18 02A Topic 33 Change of variable Author Jeremy Orloff 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 I 12 u du 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 We will show two slightly different ways of finding the u v limits 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 u v ux uy 2 1 4 vx vy 2 1 x y yO x y 1 2 dx dy 41 du dv u v 4 2x y 2 C Limits method 1 stripe the region R by level 1 R curves for u or v 2x y 0 The triangle is bounded by x 0 u v x 2x y 0 u 0 1 2 2x y 2 v 2 inner u v 2 outer 2 u 0 Z 0 Z 2 dv du I u4 v 4 Easy to compute 4 u 2 v u B Method 2 for finding u v limits is on the next page continued 1 yO u 22 v 2 v u u 0 1 2 x 18 02A topic 33 2 C Limits method 2 convert R to a region in the uv plane yO vO v 2 2 2x y 2 1 R v u R u 0 2x y 0 1 2 x u The boundary lines of R in the xy plane map to the uv plane as follow The line x 0 u y and v y v u The line 2x y 0 u 0 The line 2x y 2 v 2 The result is the region R in the uv plane which has limits u from 2 to 0 and v from u to 2 Same as with method 1 There are more change of variables examples on the last page 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 u v A abs u abs u v u v u v yu yv yu u yv v x y du dv u v 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 dA v v0 v S A O u u0 R o Q v v0 Q y u u0 u P continued 9 P x Detail from picture at left 18 02A topic 33 3 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 Examples taken from supplementary notes CV 2 Z Z x y dx dy where R is the region shown Example 1 Compute x y 2 R Since the region is bounded by the lines x y 1 and x y 1 we make a change of variable u x y v x y u v x y Computing the Jacobian 2 1 2 x y u v 1 Thus dx dy du dv 2 yO vO yO v 1 1 v 1 R 1 1 R x u 1 u 1 u 1 R u 1 u x v 1 1 v 1 Using either method 1 or method 2 we see the boundaries are given by u 1 Z Z Z 1Z 1 2 2 x y v 1 v 1 the integral is dx dy du dv x y 2 u 2 2 R 1 1 Inner integral v2 2 u 2 u 1 u 1 v2 3 Outer integral v3 9 1 1 2 9 See the other examples in supplementary notes section CV Especially examples 3 4 5
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