Answer
The region is bounded by $y=1+x^2$ and the line $x=1$.
Work Step by Step
Let us consider $y=u(1+v^2) \implies u=\dfrac{y}{1+x^2}$ and $v=x$
Consider the given inequalities $0 \leq u \leq 1$ and $0 \leq v \leq 1$
This can be re-written as:
$0 \leq \dfrac{y}{1+x^2} \leq 1$ and $0 \leq x \leq 1$
This gives: $0 \leq y \leq 1+x^2$ and $0 \leq x \leq 1$
Hence, the region is bounded by $y=1+x^2$ and the line $x=1$.
