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we are asked to use the substitution, $x=3u+v$ and $y=u+3v$. OK so this is what i did....first i plot the region Renter image description here

I made a mistake, @Kavi Rama Murthy you are right, the area needs to be divided into 2 region, Region 1 can be $\ \int _0^1\ \int _{\frac{y}{3}}^{3y}\ f\left(x,y\right)dxdy\ $ and Region 2 being $\ \int _1^3\int _{\frac{y}{3}}^{4-y}\ f\left(x,y\right)dxdy$

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    $\begingroup$ once you substitute, there will not be a need to divide the integral into two parts. $\endgroup$ – Math Lover Oct 29 at 12:51
  • $\begingroup$ @MathLover thank you!, next time when i see question like these, I'll directly substitute into the given bound $\endgroup$ – RiRi Oct 29 at 13:29
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The equations for the lines in terms of $u$ and $v$ are $u=0,v=0$ and $u+v=1$. So the limits are $0 < v < 1-u$ and $0<u<1$.

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  • $\begingroup$ can u please elaborate how you got $u=0, v= 0$ and I assumed you got $u+v=1$ because you replaced the expression $x= 3u+v$ and $y= y+3v$ into $x+y=4$ $\endgroup$ – RiRi Oct 29 at 12:46
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    $\begingroup$ You can solve for $u$ and $v$ in terms of $x$ and $y$. You get $u=\frac {3x-y} 8$ and $v=\frac {3y-x} 8$. Now you see that the lines $x=3y$ and $y=3x$ are nothing but $v=0$ and $u=0$ respectively. The line $x+y=4$ becomes $u+v=1$. @RiRi $\endgroup$ – Kavi Rama Murthy Oct 29 at 12:52
  • $\begingroup$ thank you sir!! $\endgroup$ – RiRi Oct 29 at 12:56

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