The two curves [tex]y=x[/tex] and [tex]y=x^{1/15}[/tex] intersect at [tex]x=0[/tex] and [tex]x=1[/tex], with [tex]x^{1/15}\ge x[/tex] over [tex]0\le x\le1[/tex].
[tex]\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}[/tex]
We have [tex]y=x^{1/15}\implies x=y^{15}[/tex]. The curves [tex]x=y[/tex] and [tex]x=y^{15}[/tex] intersect at [tex]y=0[/tex] and [tex]y=1[/tex], with [tex]y\ge y^{15}[/tex] over [tex]0\le y\le1[/tex].
[tex]\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}[/tex]