# changing limit of this double integral $\int \int _R^{ }f\left(x,y\right)dA$ with region R bounded by { $y=3x$, $x=3y$, $x+y=4$}

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 R 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$$

• once you substitute, there will not be a need to divide the integral into two parts. – Math Lover Oct 29 at 12:51
• @MathLover thank you!, next time when i see question like these, I'll directly substitute into the given bound – RiRi Oct 29 at 13:29

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.
• 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 $x+y=4$ – RiRi Oct 29 at 12:46
• 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 – Kavi Rama Murthy Oct 29 at 12:52