Why are there factors of $2 \pi r$ in this volume integral?

The energy of the electric field in a volume $V$ is given by

$$\mathcal{E} = \frac{\epsilon_0}{2}\int_{V} E^2 dV$$

Let $V$ be a cylinder along the rod from $0$ to $\ell$ of infinite radius. So here you have

\begin{align} \mathcal{E} &= \frac{\epsilon_0}{2}\int_{V} E^2 dV \\ &=\frac{\epsilon_0}{2}\int_{0}^\ell\int_{0}^{2\pi}\int_0^\infty E^2(r)rdrd\theta dz \\ &=\frac{\epsilon_0}{2}\int_{0}^\ell\int_{0}^{2\pi}\int_0^b E^2(r)rdrd\theta dz \\ &=\frac{\epsilon_0}{2}\int_0^bE^2(r) [2\pi r \ell] dr \end{align} Since $E = 0$ for $r>b$. Therefore the energy per length $\ell$ is $$\mathcal{E}/\ell = \frac{\epsilon_0}{2}\int_0^bE^2(r) [2\pi r] dr$$ by symmetry this holds for any cylinder with the same orientation.