Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the lemniscate $(x^2 + y^2)^2 = a^2(x^2 - y^2)$ about the $x$-axis.

We convert to polar form, carefully identify which portion of the curve to integrate, and cross-check the result directly in Cartesian coordinates.

Step 1: Convert to polar coordinates

With $x=\rho\cos\varphi$, $y=\rho\sin\varphi$: $(\rho^2)^2=a^2\rho^2\cos2\varphi \implies \rho^2=a^2\cos2\varphi$.

Step 2: Identify the loops

$\rho^2\ge0$ requires $\cos2\varphi\ge0$, i.e. $\varphi\in\left[-\frac\pi4,\frac\pi4\right]$ (right loop) or $\varphi\in\left[\frac{3\pi}{4},\frac{5\pi}{4}\right]$ (left loop) — the lemniscate's characteristic figure-eight.

Step 3: Set up the volume for the right loop

Since the right loop is symmetric about the $x$-axis, revolving just its upper half ($\varphi\in\left[0,\frac\pi4\right]$, where $y\ge0$) alone sweeps out the complete solid for that loop:

\[V_{\text{right}} = \frac{2\pi}{3}\int_0^{\pi/4}\rho^3\sin\varphi\,d\varphi = \frac{2\pi a^3}{3}\int_0^{\pi/4}(\cos2\varphi)^{3/2}\sin\varphi\,d\varphi\]

Step 4: Evaluate via a substitution

Let $w=\cos\varphi$ (so $\cos2\varphi=2w^2-1$) and $dw=-\sin\varphi\,d\varphi$. When $\varphi=0$: $w=1$; when $\varphi=\frac\pi4$: $w=\frac{1}{\sqrt2}$.

\[\int_0^{\pi/4}(\cos2\varphi)^{3/2}\sin\varphi\,d\varphi = \int_{1/\sqrt2}^1(2w^2-1)^{3/2}\,dw\]

A further substitution $v=\sqrt2\,w$ (so $2w^2-1=v^2-1$) and then $v=\cosh\theta$ reduces this to a standard hyperbolic power integral, giving the closed form:

\[\int_{1/\sqrt2}^1(2w^2-1)^{3/2}\,dw = -\frac18+\frac{3\sqrt2}{16}\ln(1+\sqrt2) \approx 0.10871\]

(this value is independently confirmed below by direct Cartesian integration).

Step 5: Volume of the right loop

\[V_{\text{right}} = \frac{2\pi a^3}{3}\left[-\frac18+\frac{3\sqrt2}{16}\ln(1+\sqrt2)\right] = a^3\pi\left[-\frac{1}{12}+\frac{\sqrt2}{8}\ln(1+\sqrt2)\right] \approx 0.2277\,a^3\]

Step 6: Cross-check directly in Cartesian coordinates

Solving $(x^2+y^2)^2=a^2(x^2-y^2)$ for $y^2$ as a quadratic gives $y^2=\dfrac{a\sqrt{a^2+8x^2}-(2x^2+a^2)}{2}$; using the disk method, $V=\pi\int_0^a y^2\,dx$ numerically reproduces $0.2277\,a^3$ exactly, confirming Step 5.

Step 7: Total volume (both loops)

The left loop is the exact mirror image of the right loop through the origin and occupies the disjoint range $x\in[-a,0]$, so the two loops' solids of revolution do not overlap; the total is twice the right-loop volume:

\[V_{\text{total}} = 2V_{\text{right}} = \pi a^3\left[\frac{\sqrt2}{4}\ln(1+\sqrt2)-\frac16\right]\]

\[\boxed{V_{\text{total}} = \pi a^3\left[\frac{\sqrt2}{4}\ln(1+\sqrt2)-\frac16\right] \approx 0.4554\,a^3 \text{ cubic units}}\]

Summary

\[\boxed{V_{\text{total}} = \pi a^3\left[\frac{\sqrt2}{4}\ln(1+\sqrt2)-\frac16\right] \approx 0.4554\,a^3 \text{ cubic units (each loop} \approx 0.2277\,a^3\text{)}}\]

Key insight: The lemniscate's polar form $\rho^2=a^2\cos2\varphi$ makes the region's extent transparent, but evaluating the resulting $(\cos2\varphi)^{3/2}$ integral genuinely requires a hyperbolic (not merely trigonometric) substitution, producing a volume expressed in terms of $\ln(1+\sqrt2)=\operatorname{arcsinh}1$ rather than a bare multiple of $\pi$ or $\sqrt2$.