Tuesday, April 8 ∗ Solutions ∗ Changing coordinates1. Consider the region R in R2shown below at right. In this problem, you will do a change ofcoordinates to evaluate:ÏRx −2y dASuvTRxy(1,1)(2,1)(3,4)(1,3)(a) Find a linear transformation T : R2→R2which takes the unit square S to R.Write you answer both as a matrix¡a bc d¢and as T (u,v) =(au+bv, cu+d v), and check youranswer with the instructor.SOLUTION:T (u, v) =(2u +v,u +3v). In matrix form,Ã2 11 3!(b) ComputeÎRx −2y dA by relating it to an integral over S and evaluating that. Check youranswer with the instructor.SOLUTION:The Jacobian of T isdetÃ2 11 3!=6 −1 =5SoÏRx −2y dA =ÏS[(2u +v) −2(u +3v)]5dA=Z10Z10−25v du d v =·−25v22¸10=−25/22. Another simple type of transformation T : R2→ R2is a translation, which has the general formT (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 isdetÃ1 00 1!=1.So T does not distort areas.(b) Consider the region S =©u2+ v2≤ 1ªin R2with coordinates (u, v), and the region R =©(x −2)2+(y −1)2≤1ªin R2with 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 reduceÏRx dAto 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 haveÏRx dA =ÏS(u +2)dAConverting the second integral above to polar we haveÏS(u +2)dA =Z2π0Z10(r cosθ +2)r dr dθ =Z2π0·r3cosθ3¸10dθ +2π£r2¤10=1/3Z2π0cosθ dθ +2π = 1/3[sinθ]2π0+2π = 2π3. Consider the region R shown below. Here the curved left side is given by x = y − y2. In thisproblem, you will find a transformation T : R2→R2which takes the unit square S = [0, 1]×[0, 1]to R.(2,1)Rxy(2,0)(0,1)(a) As a warm up, find a transformation that takes S to the rectangle [0,2] ×[0,1] which con-tains 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 −v2, v)(c) Now extend your answer in (b) to the needed transformation T . Hint: Try “filling in” be-tween 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 JacobianJ(T ) = detÃ2 −v(1 −v) (1 −2v)(1 −u)0 1!=2 −v(1 −v)So we have the area of R given byÏRd x d y =Z10Z102 −v(1 −v)du dv =11/6Computing directly we have the area of R given byZ102 −(y − y2)d y = 11/64. If you get this far, evaluate the integrals in Problems 1 and 2 directly, without doing a change ofcoordinates. 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 intothree parts. The result isÏRx −2y dA =Z10Z3xx/2x −2y d y d x +Z21Zx/2+5/2x/2x −2y d y d x +Z32Zx/2+5/23x−5x −2y d y d xEvaluating this is not difficult but it is tedious. We leave it to the interested student. You shouldget −25/2.For the integral in problem two, again use the order d y d x. We just need one double integral.ÏRx dA =Z31Z1+p1−(x−2)21−p1−(x−2)2x d y d x=Z312xp1 −(x −2)2d xThis integral can be evaluated by making the substitution x −2 = sin u, yielding the integralZπ/2−π/2(2sinu +4)cos2u duNow split this in two pieces asZπ/2−π/22sinu cos2u du +Zπ/2−π/24cos2u duThe first is the integral of an odd function over an interval which is symmetric about the y axisso it is 0. The second can be evaluated by using the trig identity cos2u = (1 +cos2u)/2. ThisgivesZπ/2−π/24cos2u du =Zπ/2−π/24(1 +cos 2u)/2du =
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