Calculate the integral over a triangle of vertices $ (0,0), (1,0) $ and $ (0,1) $

Question

$\int \int_B$ $\frac{\sqrt[3]{y-x}}{1+y+x}dx.dy$ where $ B $ is a triangle of vertices $ (0,0), (1,0) $ and $ (0,1) $

I tried to replace it, so it was $\sqrt [3] {\frac {u} {v}}$

But I couldn't find the limits of integration

I also used the "jacobi correction" there was this replacement there times $\frac {1} {2}$

Answer 1

As @ajotatxe has pointed out, symmetry considerations show that the integral over $B$ must be zero.

However, to demonstrate the use of transformations, let $u=y-x,v=y+x$, then $\frac{\partial (u,v)}{\partial (x,y)}=\begin{vmatrix}-1&1 \\1&1\\\end{vmatrix}=-2.$

On $B, 0\le v\le 1$ and $-v\le u\le v.$ The integral then becomes

$$\int_0^1 \int_{-v}^v -\frac{1}{2}\frac{\sqrt[3]u}{1+v}dudv.$$

Now $\int_{-v}^v \sqrt[3]udu=[\frac{3}{4}u^\frac{4}{3}]_{-v}^v=0$ and so the entire integral is $0$.

Answer 2

Hint:

Note that $(x,y)\in B\iff (y,x)\in B$ and $$f(x,y)=-f(y,x)$$

$f$ is the integrand, of course.