Tuesday November 6 Solutions Changing coordinates 1 Consider the region R in R2 shown below at right In this problem you will do a change of coordinates to evaluate x 2y dA R y v 3 4 T 1 3 R 1 1 2 1 S x u a Find a linear transformation T R2 R2 which takes the unit square S to R Write you answer both as a matrix ac db and as T u v au bv cu d v and check your answer with the instructor SOLUTION T u v 2u v u 3v In matrix form 2 1 1 3 b Compute R x 2y dA by relating it to an integral over S and evaluating that Check your answer with the instructor SOLUTION The Jacobian of T is d et So 2 1 1 3 6 1 5 R x 2y dA Z 1Z 0 1 0 S 2u v 2 u 3v 5 dA 25v 2 25v d u d v 2 1 25 2 0 2 Another simple type of transformation T R2 R2 is a translation which has the general form T u v u a v b for a fixed a and b a If T is a translation what is its Jacobian matrix How does it distort area SOLUTION If T u v u a v b where a and b are constants then the Jacobian is d et 1 0 0 1 1 So T does not distort areas b Consider the region S u 2 v 2 1 in R2 with coordinates u v and the region R x 2 2 y 1 2 1 in R2 with coordinates x y Make separate sketches of S and R SOLUTION c Find a translation T where T S R SOLUTION T u v u 2 v 1 d Use T to reduce x dA R to an integral over S and then evaluate that new integral using polar coordinates SOLUTION The Jacobian of T is just 1 as noted in part a So we have R x dA S u 2 dA Converting the second integral above to polar we have 2 Z 1 Z S u 2 dA 0 0 r cos 2 r d r d Z 0 2 r 3 cos 1 3 0 1 d 2 r 2 0 2 Z 1 3 0 cos d 2 1 3 sin 2 0 2 2 3 Consider the region R shown below Here the curved left side is given by x y y 2 In this problem you will find a transformation T R2 R2 which takes the unit square S 0 1 0 1 to R y 2 1 0 1 R x 2 0 a As a warm up find a transformation that takes S to the rectangle 0 2 0 1 which contains R SOLUTION L u v 2u v b Returning to the problem of finding T taking S to R come up with formulas for T u 0 T u 1 T 0 v and T 1 v Hint For three of these use your answer in part a SOLUTION T u 0 2u 0 T u 1 2u 1 T 1 v 2 v T 0 v v v 2 v c Now extend your answer in b to the needed transformation T Hint Try filling in between T 0 v and T 1 v with a straight line SOLUTION T u v 2u v 1 v 1 u v d Compute the area of R in two ways once using T to change coordinates and once directly SOLUTION To change coordinates we compute the Jacobian J T d et 2 v 1 v 1 2v 1 u 0 1 2 v 1 v So we have the area of R given by Z 1Z R dx dy 0 1 0 2 v 1 v d u d v 11 6 Computing directly we have the area of R given by 1 Z 0 2 y y 2 d y 11 6 4 If you get this far evaluate the integrals in Problems 1 and 2 directly without doing a change of coordinates It s a fun filled task SOLUTION For the integral in problem one use the order d y d x We need to split the double integral into three parts The result is Z 1Z R x 2y dA 0 3x x 2 Z 2Z x 2y d y d x x 2 5 2 x 2y d y d x x 2 1 Z 3Z 2 x 2 5 2 3x 5 x 2y d y d x Evaluating this is not difficult but it is tedious We leave it to the interested student You should get 25 2 For the integral in problem two again use the order d y d x We just need one double integral Z 3Z R x dA Z 1 x 2 2 p 1 1 3 2x 1 p 1 1 x 2 2 x dy dx p 1 x 2 2 d x This integral can be evaluated by making the substitution x 2 sin u yielding the integral Z 2 2 2 sin u 4 cos2 u d u Now split this in two pieces as Z 2 2 2 2 sin u cos u d u Z 2 4 cos2 u d u 2 The first is the integral of an odd function over an interval which is symmetric about the y axis so it is 0 The second can be evaluated by using the trig identity cos2 u 1 cos 2u 2 This gives Z 2 Z 2 2 4 cos u d u 4 1 cos 2u 2 d u 2 2 2
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