# Change of Variables

by shards5
Tags: variables
 P: 38 1. The problem statement, all variables and given/known data Suppose D is the parallelogram in the xy-plane with vertices P(-1,5), Q(1,-5), R(5,-1), S(3,9) $$\int\int ^{}_{D} (6x+12y) dA$$ HINT: Use transformation x = $$\frac{1}{6}$$(u+v) and y = $$\frac{1}{6}$$ (-5u+v). 2. Relevant equations 3. The attempt at a solution Calculating the Jacobian I get dx/du = 1/6 dx/dv = 1/6 dy/du = -5/6 dy/dv = 1/6 which gives me 1/36 - (-5/36) = 1/6 From there I need the new intervals of integration. P-Q $$\frac{5-(-5)}{-1-(1)}$$ = 10/-2 = -5 y = 5x + 5 -> plugging in . . . -5/6u+1/6v = -5/6u - 5/6v + 5 -> u cancels out and then we get 6/6v = 5 which means v = 5 S-R $$\frac{9-(-1)}{3-5}$$ = -5 y = -5x + 9 -5/6u+1/6v = -5/6u+5/6v + 9 -> v = 9 R-Q $$\frac{(-1)-(-5)}{5-1}$$ = 1 y = x-1 -> -5/6u + 1/6v = 1/6u +1/6v -1 -> u = 1 S-P $$\frac{(9-(5)}{3-(-1)}$$ = 1 y = x-9 -> -5/6u +1/6v = 1/6u + 1/6v - 9 -> u = 9 Plugging in the x and y values and the Jacobian into the integral we get the following new integral. $$\int^{9}_{5}\int^{9}_{1} (u+v + (-10u+2v)) * 1/6 dudv$$ which simplifies to . . . $$\int^{9}_{5}\int^{9}_{1} (-4/3u+1/2v) dudv$$ After the first integration I get . . . $$\int^{9}_{5} (-2/3u^2+1/2v*u) dudv$$ Plugging in 9 and 1 I get -54-9/2v+50/3-5/2v = -37.33333 - 7v Integrating what I got I get -37.33333v - 7/2v2 And after plugging in 9 and 5 I get -345.333 which is wrong. So what am I doing wrong?

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