18 02A Topic 33 Change of variable Read SN CV R 18 01 example Compute I x 1 x2 dx Direct substitution u 1 x2 du 2x dx x dx 21 du R u du I 12 R Inverse substitution x sin u dx cos u du I sin u cos2 u du RR Double integration f x y dA R Direct substitution u u x y v v x y Inverse substitution x x u v y y u v Example x r cos y r sin Finding dA dx dy Jacobian x y u v dA dx dy x y u v xu xv yu yv du dv u v x y 1 u v x y ux uy vx vy du dv Reason soon RR Change of variable for f x y dx dy R A Change f x y to u v coordinates B Express dA in terms of du dv C Find u v limits on R RR Example Given the region R shown compute I 4x2 y 2 4 dy dx R Change of variable u 2x y v 2x y A 4x2 y 2 uv f x y u4 v 4 2 1 ux uy u v 2 B x y 4 u 22 2 1 vx vy 2x y 2 v 2 x y 1 1 u v 4 dx dy 4 du dv 1 R C Limits 2x y 0 v u u 0 x 0 u y v y v u u 2 to 0 fix u v u to 2 1 Z 0 Z 2 1 2 2 4 4 dv du I uv 4 u 2 v u Easy to compute continued 1 18 02A topic 33 2 Reason for change of variables formula Draw grid lines u c1 v c2 in xy plane A area P QRS parallelogram In a moment we will see that PQ hxu u yu ui and PS hxv v yv vi Using this we have x u xv v x x x y A u u v u v u v u v yu u yv v yu yv dA x y u v du dv Now we show PQ hxu u yu ui Let PQ h x yi see picture The approximation formula says x xu u xv v and y yu u yv v Going from P to Q we have v 0 x xu u and y yu u PQ hxu u yu ui Likewise for PS hxv v yv vi v v0 v S A u u0 Q R o O P v v0 u u0 u Q P 9 tt tt t t tt y tt t t t t x Detail from picture at left Remark on the chain rule x y u v xu xv ux uy 1 or I yu yv vx vy u v x y x x u x v 1 etc x u x v x
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