= Solution
{c}
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 $t=y^{1/3}\geq0$. Its <Cartesian coordinates> satisfy
$$
x=\frac{t^2-t^4}{2},\qquad y=t^3,\qquad
\sqrt{x^2+y^2}=\frac{t^2+t^4}{2}.
$$
Hence $u=t$ and $v=t^2$, so this boundary is $v=u^2$. The condition $x\geq0$ on this <curve> requires $0\leq t\leq1$. Its endpoints are $(0,0)$ and $(0,1)$, while $u=1$ meets $v=0$ at $(1/2,0)$. These three arcs enclose exactly
$$
0\leq u\leq1,\qquad0\leq v\leq u^2.
$$
Every point in this region has $v\leq u$, and therefore $x=(u^2-v^2)/2\geq0$, as required. Using the <Jacobian determinant>, the <area> is
$$
\begin{aligned}
A&=\int_0^1\int_0^{u^2}(u^2+v^2)\,dv\,du\\
&=\int_0^1\left(u^4+\frac{u^6}{3}\right)\,du
=\frac15+\frac1{21}
=\boxed{\frac{26}{105}}.
\end{aligned}
$$
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