MATH1813 Problem Set 4 Ans

MATH1813 Problem Set 4 Ans - University of Hong Kong Fall...

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University of Hong Kong Fall 2011 Math 1813 Mathematical Methods for Actuarial Science Instructor: Siye Wu Solution 4 I. In cylindrical polar coordinates: Z = { ( r cos θ,r sin θ,z ) T R 3 : 0 6 θ < 2 π,r = 2 ,z R } S = { ( r cos θ,r sin θ,z ) T R 3 : 0 6 θ < 2 π,r > 0 ,z R ,r 2 + z 2 = 9 } C = { ( r cos θ,r sin θ,z ) T R 3 : 0 6 θ < 2 π,r > 0 ,z R ,r 2 = z 2 } In spherical polar coordinates: Z = { ( r sin θ cos φ,r sin θ sin φ,r cos θ ) T R 3 : 0 < θ < π, 0 6 φ < 2 π,r sin θ = 2 } S = { ( r sin θ cos φ,r sin θ sin φ,r cos θ ) T R 3 : 0 6 θ 6 π, 0 6 φ < 2 π,r = 3 } C = { ( r sin θ cos φ,r sin θ sin φ,r cos θ ) T R 3 : θ = π 4 or 3 π 4 , 0 6 φ < 2 π,r > 0 } II. We have J f ( x,y ) = ± - y sin( xy ) - x sin( xy ) cos( x + y ) cos( x + y ) , so J f (0 ) = ± 0 0 - 1 - 1 . (a) One has J F (2 , 3) = J f (0 ) J g (2 , 3) = ± 0 0 - 1 - 1 ¶± 3 7 - 2 0 = ± 0 0 - 1 - 7 . (b) Note that
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This note was uploaded on 12/20/2011 for the course MATH 1813 taught by Professor Wu during the Fall '11 term at HKU.

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MATH1813 Problem Set 4 Ans - University of Hong Kong Fall...

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