How to find out the limits of integral after change of variables? I am given with the integral $$\int_0^1 \int_0^{1-x}(xy(1-x-y))^{1/3}\,dy\,dx$$
I have to find out the value of above integral by making the transformation $$x+y=u, y = uv.$$
After calculating the jacobian which is $$J\bigg(\frac{(x,y)}{(u,v)}\bigg) = u$$ 
we get the integral $$\int \int (u^2v(1-v)(1-u))^{1/3}u\,du\,dv$$ 
1) I am not getting how to find out the limits?
2) What is the value of integral?
 A: You have $\displaystyle\int\left( \int \cdots\cdots \, du \right) dv,$ i.e. given a fixed value of $v$, the other variable $u$ goes from something to something. You can also write this as $\displaystyle \int\left( \int \cdot\cdots\,dv\right) du,$ so that for a given value of $v$, the other variable goes from something to something.
Since the sum $x+y$ goes from $0$ up to $1$ (i.e. for any fixed value of $x$ between $0$ and $1$, the other variable $y$ increases from $0$ until the sum of $x$ and $y$ reaches $1$) you have $u$ going from $0$ up to $1,$ thus $\displaystyle\int_0^1 \left( \int \cdots\cdots\,dv \right) du.$
Now with $u$ fixed at some point between $0$ and $1,$ what does $v$ do? On the line $x+y=u,$ with $u$ fixed, the pair $(x,y)$ moves from $(u,0)$ to $(0,u),$ staying in the first quadrant, so $y$ goes from $0$ up to $u,$ so $v$ goes from $0$ to $1.$ Thus you have $\displaystyle\int_0^1 \left( \int_0^1 \cdots\cdots\,dv\right) du.$
Then
\begin{align}
& \int_0^1 \left(\int_0^1 (u^2 v(1-v)(1-u))^{1/3}\,dv\right) du \\[10pt]
= {} & \int_0^1 \left( (u^2(1-u))^{1/3} \int_0^1 (v(1-v))^{1/3} \, dv \right) du \\
& \quad \text{because $(u^2(1-u))^{1/3}$ does not change as $v$ goes from $0$ to $1$} \\[12pt]
= {} & \int_0^1 (u^2(1-u))^{1/3} \cdot A \, du \\
& \quad \text{where $A$ is something that does not change as $u$ goes from $0$ to $1$} \\[12pt]
= {} & \int_0^1 (u^2(1-u))^{1/3} \,du \cdot A \\[10pt]
= {} & \int_0^1 (u^2(1-u))^{1/3} \, du \cdot \int_0^1 (v(1-v))^{1/3} \,dv
\end{align}
and the two integrals can be treated separately.
