Transformation of a square Having fun with some integrals, I caught myself thinking about transforming of regions. So I have the following questions.
Suppose we have the square determined by inequalities $0<x<1, 0<y<1$ and a transformation rule $u=xy,v=x+y$.

The question is: what form will this square have in new coordinates $(u,v)$?
I tried to express $x$ and $y$ in terms of $u$ and $v$ and got the following: $$x=\frac{v-\sqrt{v^2-4u}}{2}$$ $$y=\frac{v+\sqrt{v^2-4u}}{2}$$
And I don't know what my next step should be.

Perhaps you can get some insight from mapping some specific sets

$(x, y_0)\mapsto (x y_0, x + y_0)$

where $0\leq y_0  \leq 1$ is a constant number. In the $xy$-plane this a horizontal line. In the $uv$-plane a couple of things can happen, if $y_0 =0$, we obtain a vertical line running through the origin. If $y_0\not = 0$ then $u=x y_0$ and $v = x + y_0$, equivalently 
$$
u = y_0(v-y_0)
$$
These are straight lines with slope $1/y_0$ and intercept $y_0$.
From this we then know we are bounded in the $uv$-plane from above by the line $v = y + 1$ which is the result of setting $y_0 = 1$. A simlar analysis can be done for points of the form $(x_0,y)$.

$(x,\alpha x)\mapsto (\alpha x^2, (\alpha + 1)x)$

In the $xy$-plane these are straight lines going through the origin with slope $\alpha > 0$. Since we want to be inside the unit square we need to put some constraints on the domain of $x$: $ x \leq 1 / (\alpha+ 1)$. In the $uv$-plane this is mapped to $u = \alpha x^2$ and $v = (\alpha + 1) x$, or equivalently 
$$u = \frac{\alpha}{(1 + \alpha)^2} v^2$$
These are parabolas opening along the $u$-axis. The maximum value of the coefficient happens with $\alpha=1$, in that case the parabola has the form $v=2u^{1/2}$.
From this we then learn we are bounded in the $uv$-plane from below by the parabola $v=2u^{1/2}$

