F dr = f dx + g dy + h dz. Using that dz = q x dx + q y dy we get. (g + hq y ) x (f + hq x ) y da.

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1 Math 55 - Vector alculus II Notes 14.7 tokes Theorem tokes Theorem is the three-dimensional version of the circulation form of Green s Theorem. Let s quickly recall that theorem: Green s Theorem: Let be a positively oriented (parameterized counterclockwise) piecewise smooth closed simple curve in and be the region enclosed by. uppose P (x, y) and Q(x, y) have continuous first order partial derivatives on. Then ( Q P dx + Q dy x P ) da. y Now we present tokes Theorem: tokes Theorem: Let be an oriented surface in 3 with a piecewise-smooth closed boundary whose orientation (by the righthand rule) is consistent with. Let F f, g, h and assume f, g, h have continuous partial derivatives on. Then F dr ( F) d. Proving tokes Theorem in general is very time-consuming. We just consider a special case: We consider the case when is (part of) the graph of z q(x, y) (using the upward orientation). Let be the curve that bounds with a usual counterclockwise orientation. Let be the projection of into the xy-plane and be the projection of into the xy-plane. Then F dr f dx + g dy + h dz. Using that dz q x dx + q y dy we get F dr f dx + g dy + h(q x dx + q y dy) (f + hq x ) dx + (g + hq y ) dy. By applying Green s Theorem above, F dr (g + hq y ) x (f + hq x ) y da. We calculate the partials (where f, g, h are functions of (x, y, z) and z q(x, y)): (g + hq y ) x g x + g z q x + h x q y + h z q x q y + hq yx and (f + hq x ) y f y + f z q y + h y q x + h z q y q x + hq xy. When we take the difference it simplifies to (g + hq y ) x (f + hq x ) y q x (g z h y ) + q y (h x f z ) + (g x f y ).

2 o F dr q x (g z h y ) + q y (h x f z ) + (g x f y ) da. Now let s deal with the surface integral side: F h y g z, f z h x, g x f y and we use the graph parametrization. o ( F) d ( F) n d h y g z, f z h x, g x f y q x, q y, 1 da, which is the same as the double integral at the top of this page! The intuition behind tokes Theorem is the same as for the circulation form of Green s Theorem: The accumulated rotation of a vector field over a surface is the net circulation on the boundary of, meaning all the internal circulation cancels! Let s verify tokes Theorem for F z y, x z, y x where is the cap of x +y +z 4 above the plane z 3 (oriented upwards). The boundary of is the circle {(x, y, z) x + y 1, z 3}. A parametrization of that circle (consistent with the upward orientation of is r(t) cos (t), sin (t), 3. o F dr π. F(r(t)) r (t) dt 3 sin (t), cos (t) 3, sin (t) cos (t) sin (t), cos (t), dt 3 sin (t) + sin (t) + cos (t) 3 cos (t) dt F z y, x z, y x,,. Let s use the parametrization r(u, v) sin (u) cos (v), sin (u) cos (v), cos (u) over the rectangle given by u π and v π. This gives 6 t u t v 4 sin (u) cos (v), sin (u) sin (v), sin (u) cos (u)

3 o ( F) d 8 8,, 4 sin (u) cos (v), sin (u) sin (v), sin (u) cos (u) da π 6 π π 6 8π sin (u) π 6 π. sin (u) cos (v) + sin (u) sin (v) + sin (u) cos (u) dv du π sin (u) cos (u) du And that verifies tokes Theorem in this instance! Now let s see the usefulness of tokes Theorem: Let F y, x, x+y +z. Let s evaluate F dr where is the boundary of the square in the xy-plane given by x 1 and y 1 oriented consistent with an upward orientation on the square. oing this directly would require parameterizing the four line segments (in a counterclockwise orientation) that make up the square and evaluating the line integral on each. By tokes Theorem we can make this much easier: F 1, 1, x y. Using the parametrization s(x, y) x, y, over the square [ 1, 1] [ 1, 1] yields d,, 1 da (obviously?) and hence F dr ( F) d 1, 1, x y,, 1 da x y dy dx. If you are bored, come up with the same answer by evaluating F dr directly.

4 Let F z, x, y. Let s evaluate ( F) d where is the hyperboloid z x + y, for z (with an upward orientation). Let be the boundary of in the plane z : o is the circle of radius 3 11 in the plane z centered at the origin. Then r(t) 3 11 cos (t), sin (t), for t π is a parametrization of (counterclockwise). Hence ( F) d F dr 99 cos (t) dt 99, 3 11 cos (t), 3 11 sin (t) 3 11 sin (t), cos (t), dt (1/)(1 + cos (t)) dt 99π. Just for fun, let s try to do this calculation without tokes Theorem: F 1, 1, 1 We could use the graph parametrization, but alternatively consider having x + y v. o let s use the parametrization r(u, v) v cos (u), v sin (u), 1 over the rectangle determined by u π and v Then t u v sin (u), v cos (u), v and t v cos (u), sin (u), o t u t v. v cos (u), v sin (u), v. ince this points downward, we will use t u t v : ( F) d 1, 1, π. v cos (u), v sin (u), v da v (cos (u) + sin (u)) + v du dv + πv dv

5 Let be the tilted disk enclosed by the curve given by r(t) cos (φ) cos (t), sin (t), sin (φ) cos (t) where φ π/ is a fixed angle (between the disk and the positive x-axis). Let s find the area of : Area() Let s find a normal vector to : r() r () cos (φ),, sin (φ), 1, sin (φ),, cos (φ). We set this (unit vector) equal to n. 1 d. By finding a vector field F such that F sin (φ),, cos (φ) we will have ( F) n 1. Well, F y cos (φ),, y sin (φ) does the trick! Thus Area() Of course! π. 1 d ( F) n d F dr sin (t) cos (φ),, sin (t) sin (φ) cos (φ) sin (t), cos (t), sin (φ) sin (t) dt sin (t) cos (φ) + sin (t) sin (φ) dt sin (t) dt (1/)(1 cos (t)) dt

6 [ONT. FOM PEVIOU PAGE] Now let s find the circulation F dr for F y, x, as a function of φ. Then let s determine the value of φ for which the circulation is at a maximum. F dr ( F) n d,, sin (φ),, cos (φ) d cos (φ) 1 d π cos (φ). This is at a maximum when φ and is the unit disk in the xy-plane. onsider the vector field F y x + y, x x + y, z. A quick calculation, which is left as an exercise for you, shows that F. This would seem to imply that for any closed simple curve in the xy-plane encircling a region containing the origin, F dr ( F) d. However, let s see what happens when we calculate F dr directly for the unit circle in the xy-plane, which is parameterized by r(t) cos (t), sin (t), for t π: F dr sin (t), cos (t), sin (t), cos (t), dt π. What s wrong? Is this a violation of tokes Theorem!? Of course it isn t. tokes Theorem doesn t apply here because the component functions of F aren t continuous at the origin, let alone have continuous partial derivatives at the origin.