Find the volume generated by rotating the astroid x^(2/3)

Find the volume generated by rotating the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ about the $x$-axis.

We use the disk method, converting the fractional-power integrand into a polynomial via $u=x^{1/3}$.

Step 1: Solve for $y$

\[y = \left(a^{2/3}-x^{2/3}\right)^{3/2}, \qquad 0\le x\le a\]

Step 2: Disk method

\[A(x) = \pi y^2 = \pi\left(a^{2/3}-x^{2/3}\right)^3\]

Step 3: Set up the integral (using symmetry to double the $[0,a]$ half)

\[V = 2\pi\int_0^a\left(a^{2/3}-x^{2/3}\right)^3\,dx\]

Step 4: Substitute $u=x^{1/3}$

Then $x=u^3$, $dx=3u^2\,du$, and $x^{2/3}=u^2$. Writing $b=a^{1/3}$:

\[V = 2\pi\int_0^{b}(b^2-u^2)^3\cdot3u^2\,du = 6\pi\int_0^b u^2(b^2-u^2)^3\,du\]

Step 5: Expand and integrate

\[(b^2-u^2)^3 = b^6-3b^4u^2+3b^2u^4-u^6\]

\[V = 6\pi\int_0^b\left(b^6u^2-3b^4u^4+3b^2u^6-u^8\right)du = 6\pi\left[\frac{b^6u^3}{3}-\frac{3b^4u^5}{5}+\frac{3b^2u^7}{7}-\frac{u^9}{9}\right]_0^b\]

\[= 6\pi b^9\left[\frac13-\frac35+\frac37-\frac19\right]\]

Step 6: Combine the fractions (common denominator 315)

\[\frac13-\frac35+\frac37-\frac19 = \frac{105-189+135-35}{315} = \frac{16}{315}\]

Step 7: Final volume

\[V = 6\pi b^9\cdot\frac{16}{315} = \frac{32\pi b^9}{105}\]

Since $b=a^{1/3}$, $b^9=a^3$:

\[\boxed{V = \frac{32\pi a^3}{105} \text{ cubic units}}\]

Summary

\[\boxed{V = \frac{32\pi a^3}{105} \text{ cubic units}}\]

Key insight: Compared with a sphere of the same radius $a$ (volume $\frac{4\pi a^3}{3}=\frac{140\pi a^3}{105}$), the astroid's revolution sweeps out only $\frac{32}{105}\approx30.5\%$ as much volume — a direct consequence of the astroid's concave, cusped profile pulling the solid's boundary inward compared to a sphere.