Single Variable Calculus, Chapter 6, 6.3, Section 6.3, Problem 30

$\displaystyle 2 \pi \int^2_0 \frac{y}{1 + y^2} dy$ represents a volume of a solid. Describe the solid.

We can see from the equation that shell method was used with horizontal strips. The distance of these strips from a line that it will be revolved in is $y$. If you revolve this length about such line, you'll get a circumference of $C = 2 \pi y$. By these data, we assume that the line we are talking is $x$-axis. Also, the height of these strips resembles the height of the cylinder by $\displaystyle H = x_{\text{right}} - x_{\text{left}} = \frac{1}{1 + y^2} - 0$. Thus,


$\begin{equation} \begin{aligned} V =& \int^2_0 C(y) H(y) dy \\ \\ V =& \int^2_0 (2 \pi y) \left( \frac{1}{1 + y^2} \right) dy \\ \\ V =& 2 \pi \int^2_0 \frac{y}{1 + y^2} \end{aligned} \end{equation}$


In short, this expression is obtained by rotating the region bounded by $\displaystyle x = \frac{1}{1 + y^2}$ and $x = 0$ about $x$-axis using shell method.