I want compute the integral of f(x,y) = 1 over $R = \{ (x,y) : 0\leq x \leq 1 \land 0\leq y \leq x^2 \} $ with the transformation $g(u,v) = (u-v+1,u+v)$ and check that is the same that compute without change of variable. i.e. $ \int_{0}^{1} \int_{0}^{x^2} dy dx = \frac{1}{3} $
I have problems for determinate the new region, I know that if I get the inverse function of $g(u,v)$, then I can map the vertices as follow
$$ g^{-1}(x,y) = \left( \frac{x}{2} + \frac{y}{2} - \frac{1}{2} , \frac{-x}{2} + \frac{y}{2} + \frac{1}{2} \right) $$
$$ g^{-1}(0,0) = \left(\frac{-1}{2}, \frac{1}{2} \right), \ g^{-1}(1,1) = \left(\frac{1}{2}, \frac{1}{2} \right), \ g^{-1}(1,0) = (0,0) $$
But, how I determine the ranges of $u$ and $v$? for the integral $\int\int J_g(u,v) du dv$
$0 \leq x \leq 1, 0 \leq y \leq x^2$
Transformation $x = u - v + 1, y = u + v$
$J = \begin{vmatrix} 1 & -1 \\ 1 & 1\end {vmatrix} = \, 2$
For the upper bound of $v$,
$y = x^2 \implies u + v = (u-v+1)^2$ (which is equation of another parabola)
$v^2 - (2u+3)v + (u^2+u+1) = 0$ which is a quadratic in $v$.
$(v - \frac{1}{2}\big(2u+3)\big)^2 = \big(\frac{2u+3}{2}\big)^2 - (u^2+u+1)$
$v = \frac{1}{2} (2u+3 \, \pm \sqrt{8u+5})$. As we have to find area below the parabola, we take the lower part of the curve.
So your integral becomes,
$\displaystyle \int_{-1/2}^0 \int_{-u}^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du + \int_0^{1/2} \int_u^{\frac{1}{2} (2u+3 - \sqrt{8u+5})} 2 \,dv \, du = \frac{1}{3}$