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A Note About Orientation the Divergence Theorem Jim Lambers MAT 280 Spring Semester 2009-10 Lecture 24 Notes These notes correspond to Section 13.9 in Stewart and Section 8.4 in Marsden and Tromba. A Note About Orientation Recall Stokes' Theorem, Z ZZ F ⋅ dr = curl F ⋅ dS; C S where C is a simple, closed, positively oriented, piecewise smooth curve and S is a oriented surface enclosed by C. If C is parameterized by a function r(t), where a ≤ t ≤ b, and S is parameterized by a function g(u; v), where (u; v) 2 D, then Stokes' Theorem becomes Z b ZZ 0 F(r(t)) ⋅ r (t) dt = curl F(g(u; v)) ⋅ (gu × gv) du dv: a D It is important that the parameterizations r and g have the proper orientation for Stokes' Theorem to apply. This is why it is required that C have positive orientation. It means, informally, that if one were to \walk" along C, in such a way that n, the unit normal vector of S, can be viewed, then S should always be \on the left" relative to the path traced along C. It follows that the parameterizations of C and S must be consistent with one another, to ensure that they are oriented properly. Otherwise, one of the parameterizations must be reversed, so that the sign of the corresponding integral is corrected. The orientation of a curve can be reversed by changing the parameter to s = a + b − t. The orientation of a surface can be reversed by interchanging the variables u and v. The Divergence Theorem Let F be a vector field with continuous first partial derivatives. Recall a statement of Green's Theorem, Z ZZ F ⋅ n ds = div F dA; C D where n is the outward unit normal vector of D. Now, let E be a three-dimensional solid whose boundary, denoted by @E, is a closed surface S with positive orientation. Then, if we consider two- dimensional slices of E, each one being parallel to the xy-plane, then each slice is a region D with positively oriented boundary C, to which Green's Theorem applies. If we multiply the integrals on 1 both sides of Green's Theorem, as applied to each slice, by dz, the infinitesimal \thickness" of each slice, then we obtain ZZ ZZZ F ⋅ n dS = div F dV; S E or, equivalently, ZZ ZZZ F ⋅ dS = r ⋅ F dV: S E This result is known as the Gauss Divergence Theorem, or simply the Divergence Theorem. As the Divergence Theorem relates the surface integral of a vector field, known as the flux of the vector field through the surface, to an integral of its divergence over a solid, it is quite useful for converting potentially difficult double integrals into triple integrals that may be much easier to evaluate, as the following example demonstrates. Example (Stewart, Section 13.9, Exercise 6) Let S be the surface of the box with vertices (±1; ±2; ±3), and let F(x; y; z) = hx2z3; 2xyz3; xz4i. To compute the surface integral of F over S directly is quite tedious, because S has six faces that must be handled separately. Instead, we apply the Divergence Theorem to integrate div F over E, the interior of the box. We then have ZZ ZZZ F ⋅ dS = div F dV S E Z 1 Z 2 Z 3 2 3 3 4 = (x z )x + (2xyz )y + (xz )z dz dy dz −1 −2 −3 Z 1 Z 2 Z 3 = 2xz3 + 2xz3 + 4xz3 dz dy dx −1 −2 −3 Z 1 Z 2 Z 3 = 8xz3 dz dy dx −1 −2 −3 Z 1 Z 3 = 32 x z3 dz dx −1 −3 Z 1 " 4 3 # z = 32 x dx −1 4 −3 = 0: 2 The Divergence Theorem can also be used to convert a difficult surface integral into an easier one. 2 1 3 2 2 Example (Stewart, Section 13.9, Exercise 17) Let F(x; y; z) = hz x; 3 y + tan z; x z + y i. Let S be the top half of the sphere x2 + y2 + z2 = 1. To evaluate the surface integral of F over S, we note 2 2 that if we combine S with S1, the disk x + y ≤ 1, with downward orientation. We then obtain a 2 2 2 2 new surface S2 that is the boundary of the top half of the ball x + y + z ≤ 1, which we denote by E. By the Divergence Theorem, ZZ ZZ ZZZ F ⋅ dS + F ⋅ dS = div F dV: S S2 E We parameterize S2 by x = u sin v; y = u cos v; z = 0; 0 ≤ u ≤ 1; 0 ≤ v ≤ 2: This parameterization is used instead of the usual one arising from polar coordinates, due to the downward orientation. It follows from ru = hsin u; cos u; 0i; rv = hu cos v; −u sin v; 0i that 2 2 ru × rv = h0; 0; −u sin v − u cos vi = uh0; 0; −1i; which points downward, as desired. From 2 1 3 2 2 2 2 2 div F(x; y; z) = (z x)x + y + tan z + (x z + y )z = x + y + z ; 3 y which suggests the use of spherical coordinates for the integral over E, we obtain ZZ ZZZ ZZ F ⋅ dS = div F dV − F ⋅ dS S E S2 ZZZ Z 1 Z 2 = (x2 + y2 + z2) dV − F(x(u; v); y(u; v); z(u; v)) ⋅ uh0; 0; −1i dv du E 0 0 Z 1 Z 2 Z =2 Z 1 Z 2 = 22 sin d d d + u(u2 cos2 v) dv du 0 0 0 0 0 Z 1 Z =2 Z 1 Z 2 = 2 4 sin d d + u3 cos2 v dv du 0 0 0 0 Z 1 Z 1 Z 2 4 h =2i 3 1 + cos 2v = 2 − cos j0 d − u dv du 0 0 0 2 Z 1 Z 1 2 4 3 v sin 2v = 2 d + u + du 0 0 2 4 0 5 1 Z 1 3 = 2 + u du 5 0 0 4 1 2 u = + 5 4 0 3 2 = + 5 4 13 = : 20 2 Suppose that F is a vector field that, at any point, represents the flow rate of heat energy, which is the rate of change, with respect to time, of the amount of heat energy flowing through that point. By Fourier's Law, F = −KrT , where K is a constant called thermal conductivity, and T is a function that indicates temperature. Now, let E be a three-dimensional solid enclosed by a closed, positively oriented, surface S with outward unit normal vector n. Then, by the law of conservation of energy, the rate of change, with respect to time, of the amount of heat energy inside E is equal to the flow rate, or flux, or heat into E through S. That is, if (x; y; z) is the density of heat energy, then d ZZZ ZZ dV = F ⋅ (−n) dS; dt E S where we use −n because n is the outward unit normal vector, but we need to express the flux into E through S. From the definition of F, and the fact that = c0T , where c is the specific heat and 0 is the mass density, which, for simplicity, we assume to be constant, we have d ZZZ ZZ c0T dV = KrT ⋅ n dS: dt E S Next, we note that because c, 0, and E do not depend on time, we can write d ZZZ dT ZZ c0 dV = KrT ⋅ n dS: dt E dt S Now, we apply the Divergence Theorem, and obtain ZZZ ZZZ ZZZ d dT 2 c0 dV = Kr ⋅ (rT ) dV = Kr T dV: dt E dt E E That is, ZZZ dT 2 c0 − Kr T dV = 0: E dt Since the solid E is arbitrary, it follows that dT K = r2T: dt c0 This is known as the heat equation, which is one of the most important partial differential equations in all of applied mathematics. 4 Practice Problems Practice problems from the recommended textbooks are: ∙ Stewart: Section 13.9, Exercises 1-13 odd ∙ Marsden/Tromba: Section 8.4, Exercises 1-9 odd 5.
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