Use an appropriate change of variables to evaluate the double integral
$$
\iint_\Omega \frac{1}{4}\left (y^2-x^2\right )dA
$$
where $\Omega$ is the triangle with vertices $(0,0), \; (2,2),$ and $(1,3)$.
I know that the new $(u,v)$ region should be the triangle with vertices $(0,0)$,$(2,0)$ and $(2,1)$ so that $(0,0)$ maps to $(0,0), \; (2,2)$ maps to $(2,0)$ and $(1,3)$ maps to $(2,1)$, but I don't know how to find $x$ and $y$ in terms of $u$ and $v$ and hence find the corresponding transformation.
Best Answer
Let the variable changes (equivalent to 45-degree rotation),
$$x = \frac 1{\sqrt2} (u-v),\>\>\>\>\>y=\frac 1{\sqrt2}(u+v)$$
The integrand becomes,
$$\frac{1}{4}(y^2-x^2)=\frac12 uv$$
and the three vertexes are instead $(0,0)$, $(2\sqrt2,0)$ and $(2\sqrt2,\sqrt2)$. The corresponding area in the $uv$-plane is confined by $v=0$, $u=2\sqrt2$ and the line $v= \frac12 u$. The area integral becomes,
$$\frac12\int_0^{2\sqrt2}\int_0^{\frac12u}uv\>dvdu=1$$