# 15_6_13 - 2 θ 1 ² = 48 m ± π 4-Z π 4 cos θ d θ √...

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13. F = m r | r | 3 = m ( x i + y j + z k ) ( x 2 + y 2 + z 2 ) 3 / 2 . By symmetry, the flux of F out of the cube - a x , y , z a is 6 times the flux out of the top face, z = a , where ˆ N = k and dS = dx dy . The total flux is y x a a R - a - a Fig. 6-13 6 ma Z - a x a - a y a dx dy ( x 2 + y 2 + a 2 ) 3 / 2 = 48 ma ZZ R r dr d θ ( r 2 + a 2 ) 3 / 2 ( R as shown in the figure) = 48 ma Z π/ 4 0 d θ Z a sec θ 0 r dr ( r 2 + a 2 ) 3 / 2 Let u = r 2 + a 2 du = 2 r dr = 24 ma Z π/ 4 0 d θ Z a 2 ( 1 + sec 2 θ) a 2 du u 3 / 2 = 48 ma Z π/ 4 0 ± 1 a - 1 a 1 + sec 2 θ ² d θ = 48 m ± π 4 - Z π/ 4 0 cos θ d θ cos
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Unformatted text preview: 2 θ + 1 ² = 48 m ± π 4-Z π/ 4 cos θ d θ √ 2-sin 2 θ ² Let √ 2 sin v = sin θ √ 2 cos v d v = cos θ d θ = 48 m π 4-Z π/ 6 √ 2 cos v d v √ 2 cos v ! = 48 m ³ π 4-π 6 ´ = 4 π m ....
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## This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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