(a) Consider a hollow cylinder of length $l$, outer radius $r+\Delta r$ inner radius $r$, fixed at one end and twisted at the other by means of a couple of moment $N$. The angular displacement
$\varphi$, at a distance $l$ from the fixed end, is proportional to both $l$ and $N$. Consider an element of length $d x$ at the twisted end. It is moved by an angle $\varphi$ as shown. A vertical section is also shown and the twisting of the parallelopipe of length $l$ and area $\Delta r d x$ under the action of the twisting couple can be discussed by elementary means. If $f$ is the tangential force generated then shearing stress is $f / \Delta r d x$ and this must equal
Hence,
$$
\begin{gathered}
G \theta=G \frac{r \varphi}{l}, \text { since } \theta=\frac{r \varphi}{l} \\
f=G \Delta r d x \frac{r \varphi}{l}
\end{gathered}
$$
The force $f$ has moment $f r$ about the axis and so the total moment is
$$
N=G \Delta r \frac{\varphi}{l} r^{2} \int d x=\frac{2 \pi r^{3} \Delta r \varphi}{l} G
$$