Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 72

Find the volume of the region under the curve $\displaystyle y = \frac{1}{\sqrt{x^2+4}}$ from $x = 0$ to $x = 2$ that is rotated about the $x$-axis.

By using vertical strips, notice that if you slice the curve, its cross section forms a circle with radius $\displaystyle \frac{1}{\sqrt{x^2+4}}$. Hence, its cross sectional area is as $\displaystyle A = \pi \left( \frac{1}{\sqrt{x^2 +4}} \right) = \frac{\pi}{x^2+4}$

Thus, the volume is...

$
\begin{equation}
\begin{aligned}
V &= \int^2_0 A(x) dx\\
\\
V &= \int^2_0 \frac{\pi}{x^2+4} dx\\
\\
V &= \pi \int^2_0 \frac{dx}{x^2 + 4}\\
\\
\text{We can rewrite it as } V &= \pi \int^2_0 \frac{\left( \frac{1}{4} \right)}{\frac{x^2}{4}+1} dx\\
\\
&= \pi \int^2_0 \frac{\left( \frac{1}{4} \right)}{\left(\frac{x}{2}\right)^2+1} dx
\end{aligned}
\end{equation}
$


If we let $\displaystyle u = \frac{x}{2}$, then
$\displaystyle du = \frac{dx}{2}$
Make sure that the upper and lower limits are also in terms of $u$ so...
$\displaystyle V = \pi \int^{\frac{2}{2}}_{\frac{0}{2}} \frac{2 du}{u^2 + 1} \left( \frac{1}{4} \right)$


Recall that $\displaystyle \frac{d}{dx}\left( \tan^{-1} x \right) = \frac{dx}{1 + x^2}$

$
\begin{equation}
\begin{aligned}
V &= \frac{\pi}{2} \int^1_0 \frac{du}{u^2 \pi}\\
\\
V &= \frac{\pi}{2} \left[ \tan^{-1} u \right]^1_0\\
\\
V &= \frac{\pi}{2} \left[ \tan^{-1} (1) - \tan^{-1} (0) \right] = \frac{\pi}{2} \left[ \frac{\pi}{4} -0 \right] = \frac{\pi^2}{8} \text{cubic units}
\end{aligned}
\end{equation}
$