2011_Solution_9 - MA 1505 Mathematics I Tutorial 9 Solutions 1 Let z = 22 x2 y 2 Then zx = x(4 x2 y 2)1/2 and zy = y(4 x2 y 2)1/2 Substitute z = 1

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MA 1505 Mathematics I Tutorial 9 Solutions 1. Let z = ° 2 2 x 2 y 2 .Then z x = x (4 x 2 y 2 ) 1 / 2 and z y = y (4 x 2 y 2 ) 1 / 2 . Substitute z =1into x 2 + y 2 + z 2 = 4 gives x 2 + y 2 +1=4 x 2 + y 2 =3 which is the equation of a circle of radius 3. This means the plane z = 1 intersects the sphere at a circle of radius 3. Hence the projected region R of the part of the sphere is a disk of radius 3. In polar coordinates, this is given by 0 r 3 , 0 θ 2 π. Thus, A ( S )= ±± R ² x 2 + y 2 4 x 2 y 2 +1 dA = ± 2 π 0 ± 3 0 ³ r 2 4 r 2 +1 ´ 1 2 rdrdθ = ± 2 π 0 ± 3 0 2 r (4 r 2 ) 1 2 drdθ = ± 2 π 0 µ 2(4 r 2 ) 1 2 r = 3 r =0 =(2 π )[ 2[(4 3) 1 2 + 2[(4) 1 2 ]=4 π . 2. We use the criteria of conservative Feld: ∂y (2 xy )=2 x = ∂x ( x 2 +2 yz ) , ∂z (2 xy )=0= ∂x ( y 2 ) , ∂z ( x 2 +2 yz )=2 y = ∂y ( y 2 ) . So F is conservative, and hence there exists a function f such that f = F .
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This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

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2011_Solution_9 - MA 1505 Mathematics I Tutorial 9 Solutions 1 Let z = 22 x2 y 2 Then zx = x(4 x2 y 2)1/2 and zy = y(4 x2 y 2)1/2 Substitute z = 1

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