# Sionepaeauspacfj ma211 week 9 tutorial solution 12017

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[email protected] MA211 Week 9 Tutorial Solution 1/2017 2 2 1 1 2 2 2 2 1 1 2 2 2 2 1 1 ( , ) 1 2 2 2 2 1 1 ( , ) 4 2 2 2 2 x y u u u v u v x y v v v u v u x y u v u v u v v u v u v u   9-12) Find the Jacobian ( , , ) / ( , , ) x y z u v w 9) 3 , 2 , x u v y u w z v w The Jacobian is 3 1 0 1 0 1 0 2 1 3 1 0 0 2 1 2 ( , , ) 1 0 2 3 1 0 3(0 2) 1(1 0) 0 6 1 5 1 1 0 1 ( , , ) 0 1 1 x y z u u u x y z v v v x y z w w w x y z u v w   11) , , u xy v y w x z Solve for x, y and z u xy v y w x z u x y v z w x y u u w v v Then the Jacobian is 2 2 1 1 0 1 0 0 1 x y z u v u u v x u y z u v v v v v x y z w w w    
2 2 2 2 1 0 1 0 0 0 ( , , ) 1 1 1 0 1 0 0 0 0 1 ( , , ) 1 1 1 1 u v v x y z u u u v w v v v v u v v v v   21) Use the transformation 2 , 2 to find u x y v x y 2 2 R x y dA x y  where R is the rectangular region enclosed by the lines 2 1, 2 4, 2 1, 2 3 x y x y x y x y
4 Contact: [email protected] MA211 Week 9 Tutorial Solution 1/2017
5 Contact: [email protected] MA211 Week 9 Tutorial Solution 1/2017 The integral is then 4 2 3 4 3 3 3 1 1 1 1 1 1 2 1 1 1 3 1 3 3 ln ln3 2 5 5 5 2 2 2 2 R S x y u u u dA dA dudv dv dv v x y v v v v 