Good night, i have a serious problem changing the integration limits, i read two books but i don't understand, i put an example... $$\int_0^1\int_0^{1-x}(y-2x)^{2}\sqrt{x+y}\,dy\,dx$$ I use the change: $\begin{cases} u=x+y\\ v=y-2x \end{cases}$
Then: $\begin{cases} x=\frac{u}{3}-\frac{v}{3}\\ y=\frac{2u}{3}-\frac{v}{3} \end{cases}$
And evaluating: $\begin{cases} y+x=1\rightarrow u=1\\ y=0\rightarrow2u=-v\\ x=0\rightarrow u=v\\ x=1\rightarrow v=-3+u \end{cases}$
And the new integral: $$\int_{-3+u}^{-2u}\int_v^1\frac13v^2u^{1/2}\,du\,dv$$
But, this not the answer, what am I doing wrong?
Your expression for $y$ in terms of $v$ are not correct. The expressions should be
$$x=\frac u2-\frac v3, \qquad y=\frac{2u}3+\frac v3$$
The original, source integral has the outer bounds $$0\le x\le 1$$ and the inner bounds $$0\le y\le 1-x$$
Substituting the correct expressions for $x$ and $y$ give $$0\le\frac u3-\frac v3\le 1, \qquad 0\le\frac{2u}3+\frac v3\le 1-\left(\frac u3-\frac v3\right)$$
Multiplying by three gives $$0\le u-v\le 3, \qquad 0\le 2u+v \le 3-u+v$$
You should graph those inequalities in the $u$-$v$ plane to see what kind of region you get.
You see that is a triangle. Now express overall limits on $u$, then limits on $v$ based on the value of $u$.