Math13S09final_sol

Divf 1 2 3 therefore we have by the divergence

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 =< 0, 0, 1 > and projection R: x2 + y 2 ≤ 4 in the xy -plane. We have curl F =< 0, 2x, x2 > 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...
View Full Document

This note was uploaded on 01/20/2014 for the course MATH 13 taught by Professor Weiss during the Spring '07 term at Tufts.

Ask a homework question - tutors are online