 Solution

The first boundary is $u=1$. On $y=0$ in the first quadrant, the inverse parabolic polar coordinates give $v=0$. To identify the remaining curve, put $t=y^{1/3}\ge 0$. Its Cartesian coordinates satisfyHence $u=t$ and $v=t^2$, so this boundary is $v=u^2$. The condition $x\ge 0$ on this curve requires $0\le t\le 1$. Its endpoints are $(0,0)$ and $(0,1)$, while $u=1$ meets $v=0$ at $(1/2,0)$. These three arcs enclose exactlyEvery point in this region has $v\le u$, and therefore $x=(u^2-v^2)/2\ge 0$, as required. Using the Jacobian determinant, the area is