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:
$$\frac{3}{8}\int_1^2\frac{{(v-1)}^{4/3}-{(1-v)}^{4/3}}{v}dv$$
Best Answer
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?