Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the region bounded by the curves $y = \frac{1}{1 + x^2}$ and $y = 0$ about the $x$-axis.

We use the disk method over the whole curve, reducing to a standard improper integral.

Step 1: Describe the region

$y=\frac{1}{1+x^2}>0$ for all $x$, with maximum $1$ at $x=0$ and $y\to0$ as $x\to\pm\infty$; the region extends over the whole real line.

Step 2: Disk method

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

Step 3: Set up the integral using symmetry

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

Step 4: Evaluate via the standard antiderivative

\[\int\frac{1}{(1+x^2)^2}\,dx = \frac12\left(\arctan x+\frac{x}{1+x^2}\right)+C\]

As $b\to\infty$: $\arctan b\to\frac\pi2$ and $\frac{b}{1+b^2}\to0$, so:

\[\int_0^{\infty}\frac{1}{(1+x^2)^2}\,dx = \frac12\left(\frac\pi2+0\right)-0 = \frac{\pi}{4}\]

Step 5: Calculate the volume

\[V = 2\pi\cdot\frac{\pi}{4} = \frac{\pi^2}{2}\]

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

Step 6: Cross-check via trigonometric substitution

With $x=\tan\theta$: $\int_0^{\infty}\frac{dx}{(1+x^2)^2} = \int_0^{\pi/2}\cos^2\theta\,d\theta = \frac12\left[\theta+\frac{\sin2\theta}{2}\right]_0^{\pi/2} = \frac{\pi}{4} \quad\checkmark$

Step 7: Numerical value

\[V = \frac{\pi^2}{2} \approx 4.935 \text{ cubic units}\]

Summary

\[\boxed{V = \frac{\pi^2}{2} \approx 4.935 \text{ cubic units}}\]

Key insight: Even though the region extends over the whole real line, the curve decays like $1/x^2$, fast enough for the disk-method integral (which involves $1/x^4$) to converge, producing the compact result $\pi^2/2$ — a hint of the same $\arctan$ machinery seen throughout this family of problems.