Math13S09final_sol

# Divf 1 2 3 therefore we have by the divergence

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Unformatted text preview: − x2 − y 2 + x2 + y 2 dA = dA so, 2 z −z Using (a): 2π z dS = S 2π = 0 1 1 − dθ = 2π · − 2 2 2π π cos φ sin φ dφdθ = θ =0 φ=π/2 0 1 sin2 φ 2 π dθ π/2 = −π 9. divF = 1 + 2 = 3 Therefore we have by the divergence theorem E 3 dV = 3 · (volume of box) = 3 · 12 = 36 10. Note that the boundary is the same for the disk S1 : x2 + y 2 ≤ 4, z = 5, which has unit normal of n =&lt; 0, 0, 1 &gt; and projection R: x2 + y 2 ≤ 4 in the xy -plane. We have curl F =&lt; 0, 2x, x2 &gt; so by Stoke’s Theorem and with changing to polar coordinates 2 2π 2π 2 14 x2 dA = r 3 cos2 θdrdθ = curl F · ndS = F dr = r cos2 θ dθ 4 θ =0 r =0 0 R S1 C 0 2π 2π 2π 1 = 4π 1 + cos 2θ dθ = 2 θ + sin 2θ r cos2 θdθ = 2 = 2 0 0 0 3...
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## This note was uploaded on 01/20/2014 for the course MATH 13 taught by Professor Weiss during the Spring '07 term at Tufts.

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