Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 24

Find an expression, but do not evaluate an integral for the volume of the solid obtained by rotating the region bounded by the curves $\displaystyle y = \frac{1}{1 + x^2}, y = 0, x = 0, x = 2$ about $x = 2$.

If we use vertical strips, notice that the distance of these strips from the line $x = 2$ is $2 - x$. If you revolve this length about $x = 2$, you'll get a circumference of $C = 2 \pi (2 - x)$. Also, notice that the height of the strips resembles the height of the cylinder as $H = y_{\text{upper}} - y_{\text{lower}} = \displaystyle \frac{1}{1 + x^2} - 0$. Thus, the expression for the volume is

$\displaystyle V = \int^b_a C(x) H(x) dx$


$
\begin{equation}
\begin{aligned}

& V = \int^2_0 2 \pi (2 - x) \left( \frac{1}{1 + x^2} \right) dx
\\
\\
& V = 2 \pi \int^2_0 \left( \frac{2 - x}{1 + x^2}\right) dx

\end{aligned}
\end{equation}
$