# If Σ is an oriented surface bounded by a curve C, then the orientation of Σ induces an orientation for C, based on the Right-Hand-Rule. - PDF

Description
Oriented Surfaces and Flux Integrals Let be a surface that has a tangent plane at each of its nonboundary points. At such a point on the surface two unit normal vectors exist, and they have opposite directions.

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Information
Category:

## Self Improvement

Publish on:

Views: 26 | Pages: 9

Share
Transcript
Oriented Surfaces and Flux Integrals Let be a surface that has a tangent plane at each of its nonboundary points. At such a point on the surface two unit normal vectors exist, and they have opposite directions. If it is possible to select one normal at each nonboundary point in such a way that the chosen normal varies continuously on the whole of, then the surface is said to be orientable, or two-sided, and the selection of the normal gives an orientation to and thus makes an oriented surface. In such case, there are two possible orientations. Some surfaces are not orientable. For example, the Möbus band; it is one-sided. Induced Orientation If is an oriented surface bounded by a curve C, then the orientation of induces an orientation for C, based on the Right-Hand-Rule. Flux Integrals Let F be a vector field and ˆn be the unit normal to the oriented surface, the flux integral over is F ˆn ds. 1 This integral gives the net flux through. The field strength (i.e. F ) can be measured as the amount of flux per unit area perpendicular to the local direction of F. When is the graph of a function f with continuous partials on a region R in the xy plane that is composed of vertically or horizontally simple regions, and its orientation is chosen to be directed upward (i.e. the k component of the unit normal is positive), then F ˆndS = [ F 1 f x F 2 f y + F 3 ]da R for F = F 1 i + F 2 j + F 3 k. Note Another way to remember this is F ˆndS = F NdA R where N = f x i + f y j + k is the upward normal to with the z component equal to 1. 2 Example Suppose is the part of the paraboloid z = 1 x 2 y 2 that lies above the xy plane and is oriented by the unit normal directed upward. Assume that the velocity of a fluid is v = x i + y j + 2z k. Determine the flow rate (volume/time) through. 2π Flux integrals can be defined for a surface composed of several oriented surfaces 1, 2, n, as F ˆndS = 1 F ˆndS + + n F ˆndS Example Let be the unit sphere x 2 + y 2 + z 2 = 1, oriented with the unit normal directed outward, and let F(x,y,z) = z k. Find F ˆndS. 4π/3 3 The Divergence Theorem Definition A solid region D is called a simple solid region if D is the solid region between the graphs of two functions f(x, y) and g(x, y) on a simple region R in the xy plane and if D has the corresponding properties with respect to the xz plane and the yz plane. Theorem Let D be a simple solid region whose boundary surface is oriented by the normal ˆn directed outward from D, and let F be a vector field whose component functions have continuous partial derivatives on D. Then F ˆndS = FdV. D pf: Let F = F 1 i + F 2 j + F 3 k, then F 1 F 1 i ˆndS = x dv F 2 j ˆndS = D D F 3 k ˆndS = D F 2 y dv F 3 z dv 4 Note The interpretation of flux through the boundary of D. F ˆndS is net outward Example Let D be the region bounded by the xy plane and the hemisphere x 2 + y 2 + z 2 = 4 with z 0, and let F(x,y,z) = 3x 4 i + 4xy 3 j + 4xz 3 k. Evaluate F ˆndS, where is the boundary of D. 0 Note If the domain of integration R is symmetry with respect to certain thing and the integrand is anti-symmetric w.r.t. the same thing, then the integral is 0. For example, if f( x,y,z) = f(x,y,z), the function f is anti-symmetric w.r.t. the yz-plane. 5 The theorem can be applied to a not so simple solid region by considering the division of the region into a number of simple solid regions. Example Let F = q r 2 ˆr. Show that for any suface enclosing the origin, F ˆndS = 4πq. 6 Stokes Theorem Theorem Let be an oriented surface with normal ˆn (unit vector) and finite surface area. Assume that is bounded by a closed, piecewise smooth curve C whose orientation is induced by. Let F be a continuous vector field defined on, and assume that the component functions of F have continuous partial derivatives at each nonboundary point of. Then F d r = ( F) ˆndS C Example Let C be the intersection of the paraboloid z = x 2 +y 2 and the plane z = y, and give C a counterclockwise direction as viewed from the positive z axis. Evaluate 2xydx + x 2 dy + z 2 dz. C What about (2xy y)dx + x 2 dy + z 2 dz? C 0; π/4 7 Example Verify Stokes Theorem for F = 3y i xz j+ yz 2 k where is the surface of the paraboloid 2z = x 2 +y 2 bounded by z = 2, and C is its boundary. F=(z 2 +x) i (z+3) k 20π Problem Set 12 Proof of Stokes theorem ( F) N = ( y F 3 z F 2 )( f x )+( z F 1 x F 3 )( f y ) + ( x F 2 y F 1 ) = ( x F 2 +f x z F 2 +f y x F 3 ) ( y F 1 +f y z F 1 +f x y F 3 ) = x (F 2 (x,y,f(x,y)) + f y F 3 (x,y,f(x,y)) y (F 1 (x,y,f(x,y)) + f x F 3 (x,y,f(x,y)) Now apply Green s theorem on the xy-plane as following. ( x (F 2 + f y F 3 ) ) y (F 1 + f x F 3 ) da R 8 = (F 1 + f x F 3 )dx + (F 2 + f y F 3 )dy R r(t)=(x(t),y(t),z(t)) where z(t)=f(x(t),y(t)) So that r (t)=(x (t),y (t),fxx (t)+fy y (t)) and F r (t)=f 1 x +F 2 y +(F 3 fxx +F 3 fyy ) = F 1 dx + F 2 dy + F 3 dz. Therefore, ( F) ˆndS = ( F) NdA = F d r. R 9
Related Search
Similar documents

View more...