Find the volume generated by rotating the region bounded

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

We apply the disk method for the $x$-axis and the shell method for the $y$-axis.

Part 1: Rotation about the $x$-axis

Step 1: Disk method

At position $x\in[-1,1]$, the disk has radius $r(x)=\dfrac{1}{1+x^2}$, so $A(x)=\dfrac{\pi}{(1+x^2)^2}$.

Step 2: Integrate using the standard antiderivative

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

\[V_x = \pi\left[\frac12\left(\arctan x+\frac{x}{1+x^2}\right)\right]_{-1}^{1}\]

At $x=1$: $\frac12\left(\frac\pi4+\frac12\right)=\frac\pi8+\frac14$. At $x=-1$ (odd function): $-\frac\pi8-\frac14$.

\[V_x = \pi\left[\left(\frac\pi8+\frac14\right)-\left(-\frac\pi8-\frac14\right)\right] = \pi\left(\frac\pi4+\frac12\right) = \frac{\pi^2}{4}+\frac{\pi}{2}\]

\[\boxed{V_x = \frac{\pi(\pi+2)}{4} \text{ cubic units}}\]

Part 2: Rotation about the $y$-axis

Step 1: Shell method

A shell at radius $x\in[0,1]$ has height $\frac{1}{1+x^2}$; doubling by symmetry for $x\in[-1,0]$:

\[V_y = 2\int_0^1 2\pi x\cdot\frac{1}{1+x^2}\,dx = 4\pi\int_0^1\frac{x}{1+x^2}\,dx\]

Step 2: Evaluate via substitution

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

\[\int_0^1\frac{x}{1+x^2}\,dx = \frac12[\ln u]_1^2 = \frac{\ln2}{2}\]

Step 3: Calculate the volume

\[V_y = 4\pi\cdot\frac{\ln2}{2} = 2\pi\ln2\]

\[\boxed{V_y = 2\pi\ln2 \text{ cubic units}}\]

Numerical values

\[V_x = \frac{\pi(\pi+2)}{4} \approx 4.04, \qquad V_y = 2\pi\ln2 \approx 4.36 \text{ cubic units}\]

Summary

\[\boxed{V_x = \frac{\pi(\pi+2)}{4}\approx4.04 \text{ cubic units}, \qquad V_y = 2\pi\ln2\approx4.36 \text{ cubic units}}\]

Key insight: $y=\frac{1}{1+x^2}$ is the derivative of $\arctan x$, so the volume about the $x$-axis (which needs $\int\frac{1}{(1+x^2)^2}dx$) naturally produces $\pi$-terms, while the volume about the $y$-axis (via the shell method's $\int\frac{x}{1+x^2}dx$) produces a $\ln2$ term instead — two very different transcendental constants emerging from the same curve.