ) cπ + P z (,, ) ( bπ)] = bp z cp y. Identical arguments to the one above for A, show that Hence r + B = cq x aq z = ar y br x. r + r + F dr = bp z cp y + cq x aq z + ar y br x r = a (R y Q z ) + b (P z R x ) + c (Q x P y ) = curl F n. Remark. Notice that for the statement of claim, we used nothing about the fact that F was planar. In other words, the same would follow for a general vector field F in 3-variables. This, in particular, tells us exactly why curl F is defined as a vector for 3-variable vector fields. We d like a convenient way to measure the circulation of F about some point (x, y, z) in some plane with normal vector n. The above results tell us that this is easy; it s simply curl F n. Also, notice that curl F n = curl F n cos θ = curl F cos θ, (where θ is the angle between curl F and n) is maximal when F points in the same direction as n. o this also tells us exactly why curl F points in the direction of the normal vector to the plane where the circulation about a given point is maximal, what s more, the magnitude of curl F gives that maximal rate of circulation. Getting back to Green s theorem for planar regions in R 3, we now know that F dr = curl F n d = curl F d where F is a planar vector field in the planar region and the orientations of and agree. How common are planar vector fields? Well, given an arbitrary vector field F in R 3, we could always project F onto the plane P with normal vector n via F proj n F. Then for arbitrary F, we have F proj n F dr = But proj n F dr =, curl(f proj n F) d.

4 4 since proj n F is always perpendicular to tangents of, and curl( proj n F) d =, since curl(proj n F) n = (verification is left to the reader). o for an arbitrary vector field F with continuous first partial derivatives and a simply connected planar region with smooth boundary, we have F dr = curl F d where the orientations of and agree. We know that smooth surfaces look more and more planar when we look at them over smaller and smaller domains (see lecture 4). In fact, they are well approximated by planes over sufficiently small domains. o it seems reasonable to conjecture that the above theorem for planar surfaces is actually true for smooth surfaces. This is, in fact, the case: Theorem. (tokes ) Let F be a vector field in R 3 with continuous first partial derivatives. Let be a smooth, simply connected, oriented surface with smooth, oriented boundary where the orientations of and agree. Then F dr = curl F d. Remark. We should take tokes theorem to be a near direct analogue of Green s theorem for surfaces. In that respect, it equates the tendency of F to circulate about the boundary of a surface to the sum total circulation of F about the interior points of. tool. Other than being very interesting, tokes theorem is a valuable computational Example. ompute F dr where is the positively oriented parallelogram spanned by the vectors b =,, and a = 2,,, and F(x, y, z) = arctan x + y + z, x + arctan y + z, x y + arctan z. olution. We can quickly parameterize the surface of the parallelogram, say, via r(u, v) = ua + vb, i.e., x(u, v) = 2u + v y(u, v) = u + v z(u, v) = u + v, with (u, v) [, ] 2. This parameterization gives us i j k r u r v = 2 =, 2, 2.

5 5 The above vector serves to orient, and since it has positive k component we know it gives the upward orientation on, i.e., the orientation that agrees with the orientation with, in this case. We can also easily compute curl F: o, by tokes theorem, F dr = curl F =,, + =,, 2. curl F d = 2 u + v dv du = 2. Remark. Notice how much of a pain it would have been to compute F dr in the above example. We would have had to break up that integral into 4 integrals, one for each straight-edge boundary. Problem. ompute F dr where is any smooth loop in R3 avoiding the origin and F is the gravitational field F(x, y, z) = Explain your work. x (x 2 + y 2 + z 2 ) ( 3/2), y (x 2 + y 2 + z 2 ) ( 3/2), z (x 2 + y 2 + z 2 ) ( 3/2) Problem 2. Let F(x, y, z) = yi + xj + k. ompute curl F d where is (a) the part of the paraboloid z = x 2 y 2 above z = with upward orientation. (b) the disk x 2 + y 2 in the xy-plane with upward orientation. (c) the paraboloid z = x 2 + y 2 below z = with downward orientation. (d) the upper hemisphere z = x 2 y 2 with upward orientation..