1211s2 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1211 Multivariable Calculus 2010-11 1st Semester: Solution to Assignment 2 1. Using Cartesian coordinates, we have x = ρ sin ϕ cos θ,y = ρ sin ϕ sin θ,z = ρ cos ϕ and the given equation ρ sin ϕ sin θ = 2 is translated to y = 2. In Cylindrical coordinates, r = ρ sin ϕ,θ = θ,z = ρ cos ϕ and the equation is translated to r sin θ = 2. The surface is a vertical plane. 2. Since p x 2 + y 2 + z 2 = ρ , from 0 ρ 2 we obtain 0 p x 2 + y 2 + z 2 2. The solid is a portion of the sphere of radius 2. Also we have 0 ρ π 4 , hence, the solid is an ice-cream- cone-like solid in R 3 3. Assume that the centre of the sphere (that is, the completion of hemisphere) is positioned at the origin. The solid is, when using Spherical coordinates, the set: { ( ρ,ϕ,θ ) | 0 ρ 5 ,π/ 2 ρ π, 0 θ < 2 π } When using Cylindrical coordinates, the solid is the set: { ( r,θ,z ) | z 2 + r 2 5 ,z 0 , 0 θ < 2 π } 4. (a) Evaluating the limit along the line x = 0, we obtain lim ( x,y ) (0 , 0) ,x =0 2 xy x 2 + y 2 = lim y 0 0 y 2 = 0 . Evaluating the lime along the line x = y , we obtain lim ( x,y ) (0 , 0) ,x = y 2 xy x 2 + y 2 = lim x 0 2 x 2 x 2 + x 2 = 1 . Because these two limits are not equal, we conclude that lim ( x,y ) (0 , 0) ,x =0 2 xy x 2 + y 2 does not exist. (Remark : You may use polar form to solve this problem.) (b) Note that when ( x,y,z ) (0 , 0 , 0), we have y 0, y 0, and z 0 respectively. Note also 0 ± ± ± ± y 3 x 2 + y 2 + z 4 ± ± ± ± = | y | y 2 x 2 + y 2 + z 4 ≤ | y | . Here the last inequality follows because (0 ) y 2 x 2 + y 2 + z 4 1. Thus when ( x,y,z ) (0 , 0 , 0), we have x 0, and hence y 3 x 2 + y 2 + z 4 0 also, i.e., lim ( x,y,z ) (0 , 0 , 0) y 3 x 2 + y 2 + z 4 = 0. Similarly, we have
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This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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1211s2 - THE UNIVERSITY OF HONG KONG DEPARTMENT OF...

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