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

Question

I have a question regarding the solution to part of this homework question:

An infinite filled cylinder of radius $a$ contains a 3D charge density $\rho$. A thin-walled hollow cylinder of radius $b \gt a$ centered on the same axis surrounds it, and contains a charge with the same charge per unit length, but with opposite sign.

a) Compute the electric field $\vec E$ everywhere.

b) Compute the electrostatic potential $V$, defined by $\vec E = −\nabla V$, everywhere, subject to $V(r \to \infty) = 0$

c) Compute the energy in the electric field, per unit length. The 3D energy density is $U_E = \varepsilon_0 E^2/2$.

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I only have a question regarding the solution to part c). But, unfortunately for my question to make sense I will have to typeset the full solutions to a), b) and c):

The arrangement is shown above and the solution to part a) is

By symmetry, the electric field is radial everywhere. For $r \lt a$, Gauss’s theorem in a cylinder of unit length (or use a length $L$ if preferred) gives $$\oint \vec E \cdot d\vec S =\frac{Q}{\varepsilon_0}\implies E 2 \pi r=\frac{\pi r^2 \rho}{\varepsilon_0}\implies E=\frac{\rho r}{2 \varepsilon_0}$$ For $a \lt r \lt b$ the charge enclosed is $\pi a^2 \rho$, so $$E 2 \pi r=\frac{\pi a^2 \rho}{2 \varepsilon_0}\implies E=\frac{\rho a^2}{2 r \varepsilon_0}$$ For $r \gt b$ the charge enclosed is zero, so $$E=0$$

The image below is just for clarity and it shows the cylinder as viewed from its cross-section:

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The solution to part b) is

In cylindrical polars, the radial gradient is $\frac{\partial V}{\partial r}$, so $$V(r)=-\int_{\infty}^r E(r^{\prime})\,dr^{\prime}$$ Evidently $V=0$ for $r \gt b$.

For $a \lt r \lt b$, $$V(r)=-\int_{b}^r \frac{\rho a^2}{2 r^{\prime} \varepsilon_0} \, dr^{\prime}=-\frac{\rho a^2}{2 \varepsilon_0}\ln\left(\frac{r}{b}\right)$$

For $r \lt a$, $$V(r)=-\int_a^r \frac{\rho r^{\prime}}{2 \varepsilon_0} \,dr^{\prime}-\frac{\rho a^2}{2 \varepsilon_0} \ln\left(\frac{a}{b}\right) = \frac{\rho\left(a^2-r^2 \right)}{4 \varepsilon_0}-\frac{\rho a^2}{2 \varepsilon_0} \ln\left(\frac{a}{b}\right)$$

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The solution to part c) is

The energy per unit length is $$\frac{\varepsilon_0}{2} \int_0^b E^2(r) \, dr = \frac{\varepsilon_0}{2} \left(\frac{\rho}{2 \varepsilon_0}\right)^2 \left[\int_0^a {r^{\prime}}^2 \color{red}{2 \pi r^{\prime}} \, dr^{\prime}+\int_a^b \frac{1}{{r^{\prime}}^2} \color{red}{2 \pi r^{\prime}} \, dr^{\prime}\right]$$ $$=\frac{\pi \rho^2 a^4}{16 \varepsilon_0} \left[1+4 \ln\left(\frac{b}{a} \right) \right]$$

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Now, finally, my question is very simple. Why are those factors of $2 \pi r^{\prime}$ (shown in red) present in the integrands?

Since the question asked for the energy per unit length I think that the integral should be given by $$\frac{\varepsilon_0}{2}\int_0^b E^2(r) \, dr = \frac{\varepsilon_0}{2} \left(\frac{\rho}{2 \varepsilon_0}\right)^2 \left[\int_0^a {r^{\prime}}^2 \, dr^{\prime}+\int_a^b \frac{1}{{r^{\prime}}^2} \, dr^{\prime} \right]$$

I note that the formula for $U_E$ is energy per unit volume, but the $2 \pi r^{\prime} \, dr^{\prime}$ implies that the integration is performed over infinitesimal annuli from the inner radius ($r=0$) to the outer radius ($r=b$). If this is the case then dimensionally this is a surface integral (and not a volume integral).

Could someone please explain what necessitates the $2 \pi r^{\prime} \, dr^{\prime}$ factors in the integrands?

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Images shown in this question were taken from this pdf by MIT

Answer 1

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.

Answer 2

Forget about the electromagnetic details for a moment. Let $u$ denote an energy per unit volume. Across a cylinder of finite length, the stored energy is a volume integral of $u$. The energy per unit length is some quantity that integrates across the cylinder's length to give that stored energy. It is therefore a double integral of $u$, across the cylinder's cross section. Since $u$ depends only on $r^\prime\in[0,\,r]$, not on the angle $\theta$ or a distance $z$ along the cylinder's axis, this double integral's operator is$$\int_0^{2\pi}d\theta\int_0^rdr^\prime r^\prime=\int_0^r2\pi r^\prime dr^\prime.$$In particular, the energy per unit length is$$\int_0^r2\pi r^\prime u(r^\prime)dr^\prime.$$ The quoted calculation splits the integral up to $b$, the maximum $r$ for which $u\ne0$, into the $[0,\,a]$ and $[a,\,b]$ pieces, although it has an unfortunate typo in that the latter uses $\frac{1}{r^{\prime2}}$ instead of $\frac{a^4}{r^{\prime2}}$, an error exposed by dimensional analysis.