Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the cissoid $y^2 = \frac{x^3}{2a - x}$ about its asymptote.

We use the shell method, and a key algebraic cancellation reduces the integrand to a simple power of $x$.

Step 1: Find the asymptote

As $x\to2a$: $y^2=\frac{x^3}{2a-x}\to\infty$, so $x=2a$ is a vertical asymptote. The curve is defined for $0\le x < 2a$ and passes through the origin.

Step 2: Set up the shell method

A shell at position $x$ has radius (distance to the asymptote) $r=2a-x$ and height $2|y|=2\sqrt{\dfrac{x^3}{2a-x}}$ (accounting for both the $+y$ and $-y$ branches). Simplifying the product carefully:

\[(2a-x)\sqrt{\frac{x^3}{2a-x}} = (2a-x)\cdot\frac{x^{3/2}}{\sqrt{2a-x}} = x^{3/2}\cdot\frac{2a-x}{\sqrt{2a-x}} = x^{3/2}\sqrt{2a-x}\]

so the shell integrand is $dV = 4\pi x^{3/2}\sqrt{2a-x}\,dx$ — note the $\sqrt{2a-x}$ factor survives; it does not cancel away entirely.

Step 3: Integrate using a Beta-function substitution

\[V = 4\pi\int_0^{2a} x^{3/2}\sqrt{2a-x}\,dx\]

Let $x=2a\sin^2\theta$, so $2a-x=2a\cos^2\theta$ and $dx=4a\sin\theta\cos\theta\,d\theta$. When $x=0$: $\theta=0$; when $x=2a$: $\theta=\frac\pi2$.

\[x^{3/2}\sqrt{2a-x} = (2a)^{3/2}\sin^3\theta\cdot(2a)^{1/2}\cos\theta = (2a)^2\sin^3\theta\cos\theta\]

\[V = 4\pi(2a)^2\int_0^{\pi/2}\sin^3\theta\cos\theta\cdot4a\sin\theta\cos\theta\,d\theta = 64\pi a^3\int_0^{\pi/2}\sin^4\theta\cos^2\theta\,d\theta\]

Step 4: Evaluate the Wallis-type integral

Using the standard reduction (Wallis) formula, $\displaystyle\int_0^{\pi/2}\sin^4\theta\cos^2\theta\,d\theta = \frac{3\cdot1\cdot1}{6\cdot4\cdot2}\cdot\frac\pi2 = \frac{\pi}{32}$.

Step 5: Calculate the volume

\[V = 64\pi a^3\cdot\frac{\pi}{32} = 2\pi^2a^3\]

\[\boxed{V = 2\pi^2a^3 \text{ cubic units}}\]

Step 6: Numerical coefficient

\[V = 2\pi^2a^3 \approx 19.74\,a^3 \text{ cubic units}\]

Summary

\[\boxed{V = 2\pi^2a^3 \approx 19.74\,a^3 \text{ cubic units}}\]

Key insight: A tempting shortcut treats $(2a-x)\sqrt{x^3/(2a-x)}$ as if it collapsed to a bare power of $x$, but it actually simplifies only to $x^{3/2}\sqrt{2a-x}$ — the substitution $x=2a\sin^2\theta$ (standard for $\sqrt{x}\sqrt{2a-x}$-type integrands) is what's needed to finish it cleanly, producing the elegant closed form $2\pi^2a^3$.