Unformatted text preview: 2 G2 (x, y, z ) = − x2 + 1 z + g2 (x, y ) . If we apply these to the last equation we get ∂g2 − ∂g1 = y . You can try to set
∂x
∂y
g2 to be zero because this is the last equation you need to satisfy. This yields
2
g1 (x, y ) = − y2 . Thus one solution is
G (x, y, z ) = y2
z2
− 2xyz − , − x2 + 1 z , 0 .
2
2 Note that G is not unique because if I choose any C 2 function f and ta...
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This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.
 Winter '13
 Magpantay
 Math, Vector Calculus

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