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Change of Variablesby shards5
Tags: variables 
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#1
Mar2511, 05:25 PM

P: 38

1. The problem statement, all variables and given/known data
Suppose D is the parallelogram in the xyplane with vertices P(1,5), Q(1,5), R(5,1), S(3,9) [tex] \int\int ^{}_{D} (6x+12y) dA [/tex] HINT: Use transformation x = [tex]\frac{1}{6}[/tex](u+v) and y = [tex]\frac{1}{6}[/tex] (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. PQ [tex]\frac{5(5)}{1(1)}[/tex] = 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 SR [tex]\frac{9(1)}{35}[/tex] = 5 y = 5x + 9 5/6u+1/6v = 5/6u+5/6v + 9 > v = 9 RQ [tex]\frac{(1)(5)}{51}[/tex] = 1 y = x1 > 5/6u + 1/6v = 1/6u +1/6v 1 > u = 1 SP [tex]\frac{(9(5)}{3(1)}[/tex] = 1 y = x9 > 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. [tex] \int^{9}_{5}\int^{9}_{1} (u+v + (10u+2v)) * 1/6 dudv [/tex] which simplifies to . . . [tex] \int^{9}_{5}\int^{9}_{1} (4/3u+1/2v) dudv [/tex] After the first integration I get . . . [tex] \int^{9}_{5} (2/3u^2+1/2v*u) dudv [/tex] Plugging in 9 and 1 I get 549/2v+50/35/2v = 37.33333  7v Integrating what I got I get 37.33333v  7/2v^{2} And after plugging in 9 and 5 I get 345.333 which is wrong. So what am I doing wrong? 


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