A c r t cos t sin t 0 with 0 t 2 figure 43 2 curl

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Unformatted text preview: 2 − 2x − y } div F = x 2−2x 1 F • dS = 0 0 2−2x−y x dz dy dx = 0 1 6 S (c) E = {(r, θ, z ) | 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, r sin θ − 3 ≤ z ≤ 0} div F = 2y = 2r sin θ 2π F • dS = 0 0 3 0 S r sin θ−3 2r sin θ (r) dz dr dθ = − 81π 2 (d) E = (ρ, φ, θ) | 0 ≤ ρ ≤ 2, 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π 2 div F = 3 F • dS = S 3 dV = 3 E dV = 3 × (Volume of the hemisphere) E 14 =3 × × π (2)3 = 16π 23 Stoke’s Theorem 1. (a) C : r (t) = [cos t, sin t, 0] with 0 ≤ t ≤ 2π (Figure 43) 2π curl F • dS = S C F • dr = 0 3 0, cos t, ecos t sin t • [− sin t, cos t, 0] dt = π z S y C x Figure 43 (b) z = x2 + y 2 =⇒ z = 1 x2 + y 2 = 1 C : r (t) = [cos t, sin t, 1] with 0 ≤ t ≤ 2π (Figure 44) 2π curl F • dS = 0 (sin t)2 , cos t, (cos t sin t)2 • [− sin t, cos t, 0] dt = π S z C S y x Figure 44 (c) √ x = 9 − y2 − z2 =⇒ x = 5 2 2 y +z =4 C : r (t) = √ 5, 2 cos t, 2 sin t with 0 ≤ t ≤ 2π (Figure 45) 2π curl F • dS = 0 4 sin t cos t − 16 sin2 t cos2 t dt = −4π S 4 z y x Figure 45 (d) C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1). (Figure 46) Let Γ1 , Γ2 , and Γ3 , line segments, be the edges of the triangle. Γ1 : from (1, 0, 0) to (0, 1, 0), r (t) = [1 − t, t, 0], with 0 ≤ t ≤ 1 Γ2 : from (0, 1, 0) to (0, 0, 1), r (t) = [0, 1 − t, t], with 0 ≤ t ≤ 1 Γ3 : from (0, 0, 1) to (1, 0, 0), r (t) = [t, 0, 1 − t], with 0 ≤ t ≤ 1 1 1 2 Γ1 0 1 1 F • dr = [1 − t, t, 0] • [0, −1, 1] dt = − 2 Γ2 0 1 1 [0, 1 − t, t] • [1, 0, −1] dt = − F • dr = 2 Γ3 0 F • dr = curl F • dS = S C [t, 0, 1 − t] • [−1, 1, 0] dt = − F • dr = Γ1 F • dr + Γ2 F • dr + Γ3 z y x Figure 46 2. (a) curl F = i j k ∂ ∂...
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This document was uploaded on 01/23/2014.

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