Week 13 - Solution

# 0 2 c f dr 0 1 2 1 0 dr d 0 0 3 a 2r

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Unformatted text preview: ∂ ∂x ∂y ∂z xz 2xy 3xy 5 = [3x, x − 3y, 2y ] F • dr = − 3 2 r (x, y ) = [x, y, 3 − 3x − y ] with 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 − 3x 3−3x 1 C F • dr = 0 0 [3x, x − 3y, 2y ] • ([1, 0, −3] × [0, 1, −1]) dy dx = 7 2 (b) curl F = [x, 2z − y, 0] r (x, y ) = [x, y, 2 − 2x − 2y ] with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x 1−x 1 F • dr = C 0 0 [x, 2z − y, 0] • ([1, 0, −2] × [0, 1, −2]) dy dx = 4 3 (c) curl F = [5, 2, 4] r (r, θ) = [r cos θ, r sin θ, r cos θ + 4] with 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π 2π C F•dr = 0 2 [5, 2, 4]•([cos θ, sin θ, cos θ] × [−r sin θ, r cos θ, −r sin θ]) dr dθ = −4π 0 (d) r (r, θ) = r cos θ, r sin θ, 1 − r2 with 0 ≤ r ≤ 1, 0 ≤ θ ≤ π 2 curl F = [2y, −2x, 0] = [2r sin θ, −2r cos θ, 0] π 2 C F • dr = 0 1 π 2 1 = 0 dr dθ = 0 0 3. (a) [2r sin θ, −2r cos θ, 0] • ([cos θ, sin θ, −2r] × [−r sin θ, r cos θ, 0]) dr dθ 0 0 i. S is part of a sphere and it is an OPENED surface. Since we can only apply the Divergence theorem when the surface is CLOSED, then we have to add another surface at the bottom and make the surface to be closed. So, we introduce a new surface S1 for the bottom part of the CLOSED surface. E is the region inside the surface, which is upper part of the sphere and E= (ρ, φ, θ| 0 ≤ ρ ≤ 1, 0 ≤ φ ≤...
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## This document was uploaded on 01/23/2014.

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