Stokes Theorem. MATH 311, Calculus III. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Stokes Theorem

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1 tokes Theorem MATH 311, alculus III J. Robert Buchanan Department of Mathematics ummer 2011

2 Background (1 of 2) Recall: Green s Theorem, M(x, y) dx + N(x, y) dy = R ( N x M ) da y where is a piecewise smooth, positively oriented, simple closed curve in the xy-plane enclosing region R. Define the vector field F(x, y) = M(x, y), N(x, y), 0. Note that F = 0, 0, N x M y. The component of F along k is ( N ( F) k = x M ) k k = N y x M y

3 Background (2 of 2) Thus we have developed the vector form of Green s Theorem ( N M(x, y) dx + N(x, y) dy = R x M ) da y F dr = ( F) k da Today we will generalize this result to three dimensions. R

4 urve Orientation 2.0 n z x y Using the right-hand rule, curve has positive orientation if it has the same orientation as the right hand s fingers when the right thumb points in the direction of the normal n to surface. Otherwise has negative orientation.

5 tokes Theorem Theorem (tokes Theorem) uppose that is an oriented, piecewise-smooth surface with unit normal vector n, bounded by the simple closed, piecewise-smooth boundary curve having positive orientation. Let F(x, y, z) be a vector field whose components have continuous first partial derivatives in some open region containing. Then, F(x, y, z) dr = ( F) n d.

6 Interpretation Recalling that the unit tangent vector is T(t) = r (t) r (t) r (t) = r (t) T(t) and that differential arc length is defined as ds = r (t) dt, sometimes tokes Theorem is written as ( F) n d = F(x, y, z) dr = where T is the unit tangent in the direction of. F T ds Interpretation: The line integral of the tangential component of F is equal to the flux of the curl of F. This integral is the average tendency of the flow of F to rotate around path.

7 Example (1 of 4) Evaluate F dr where F(x, y, z) = y 2 i + xj + z 2 k and is the intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1. 3 z y x

8 Example (2 of 4) Note that F = (1 + 2y)k. There are many surfaces which have as a boundary, choose the elliptical disk bounded by in the plane y + z = 2. The unit normal to this surface is n = 1 2 0, 1, 1.

9 Example (3 of 4) y n 2.5 z 2 z y x x

10 Example (4 of 4) According to tokes Theorem F dr = ( F) n d 1 = (1 + 2y)k 0, 1, 1 d 2 1 = (1 + 2y) d 2 = (1 + 2y) da = = = π R 2π π (1 + 2r sin θ)r dr dθ r dr dθ

11 Example (1 of 4) Evaluate ( F) nd where F(x, y, z) = yzi + xzj + xyk and is the part of the sphere x 2 + y 2 + z 2 = 4 that lies inside the cylinder x 2 + y 2 = 1 and above the xy-plane z y 0 x

12 Example (2 of 4) urface is bounded by a circle formed by the intersection of the sphere of radius 2 and the cylinder of radius 1. We can describe using the vector-valued function with 0 t 2π. r(t) = cos t, sin t, 3,

13 Example (3 of 4) 3.0 n z z y y 1 x x

14 Example (4 of 4) According to tokes Theorem ( F) n d = F dr = 2π 0 = 3 = 3 = 0 F(cos t, sin t, 3) sin t, cos t, 0 dt 2π 0 2π 0 (cos 2 t sin 2 t) dt cos 2t dt

15 An Identity how that (f f ) dr = (f f ) dr = 0. = = = 0 ( (f f )) n d ( 0 n d f f x, f f y, f f z ) n d

16 Interpretation of the url (1 of 3) Let P = (x 0, y 0, z 0 ) be any point in a vector field F and let a be a circular disk of radius a > 0 centered at P. Let a be the boundary of a. n P a a

17 Interpretation of the url (2 of 3) Average value of ( F) n on surface a : [( F) n] Pa = 1 πa 2 ( F) n d a where πa 2 is the area of a and point P a a by the Integral Mean Value Theorem.

18 Interpretation of the url (2 of 3) Average value of ( F) n on surface a : [( F) n] Pa = 1 πa 2 ( F) n d a where πa 2 is the area of a and point P a a by the Integral Mean Value Theorem. By tokes Theorem a F dr = a ( F) n d = πa 2 [( F) n] Pa

19 Interpretation of the url (3 of 3) [( F) n] Pa = 1 πa 2 F dr a lim [( F) n] 1 a 0 + P a = lim a 0 + πa 2 1 [( F) n] P = lim a 0 + πa 2 If F describes the flow of a fluid then F dr a a F T ds F T ds is the circulation around, the average tendency of the fluid to circulate around the curve.

20 Remarks [( F) n] P attains its maximum when F is parallel to n. We can define rot F = ( F) n. This quantity is the rotation of the vector field at a point. F is an irrotational vector field if and only if F = 0.

21 Example uppose F = 3y, 4z, 6x. Find the direction of the maximum value of ( F) n.

22 Example uppose F = 3y, 4z, 6x. Find the direction of the maximum value of ( F) n. ince F = 4, 6, 3, the maximum of ( F) n will occur in the direction of n = 4, 6, 3 4, 6, 3 = 1 4, 6, 3. 61

23 Irrotational Vector Fields Theorem uppose that F(x, y, z) has continuous partial derivatives throughouta simply connected region D, then F = 0 in D if and only if F dr = 0 for every simple closed curve in D.

24 Proof (1 of 2) uppose F = 0. According to tokes Theorem F dr = ( F) n d = 0 d = 0.

25 Proof (2 of 2) uppose F dr = 0 for every simple closed curve. If at some point P, ( F) 0, then by continuity there exists a subregion of D on which ( F) 0. In the subregion choose a circular disk whose normal n is parallel to F. F dr = ( F) n d 0 which contradicts our first assumption.

26 Inclusive Result Theorem uppose that F(x, y, z) has continuous first partial derivatives throughout a simply connected region D, then the following statements are equivalent. 1 F is conservative in D, i.e. F = f. 2 F dr is independent of path in D. 3 F is irrotational in D, i.e. F = 0 in D. 4 F dr = 0 for every simple closed curve in D.

27 Example Let F(x, y, z) = y 2 i + (2xy + e 3z )j + 3ye 3z k and show that F is conservative.

28 Example Let F(x, y, z) = y 2 i + (2xy + e 3z )j + 3ye 3z k and show that F is conservative. We can accomplish this by showing that F = 0. i j k F = x y z = 0, 0, 0 y 2 2xy + e 3z 3ye 3z

29 Homework Read ection Exercises: 1 33 odd