1 1 Divergence2 2 Divergence If F = P i + Q j + R k is a vector field on and ∂P/∂x, ∂Q/∂y, and ∂R/∂z exist, then the divergence of F is the function of three variables defined by Observe that curl F is a vector field but div F is a scalar field.3 3 Divergence In terms of the gradient operator ∇ = (∂/∂x) i + (∂/∂y) j + (∂/∂z) k, the divergence of F can be written symbolically as the dot product of ∇ and F:4 4 Divergence If F is a vector field on , then curl F is also a vector field on . As such, we can compute its divergence. The next theorem shows that the result is 0.5 5 Vector Forms of Green’s Theorem6 6 Vector Forms of Green’s Theorem If C is given by the vector equation r(t) = x(t) i + y(t) j a ≤ t ≤ b then the unit tangent vector is7 7 Vector Forms of Green’s Theorem The outward unit normal vector to C is given by (See Figure 2.) Figure 28 8 Vector Forms of Green’s Theorem Then from the equation we have9 9 Vector Forms of Green’s Theorem by Green’s Theorem. But the integrand in this double integral is just the divergence of F.10 10 Vector Forms of Green’s Theorem So we have a second vector form of Green’s Theorem. This version says that the line integral of the normal component of F along C is equal to the double integral of the divergence of F over the region D enclosed by C.11 11 16.9 The Divergence Theorem12 12 The Divergence Theorem • We write Green’s Theorem in a vector version as • where C is the positively oriented boundary curve of the plane region D. • If we were seeking to extend this theorem to vector fields on we might make the guess that • where S is the boundary surface of the solid region E.13 13 The Divergence Theorem • It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem. • Notice its similarity to Green’s Theorem and Stokes’ Theorem in that it relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region. • We state the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3 and we call such regions simple solid regions. (For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.)14 14 The Divergence Theorem • The boundary of E is a closed surface, and we use the convention, that the positive orientation is outward; that is, the unit normal vector n is directed outward from E. • Thus the Divergence Theorem states that, under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.15 15 Example 1 • Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x2 + y2 + z2 = 1. • Solution: • First we compute the divergence of F: • The unit sphere S is the boundary of the unit ball B given by x2 + y2 + z2 ≤ 1.16 16 Example 1 – Solution • Thus the Divergence Theorem gives the flux as cont’d17 17 The Divergence Theorem • Let’s consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2. Let n1 and n2 be outward normals of S1 and S2. • Then the boundary surface of E is S = S1 U S2 and its normal n is given by n = –n1 on S1 and n = n2 on S2. (See Figure 3.) Figure 318 18 The Divergence Theorem • Applying the Divergence Theorem to S, we get The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.19 19 Example 3 • We considered the electric field: • where the electric charge Q is located at the origin and is a position vector. • Use the Divergence Theorem to show that the electric flux of E through any closed surface S2 that encloses the origin is20 20 Example 3 – Solution • The difficulty is that we don’t have an explicit equation for S2 because it is any closed surface enclosing the origin. The simplest such surface would be a sphere, so we let S1 be a small sphere with radius a and center the origin. You can verify that div E = 0. • Therefore Equation 7 gives21 21 Example 3 – Solution • The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere. The normal vector at x is x/| x |. • Therefore • since the equation of S1 is | x | = a. cont’d22 22 Example 3 – Solution • Thus we have • This shows that the electric flux of E is 4πεQ through any closed surface S2 that contains the origin. [This is a special case of Gauss’s Law for a single charge. The relationship between ε and ε0 is ε =1/(4πε0).] cont’d23 23 The Divergence Theorem • Another application of the Divergence Theorem occurs in fluid flow. Let v(x, y, z) be the velocity field of a fluid with constant density ρ. Then F = ρ v is the rate of flow per unit area.24 24 The Divergence Theorem • If P0(x0, y0, z0) is a point in the fluid and Ba is a ball with center P0 and very small radius a, then div F(P) ≈ div F(P0) for all points in Ba since div F is continuous. We approximate the flux over the boundary sphere Sa as follows: • This approximation becomes better as a → 0 and suggests that25 25 The Divergence Theorem • Equation 8 says that div F(P0) is the net rate of outward flux per unit volume at P0. (This is the reason for the name divergence.) • If div F(P) > 0, the net flow is outward near P and P is called a source. • If div F(P) < 0, the net flow is inward near P and P is called a sink.26 26 The Divergence Theorem • For the vector field in Figure 4, it appears that the vectors that end near P1 are shorter than the vectors that start near P1. Figure 4 The vector field F = x2 i …
View Full Document
Unlocking...