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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 =< 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...
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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.
 Spring '07
 Weiss
 Math, Calculus

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