A bijection between two sets and a multiple integral

Question

Define subsets $\Delta$ and $D$ of $\Bbb R^{2}$ by \begin{align*} \Delta &= \{ (s,t) \mid s \geq 0,\ t \geq 0,\ s+t \leq 1 \}, \\ D &= \{ (x,y) \mid x \geq 0,\ y \geq 0,\ f(x,y) \geq 0 \}. \end{align*} Let $f(x,y)=x^{2}+y^{2}+1-2xy-2x-2y$.

1. Prove that the restriction $\phi |_{\Delta}$ of the map $$ \phi : \Bbb R^{2} \ni (s,t) \mapsto (x,y)=(s(1-t),(1-s)t) \in \Bbb R^{2} $$ to $\Delta$ is a bijection from $\Delta$ to $D$.

Firstly, it's easy to see that $f(s(1-t),t(1-s))=(s+t-1)^2\ge0$ thus $\phi(\Delta) \in D$. Secondly, to prove $\phi$ is injective assume we have $(s_1,t_1),(s_2,t_2)\in \Delta$ such that $\phi(s_1,t_1)=\phi(s_2,t_2)$ i.e. $s_1(1-t_1)=s_2(1-t_2)$,$(1-s_1)t_1=(1-s_2)t_2$. What I get from this is that $s_1-t_1=s_2-t_2$ and it isn't enough to show $s_1=s_2, t_1=t_2$. Finally, to prove $\phi$ is surjective assume we have $(a,b)\in D$ and we want to find $(s,t) \in \Delta$ such that $s(1-t)=a,(1-s)t=b$. Solving this system of equations gives us, $t=s-a+b$ and $s^2-s(a-b+1)+a=0$. The quadratic equation always has roots since $(a-b+1)^2-4a=f(a,b) \ge 0$. Additionally, $$s_{1,2}=\frac{(a-b+1)\pm\sqrt{f(a,b)}}{2}.$$ It seems that we need the condition $a-b+1 \ge 0$ to obtain at least one non-negative root $s_i$, but it is not always the case. Because $a-b+1$ can be negative despite the fact that $(a,b) \in D$. I again got stuck here.

2. Compute the multiple integral $\iint_{D} \sqrt{f(x,y)} dxdy$.

It seems that results obtaining from the first part will play some roles here. However, it isn't easy to figure out.

Any help would be much appreciated.

Answer 1

For the bijection, it is easy to show that is true on the boundary. To show the bijection for the interior, we note that the Jacobian is given by $1-s-t$, which is positive. Then, by the implicit function theorem, it is bijective.

Answer 2

Because this looks like homework I'm loath to divulge too many details, but here are a couple of observations (too long for a comment, but by no means a complete answer) that may clarify the geometry, and thereby help you organize the algebraic arguments:

1. In the variables $u = x - y$ and $v = x + y$, the region $D$ is easily seen to be bounded by the $x$-axis, the $y$-axis, and the parabola $u^{2} -2v + 1 = 0$, which opens along the $v$-axis (a.k.a. the line $x = y$) and passes through the points $(x, y) = (1, 0)$, $(0, 1)$, and $(\frac{1}{4}, \frac{1}{4})$.

   This region is (fairly well-known to be) the union of segments joining $(s, 0)$ and $(0, 1 - s)$ for $0 \leq s \leq 1$.

$(s, 0)$ and $(0, 1-s)$" />

2. For each $s$ in $[0, 1]$, the coordinate curve $t \mapsto \phi(s, t) = (s(1 - t), t(1 - s))$ traces a line segment. If $0 \leq t \leq 1$, this segment joins $(s, 0)$ and $(0, 1 - s)$. If instead $0 \leq t \leq 1 - s$ (i.e., if $(s, t) \in \Delta$), the segment joins $(s, 0)$ and $(s^{2}, (1 - s)^{2})$, which lies on (and is tangent to) the parabola in item 1.

   In other words, restricted to the unit square, the map $\phi$ covers $D$ twice; restricted to the triangle $\Delta$, $\phi$ is a bijection.