Using change of variables, solve the integral and show the domain obtained by the change. I need to solve the following integral using change of variables:
$$\int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA$$
where D is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$.
I tried to change the variables to $u=y-x$ and $v=1+x+y$, but then I couldn't solve the integral.
Any tips on how to solve it (or the full solution) would be highly appreciated!
Thanks in advance!
EDIT:
I should've been more detailed in what I did. I tried the change $u=y-x$ and $v=1+x+y$, found the new domain and calculated the Jacobian. The integral then is:
$$\int_1^2\int_{1-v}^{v-1}\frac{\sqrt[3]{u}}{v}\frac{1}{2}dudv$$
That is what I can't solve.
I can solve it in u, but I don't even know if it's right:
$$\frac{3}{8}\int_1^2\frac{{(v-1)}^{4/3}-{(1-v)}^{4/3}}{v}dv$$
 A: Your triangle is bounded above by the line $y=1-x,$ meaning $y\le 1-x,$ which we can rewrite as $x+y\le1.$ This suggests using $v=x+y$ as your other variable, rather than your choice of $v$.
Since we put $$u=y-x\\v=x+y,$$ then we have $$y=\frac{u+v}2\\x=\frac{v-u}2.$$ It can be shown, then, that the Jacobian is $-\frac12,$ so that $$\iint_D\frac{\sqrt[3]{y-x}}{1+x+y}\,dA=-\frac12\iint_D\frac{\sqrt[3]{u}}{1+v}\,du\,dv.$$ It remains only to find the new limits of integration that describe $D$.
Since $x$ is bounded below by $0,$ then $$0\le\frac{v-u}2\\0\le v-u\\u\le v$$ Since $y$ is bounded below by $0,$ then $$0\le\frac{u+v}2\\0\le u+v\\-v\le u.$$ Hence, we have $-v\le u\le v$ and $v\le 1$ (by previous discussion). Also, we have $0\le v$ (why?), and so the integral becomes $$-\frac12\int_0^1\int_{-v}^v\frac{\sqrt[3]{u}}{1+v}\,du\,dv=-\frac12\int_0^1\frac1{1+v}\left(\int_{-v}^v\sqrt[3]{u}\,du\right)\,dv.$$ 
Can you take it from there?
A: with the change of variables you need to
1) Transform the Domain $D$ by using the $(x,y)\rightarrow (u,v)$ i.e. $(0,0)\rightarrow (0-0,1+0+0)=(0,1)$. Do this for all points.
2) Compute the Jacobian 
$$
 \mathrm{dA}=\left| \begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}   \end{array} \right|\partial u\partial v.
$$
Then combine them all together and solve with the appropriate limits based on the new transformed boundary of the integral.
Hope it helps.
Rob
A: Hint:
$$
\int\int_D\frac{\sqrt[3]{y-x}}{1+x+y}dA=\int_{1}^{2}\int_{A}^{B} \frac{\sqrt[3]{u}}{v}\,\,J\,\,du\, dv
$$
where $A=A(v)$ and $B=B(v)$ and $J$ is the jacobian.
