Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 76

Determine the volume of solid obtained by rotating the region under the curve $\displaystyle y = \frac{1}{x^2+1}$ from 0 to 3 about the $y$-axis.

By using vertical strips, and applying the shell method, notice that the strips have distance from $y$-axis as $x$ and if you rotate this length about $y$-axis, you'll get a circumference of $c = 2\pi x$. Also, the height of the strips resembles the height of the cylinder as $\displaystyle H = y_{\text{upper}} - y_{\text{lower}} = \frac{1}{x^2+1} - 0 = \frac{1}{x^2+1}$. Theus,



$\begin{equation} \begin{aligned} V &= \int^3_0 c(x) H (x) dx \\ \\ V &= \int^3_0 3(2 \pi x) \left( \frac{1}{x^2+1} \right) dx \end{aligned} \end{equation}$



Let $u = x^2 + 1$, then

$du = 2x dx$

Make sure that the upper and lower units are also in terms of $u$



$\begin{equation} \begin{aligned} V &= \pi \int^{(3)^2+1}_{(0)^2 +1} \frac{1}{u} du \\ \\ V &= \pi \int^{10}_1 \frac{du}{u} \\ \\ V &= \pi [ \ln u]^{10}_{1} \\ \\ V &= \pi [\ln10-\ln1] \\ \\ V &= \pi \ln(10) \text{ cubic units} \end{aligned} \end{equation}$