lest say we have the function $x^{2/3}+y^{2/3} = a^{2/3}$ and we want to Find the volume of the solid of revolution generated when the area bounded by the given curves is revolved about the x-axis hence:
$$y = (a^{2/3}-x^{2/3})^{3/2}$$
lets do the method of disk a tiny disk has volume:
$$dV= \pi (a^{2/3}-x^{2/3})^{3}dx$$
the volume of the total solid is then:
$$V = \int_0^a\pi (a^{2/3}-x^{2/3})^{3}dx$$
$$=\int_0^a \pi(a^2-3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}-x^2)dx$$
$$=a^2x-3a^{4/3}\frac{x^{5/3}}{5/3}+3\frac{a^{2/3}x^{7/3}}{7/3} - \frac{x^3}{3}$$
$$a^3-3\frac{a^{9/3}}{5/3}+3\frac{a^{9/3}}{7/3} - \frac{a^3}{3}$$
my problem is that I am not getting the correct form: $$16\frac{\pi a^3}{105}$$
note: the volume is only in the first quadrant.
You've differentiated when you needed to integrate:
$$ \int a^2 - 3a^{4/3} x^{2/3} + 3a^{2/3} x^{4/3} - x^2 \, dx = a^2 x - 3a^{4/3} \frac{x^{\color{red}{5/3}}}{\color{red}{5/3}} + 3a^{2/3} \frac{x^{\color{red}{7/3}}}{\color{red}{7/3}} - \frac{x^3}{3} + C.$$