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Divergence Theorem §17.3 11 December 2013 Fundamental Theorem, Four Ways. Rb I F 0 (x) dx = F (b) − F (a) [a, b] I If C path from P to Q, φ on boundary of C I Green’s Theorem: RR I Stokes’s Theorem: RR a R C F (x) on boundary of (∇φ) · ds = φ(Q) − φ(P) curlz (F) dA = R S R curl(F) · dS = ∂R F · ds R ∂S F · ds Every time: Integral of a derivative of something over some domain equals the integral of the original over the boundary of the domain. 3D Fundamental Theorem. Say R is a 3D region, with boundary surface ∂R, oriented with outward pointing normal. ZZZ ZZ (???) dV = F · dS R ∂R The derivative to use is the divergence: Definition Given a vector field F = hF1 , F2 , F3 i, the divergence is div F = ∇ · F = h∂x , ∂y , ∂z i · hF1 , F2 , F3 i = F1x + F2y + F3z . Divergence Theorem ZZZ ZZ F · dS. div F dV = R ∂R Example. Example: Calculate the flux of F(x, y , z) = hx + y , x − z, z 2 i through the boundary of the cube −1 ≤ x, y , z ≤ 1, oriented positively (with outward pointing normals). Example. Example: Calculate the flux of F(x, y , z) = hx + y , x − z, z 2 i through the boundary of the cube −1 ≤ x, y , z ≤ 1, oriented positively (with outward pointing normals). Remark: The surface integral is not difficult, but requires taking care of the cube’s 6 separate faces. If F would happen to be a curl vector field, then we would know the flux is zero; however F is NOT a curl vector field (can you figure out why not?). So, that shortcut is not available for this example. Example. Example: Calculate the flux of F(x, y , z) = hx + y , x − z, z 2 i through the boundary of the cube −1 ≤ x, y , z ≤ 1, oriented positively (with outward pointing normals). Remark: The surface integral is not difficult, but requires taking care of the cube’s 6 separate faces. If F would happen to be a curl vector field, then we would know the flux is zero; however F is NOT a curl vector field (can you figure out why not?). So, that shortcut is not available for this example. Answer: ZZ ZZZ F · dS = div F dV ∂C C Z 1Z 1Z 1 1 2 = (1 + 0 + 2z) dx dy dz = 4(z + z ) = 8. −1 −1 −1 −1 Why Divergence Theorem Is True. Let’s see why the Divergence Theorem works in the 2D case. I C : simple closed curve in R2 I enclosing region R with positive orientation (counterclockwise) I parametrized by c(t) = (x(t), y (t)). I I Then c0 (t) = hx 0 (t), y 0 (t)i outward-pointing normal: n(t) = hy 0 (t), −x 0 (t)i n n c0 c0 n c0 Explanation, continued. The flux of F across C is Z Z F · n ds = hF1 , F2 i · hy 0 (t), −x 0 (t)i dt C ZC = F1 dy − F2 dx C Z = h−F2 , F1 i · ds ZCZ = curlz (h−F2 , F1 i) dA (Green’s Theorem) R ZZ = (F1x + F2y ) dA. {z } R| div(F) Example. Example: Verify the Divergence Theorem for F(x, y , z) = hxz, x, z 2 i and the cylinder W given by 0 ≤ x 2 + y 2 ≤ 9, 0 ≤ z ≤ 6. Answer: First, compute the flux of F across ∂W (with outward pointing normals): I Top surface W1 : 0 ≤ x 2 + y 2 ≤ 9, z = 6, n = h0, 0, 1i. So F · n = z 2 = 36. ZZ ZZ Flux = F · dS = 36 dA = 36 Area(W1 ) W1 W1 = 36 · 9π = 324π I Bottom surface W3 : 0 ≤ x 2 + y 2 ≤ 9, z = 0, n = h0,RR 0, −1i. So F ·RR n = −z 2 = 0, flux = W3 F · dS = W3 0 dA = 0. Example, continued. Side surface W2 , parametrized by (x, y , z) = (3 cos θ, 3 sin θ, z), 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 6: Tθ = h−3 sin θ, 3 cos θ, 0i, Tz = h0, 0, 1i, n = h3 cos θ, 3 sin θ, 0i So, Z 2π Z 6 h3z cos θ, 3 cos θ, z 2 i · h3 cos θ, 3 sin θ, 0i dθ flux = 0 Z 0 2π Z = 0 6 9z cos2 θ + 9 cos θ sin θ dz dθ 0 = · · · = 162π. Therefore the total flux is 324π + 0 + 162π = 486π. Example, concluded. Alternatively, find the divergence: ∇ · F = z + 0 + 2z = 3z, so ZZ ZZZ F · dS = flux = ∇ · F dV ∂W (Divergence Theorem) W Z 6 ZZ = 3z dx dy dz 0≤x 2 +y 2 ≤9 0 Z 6 3z · Area(0 ≤ x 2 + y 2 ≤ 9) dz 0 1 2 6 = 27π · z 2 0 = 27π · 18 = 486π. = Physical Interpretation. If F represents the velocity of a fluid flow and S is a closed surface, then the flux of F across S represents the amount of fluid leaving or entering S. The divergence of F represents the flow rate (or flux) per unit volume. I If div(F)P > 0 it means there is net “creation” of the fluid near P. I If div(F)P < 0 it means there is net “destruction” of the fluid near P. Clicker Question: Moving from the picture on the left to the picture on the right, what are the signs of ∇ · F? A. B. C. D. positive, positive, negative zero, positive, negative positive, negative, zero zero, negative, positive receiver channel: 41 session ID: bsumath275 Clicker Question: True or False? The vector field F is defined everywhere in a region W bounded by a surface S. If ∇ · F > 0 at all points of W , then the vector field F points outward at all points of S. A. True, and I am very confident. B. True, but I am not very confident. C. False, but I am not very confident. D. False, and I am very confident. receiver channel: 41 session ID: bsumath275 Div, Grad, Curl. I If φ is a scalar-valued function, grad φ = ∇φ = h∂x , ∂y , ∂z iφ = ∂φ ∂φ ∂φ , , ∂x ∂y ∂z The result is a vector field. I If F is a vector field, curl F = ∇×F = h∂x , ∂y , ∂z i×hF1 , F2 , F3 i = hF3y −F2z , . . .i The result is a vector field. I If F is a vector field, div F = ∇ · F = h∂x , ∂y , ∂z i · hF1 , F2 , F3 i = F1x + F2y + F3z The result is a scalar-valued function. Clicker Question: If F(x, y , z) is a vector field and f (x, y , z) is a scalar function, which of the following is not defined? A. ∇f B. ∇ · F + f C. F + ∇f D. ∇ · F + ∇f E. More than one of the above F. None of the above receiver channel: 41 session ID: bsumath275 Clicker Question: If F(x, y , z) is a vector field and f (x, y , z) is a scalar function, which of the following quantities is a vector? A. ∇ · F B. ∇f · u~ C. ∇ · ∇f D. (∇ · F)F receiver channel: 41 session ID: bsumath275 Notice: If F is a curl vector field, F = ∇ × A, then its divergence is zero: ∇ · (∇ × A) = 0 div(curl A) = 0 This is like ∇ × (∇φ) = 0. I Just like checking cross-partials is a test for exactness, we can use divergence to test whether F is a curl vector field. I Interesting connections to further topics in math: geometry, topology Worksheet. Worksheet #1–3 Clicker Question: What is the divergence of the vector field hx 2 , y − e z , y − 2zxi? A. 0 B. 1 C. −1 D. h2x, 1, −2xi E. h1 + e z , 2z, 0i receiver channel: 41 session ID: bsumath275 Maxwell’s Equations, First Two. Maxwell’s equations describe electricity and magnetism. They revolutionized science and mathematics. I The divergence of the electric field is the charge density: ∇ · E = q. (Sources and sinks of the electric field are given by charged particles.) In other words, the flux of the electric field through any closed surfaceRR is equal to the total charge enclosed inside that surface: ∂W E · dS = charge inside W . I The divergence of the magnetic field is zero: ∇ · B = 0. There are no magnetic monopoles. The flux of the magnetic field through any closed surface is zero: RR B · dS = 0. ∂W (For simplicity, leaving out some constant factors.) Maxwell’s Equations, Last Two. I The time derivative of the magnetic field is the current plus the curl of the electric field: ∂B = J + ∇ × E. ∂t A changing magnetic field induces both current and electric field. I The time derivative of the electric field is the opposite of = −∇ × B. the curl of the magnetic field: ∂E ∂t A changing electric field induces a magnetic field. ∇·E=q ∇·B=0 ∂B =J +∇×E ∂t ∂E = −∇ × B ∂t RR ∂W E · dS = charge inside W RR B · dS = 0 ∂W