Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curve $x - (1 + x^2)^2 y^2 = 0$ about its asymptote.

We solve for $y$, apply the disk method, and reduce the integral to a simple power-rule form.

Step 1: Rewrite the curve

\[(1+x^2)^2y^2 = x \implies y = \pm\frac{\sqrt x}{1+x^2}, \qquad x\ge0\]

Step 2: Find the asymptote

As $x\to0^+$: $y\to0$. As $x\to\infty$: $y=\dfrac{\sqrt x}{1+x^2}\sim\dfrac{1}{x^{3/2}}\to0$. The horizontal asymptote is $y=0$.

Step 3: Disk method

\[A(x) = \pi y^2 = \pi\cdot\frac{x}{(1+x^2)^2}\]

Step 4: Set up the integral

\[V = \pi\int_0^{\infty}\frac{x}{(1+x^2)^2}\,dx\]

Step 5: Evaluate via substitution

Let $u=1+x^2$, $du=2x\,dx$. When $x=0$: $u=1$; when $x\to\infty$: $u\to\infty$.

\[\int_0^{\infty}\frac{x}{(1+x^2)^2}\,dx = \frac12\int_1^{\infty}u^{-2}\,du = \frac12\left[-\frac1u\right]_1^{\infty} = \frac12(0+1) = \frac12\]

Step 6: Calculate the volume

\[V = \pi\cdot\frac12 = \frac{\pi}{2}\]

\[\boxed{V = \frac{\pi}{2} \text{ cubic units}}\]

Summary

\[\boxed{V = \frac{\pi}{2} \approx 1.571 \text{ cubic units}}\]

Key insight: This bell-shaped curve (related to the Witch of Agnesi) rises from the origin, peaks near $x=\frac{1}{\sqrt3}$, and decays like $x^{-3/2}$ — fast enough that the disk-method volume converges cleanly to $\frac{\pi}{2}$.