= Solution
{c}
Use separate moving <orthonormal bases> in the reference and current <cylindrical coordinate systems>. For <pure torsion of an incompressible cylinder>, the <deformation map> is
$$
\mathbf x(R,\Theta,Z)=R\mathbf e_r(\Theta+\tau Z)+Z\mathbf e_z.
$$
The <unit vectors> depend on the angle: $\partial_\theta\mathbf e_r=\mathbf e_\theta$ and $\partial_\theta\mathbf e_\theta=-\mathbf e_r$. Consequently
$$
\partial_R\mathbf x=\mathbf e_r,\qquad
\partial_\Theta\mathbf x=R\mathbf e_\theta,\qquad
\partial_Z\mathbf x=\tau R\mathbf e_\theta+\mathbf e_z.
$$
In particular, the last derivative includes the rotation of $\mathbf e_r$ along the cylinder. The material <gradient> has the reference coordinate scale factor $1/R$:
$$
\begin{aligned}
\mathbf F
&=(\partial_R\mathbf x)\otimes\mathbf E_R
+\frac1R(\partial_\Theta\mathbf x)\otimes\mathbf E_\Theta
+(\partial_Z\mathbf x)\otimes\mathbf E_Z\\
&=\mathbf e_r\otimes\mathbf E_R
+\mathbf e_\theta\otimes\mathbf E_\Theta
+(\tau r\mathbf e_\theta+\mathbf e_z)\otimes\mathbf E_Z.
\end{aligned}
$$
The <outer products> specify the images of the three reference <unit vectors>. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,
$$
\boxed{\mathbf F=\begin{pmatrix}1&0&0\\0&1&\tau r\\0&0&1\end{pmatrix}.}
$$
Its <determinant> is one, checking <incompressibility>. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in <Cartesian coordinates>.
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