Choose the outer radius after rewriting in terms of y

The curves meet where $x^2=x^3$, namely $x=0,1$, so the enclosed region has $0\le y\le1$ in the first quadrant. There $y=x^2$ becomes $x=\sqrt y$ and $y=x^3$ becomes $x=\sqrt[3]y$. The explicit intersection derivation here is independently authored; the Teacher marks the meeting point $(1,1)$ in the sketch.

For $0\le y\le1$, $\sqrt[3]y\ge\sqrt y$. Thus outer radius is $f(y)=\sqrt[3]y$ and inner radius is $g(y)=\sqrt y$, matching the Teacher annotations. Set $V=\pi\int_0^1[(\sqrt[3]y)^2-(\sqrt y)^2]\,dy$.

The source answer is $0.314$ cubic units to three significant figures. Independently, $V=\pi\int_0^1(y^{2/3}-y)\,dy=\pi[3y^{5/3}/5-y^2/2]_0^1=\pi/10\approx0.314159265$.