MATH 311 Topics in Applied Mathematics I Lecture 34: Conservative vector fields. Area of a surface. Surface integrals.

Transcription

1 MATH 311 Topics in Applied Mathematics I Lecture 34: Conservative vector fields. Area of a surface. Surface integrals.

2 Conservative vector fields Let R be an open region in R n such that any two points in R can be connected by a continuous path. Such regions are called (arcwise) connected. efinition. A continuous vector field F : R R n is called conservative if F ds = F ds C 1 C 2 for any two simple, piecewise smooth, oriented curves C 1,C 2 R with the same initial and terminal points. An equivalent condition is that F ds = 0 for any piecewise smooth closed curve C R. C

3 Conservative vector fields Theorem The vector field F is conservative if and only if it is a gradient field, that is, F = f for some function f : R R. If this is the case, then F ds = f(b) f(a) C for any piecewise smooth, oriented curve C R that connects the point A to the point B. Remark. In the case F is a force field, conservativity means that energy is conserved. Moreover, in this case the function f is the potential energy.

4 Test of conservativity Theorem If a smooth field F = (F 1,F 2,...,F n ) is conservative in a region R R n, then the Jacobian matrix (F 1,F 2,...,F n ) (x 1,x 2,...,x n ) is symmetric everywhere in R, that is, F i x j = F j x i for i j. Indeed, if the field F is conservative, then F = f for some smooth function f : R R. It follows that the Jacobian matrix of F is the Hessian matrix of f, that is, the matrix of F i second-order partial derivatives: = 2 f. x j x j x i Remark. The converse of the theorem holds provided that the region R is simply-connected, which means that any closed path in R can be continuously shrunk within R to a point.

5 Finding scalar potential Example. F(x,y) = (2xy 3 +3y cos3x, 3x 2 y 2 +sin3x). The vector field F is conservative if F 1 / y = F 2 / x. F 1 y = F 2 6xy2 +3cos3x, x = 6xy2 +3cos3x. Thus F = f for some function f (scalar potential of F), f that is, x = f 2xy3 +3y cos3x, y = 3x2 y 2 +sin3x. Integrating the second equality by y, we get f(x,y) = (3x 2 y 2 +sin3x)dy = x 2 y 3 +y sin3x +g(x). Substituting this into the first equality, we obtain that 2xy 3 +3y cos3x +g (x) = 2xy 3 +3y cos3x. Hence g (x) = 0 so that g(x) = c, a constant. Then f(x,y) = x 2 y 3 +y sin3x +c.

6 Surface Suppose 1 and 2 are domains in R 3 and T : 1 2 is an invertible map such that both T and T 1 are smooth. Then we say that T defines curvilinear coordinates in 1. efinition. A nonempty set S R 3 is called a smooth surface if for every point p S there exist curvilinear coordinates T : 1 2 in a neighborhood of p such that T(p) = 0 and either T(S 1 ) = {(x,y,z) 2 z = 0} or T(S 1 ) = {(x,y,z) 2 z = 0, y 0}. In the first case, p is called an interior point of the surface S, in the second case, p is called a boundary point of S. The set of all boundary points of the surface S is called the boundary of S and denoted S. A smooth surface S is called complete if for any convergent sequence of points from S, the limit belongs to S as well. A complete surface with no boundary points is called closed.

7 Parametrized surfaces efinition. Let R 2 be a connected, bounded region. A continuous one-to-one map X : R 3 is called a parametrized surface. The image X() is called the underlying surface. The parametrized surface is smooth if X is smooth and, moreover, the vectors X (s s 0,t 0 ) and X (s t 0,t 0 ) are linearly independent for all (s 0,t 0 ). If this is the case, then the plane in R 3 through the point X(s 0,t 0 ) parallel to vectors X (s s 0,t 0 ) and X(s t 0,t 0 ) is called the tangent plane to X() at X(s 0,t 0 ). Example. Suppose f : R 3 R is a smooth function and consider a level set P = {(x,y,z) : f(x,y,z) = c}, c R. If f 0 at some point p P, then near that point P is the underlying surface of a parametrized surface. Moreover, the gradient ( f)(p) is orthogonal to the tangent plane at p.

8 Area of a surface Let P be a smooth surface parametrized by X : R 3. Then the area of P is area(p) = X s X t dsdt. Suppose P is the graph of a smooth function g : R, i.e., P is given by z = g(x,y). We have a natural parametrization X : R 3, X(x,y) = (x,y,g(x,y)). Then X = x (1,0,g x) and X = y (0,1,g y). Consequently, X x X y = It follows that area(p) = e 1 e 2 e g x 0 1 g y = ( g x, g y,1). 1+ g x 2 + g y 2 dx dy.

9 Scalar surface integral Scalar surface integral is an integral of a scalar function f over a parametrized surface X : R 3 relative to the area element of the surface. It can be defined as a limit of Riemann sums S(f,R,τ j ) = k j=1 f( x(τ j ) ) area ( X( j ) ), where R = { 1, 2,..., k } is a partition of into small pieces and τ j j for 1 j k. Theorem Let X : R 3 be a smooth parametrized surface, where R 2 is a bounded region. Then for any continuous function f : X() R, f ds = f ( X(s,t) ) X s X t dsdt. X

10 Vector surface integral Vector surface integral is an integral of a vector field over a smooth parametrized surface. It is a scalar. efinition. Let X : R 3 be a smooth parametrized surface, where R 2 is a bounded region. Then for any continuous vector field F : X() R 3, the vector integral of F along X is F ds = F ( X(s,t)) N(s,t)dsdt, X where N = X Equivalently, X s t X, a normal vector to the surface. F 1 F 2 F 3 X F ds = 1 X 2 X 3 s s s dsdt. X 1 t X 2 t X 3 t

11 Applications of surface integrals Mass of a shell If f is the density of a shell P, then f ds is the mass of P. P Center of mass of a shell If f is the density of a shell P, then xf(x,y,z)ds P P P f ds, yf(x,y,z)ds P P f ds, zf(x,y,z)ds f ds P are coordinates of the center of mass of P. Flux of fluid If F is the velocity field of a fluid, then F ds is the flux P of the fluid across the surface P.

12 Surface integrals and reparametrization Given two smooth parametrized surfaces X : 1 R 3 and Y : 2 R 3, we say that Y is a smooth reparametrization of X if there exists an invertible function H : 2 1 such that Y = X H and both H and H 1 are smooth. Theorem Any scalar surface integral is invariant under smooth reparametrizations. As a consequence, we can define the scalar integral of a function over a non-parametrized smooth surface.

13 Any vector surface integral can be represented as a scalar surface integral: F ds = F ( X(s,t)) N(s,t)dsdt = (F n)ds, X where n = N N is a unit normal vector to the surface. Note that n depends continuously on a point on the surface, hence determining an orientation of X. A smooth reparametrization may be orientation-preserving (when n is preserved) or orientation-reversing (when n is changed to n). Theorem Any vector surface integral is invariant under smooth orientation-preserving reparametrizations and changes its sign under orientation-reversing reparametrizations. As a consequence, we can define the vector integral of a vector field over a non-parametrized, oriented smooth surface.

14 Moebius strip: non-orientable surface M.C.Escher, 1963

15 Example Let C denote the closed cylinder with bottom given by z = 0, top given by z = 4, and lateral surface given by x 2 +y 2 = 9. We orient C with outward normals. (xe 1 +ye 2 ) ds =? C The top of the cylinder is parametrized by X top : R 3, X top (x,y) = (x,y,4), where = {(x,y) R 2 : x 2 +y 2 9}. The bottom is parametrized by X bot : R 3, X bot (x,y) = (x,y,0). The lateral surface is parametrized by X lat : [0,2π] [0,4] R 3, X lat (φ,z) = (3cosφ,3sinφ,z).

16 We have Xtop X = (1,0,0), top = (0,1,0). Hence x y X top Xtop = e x y 1 e 2 = e 3. Since X bot = X top (0,0,4), we also have X bot = e x 1, X bot = e y 2, and X bot X bot = e x y 3. X Further, lat = ( 3sinφ,3cosφ,0) and X lat = (0,0,1). φ z Therefore X lat φ X lat e 1 e 2 e 3 = z 3sinφ 3cosφ = (3cosφ,3sinφ,0). We observe that X top and X lat agree with the orientation of the surface C while X bot does not. It follows that F ds = F ds F ds+ F ds. C X top X bot X lat

17 Integrating the vector field F = xe 1 +ye 2 over each part of C, we obtain: F ds = (x,y,0) (0,0,1)dxdy = 0dx dy = 0, X top F ds = (x,y,0) (0,0,1)dxdy = 0dx dy = 0, X bot F ds = X lat = (3cosφ,3sinφ,0) (3cosφ,3sinφ,0)dφdz [0,2π] [0,4] = 9dφdz = 72π. Thus [0,2π] [0,4] C F ds = 72π.