16.5 Ex 39

# 16.5 Ex 39 - S E C T I O N 16.5 Change of Variables(ET...

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SECTION 16.5 Change of Variables (ET Section 15.5) 1013 Let 8 1 ( u ,v) = ( Au + C v, Bu + D v) .Weaskthat 8 1 ( 0 , 1 ) = ( 1 , 1 ), 8 1 ( 1 , 0 ) = ( 1 , 1 ) That is, 8 1 ( 0 , 1 ) = ( C , D ) = ( 1 , 1 ) C = 1 , D = 1 8 1 ( 1 , 0 ) = ( A , B ) = ( 1 , 1 ) A = 1 , B =− 1 Therefore, 8 1 ( u = ( u + u + . The translate 8 that maps D 0 onto R is thus (see Exercise 26) 8 1 ( u = ( 1 + u + u + We use the change of variables x 1 + u + v , y u + v to compute the integral ZZ R ( x + y ) 2 e x 2 y 2 dx dy .The function expressed in the new variables u and v is f ( u = (( 1 + u + + ( u + v)) 2 e ( 1 + u + 2 ( u + 2 = ( 2 v 1 ) 2 e ( 2 v 1 )( 2 u 1 ) We compute the Jacobian of 8 : Jac (8) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x u x ∂v y u y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 11 ¯ ¯ ¯ ¯ = 1 + 1 = 2 Using the Change of Variables Formula gives I = R ( x + y ) 2 e x 2 y 2 = D 0 ( 2 v 1 ) 2 e ( 2 v 1 )( 2 u 1 ) · 2 dud v = Z 1 0 Z 1 0 2 ( 2 v 1 ) 2 e ( 2 v 1 )( 2 u 1 ) v = Z 1 0 2 ( 2 v 1 ) 2 ± Z 1 1 e ( 2 v 1 ) t · 1 2 dt ! d v = Z 1 0 ( 2 v 1 ) 2 2 v 1 e ( 2 v 1 ) t ¯ ¯ ¯ ¯ 1 t =− 1 d v = Z 1 0 ( 2 v 1 ) e 2 v 1 d v Z 1 0 ( 2 v

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16.5 Ex 39 - S E C T I O N 16.5 Change of Variables(ET...

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