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## Homework Statement

Use the given transformation to evaluate the given integral.

[tex]\int[/tex][tex]\int(x-3y)DA[/tex]

R.

where R is the triangular region with vertices (0,0), (2,1) and (1,2) ; x = 2u + v , y = u + 2v

Trial :

Using the points given I came up with these equations for the triangle lines :

y = 2x;

y = 1/2x;

y = 3 - x;

Now just substituting x = 2u and y = u+2v I get this :

u = 0

v = 0

v = 3(1-u)

plotting this , I see that 0<= u <= 1 and 0<= v <= 3(1-u)

And from the original integral the integrand is x - 3y. Again substituting

for x and y with the given transformation coordinate for u and v I get :

(x-3y)DA = -(u+5v) dv*du

Now the integral becomes :

[tex]\int^{1}_{0} du[/tex][tex]\int^{3-3u}_{0} -u - 5v dvdu[/tex]

In which I get the answer -4/5 which is not correct. Any help?