= Solution
{c}
Write $W_i=\partial W/\partial\alpha_i$, evaluated at $(\alpha^{-1/2},\alpha^{-1/2},\alpha)$. Applying the <chain rule> to the reduced <strain energy density> gives
$$
\hat W'(\alpha)
=-\frac12\alpha^{-3/2}W_1-\frac12\alpha^{-3/2}W_2+W_3.
$$
Multiply by $\alpha$ and use $\alpha_1=\alpha_2=\alpha^{-1/2}$:
$$
\alpha\hat W'(\alpha)
=\alpha_3W_3-\frac12\alpha_1W_1-\frac12\alpha_2W_2.
$$
The transverse symmetry specified in the hint has $W_1=W_2$. Thus
$$
\boxed{\alpha\hat W'(\alpha)=\alpha_3W_3-\alpha_1W_1.}
$$
The lateral faces have zero <traction>, so their diagonal <Cauchy stress tensor> components satisfy $t_1=t_2=0$. The <hydrostatic pressure> enforcing <elastic incompressibility> is therefore $p=\alpha_1W_1=\alpha_2W_2$, and the axial component simplifies to
$$
\boxed{t_3=\alpha_3W_3-p=\alpha\hat W'(\alpha).}
$$
This elimination of the <Lagrange multiplier> $p$ is what makes the reduced <strain energy density> useful for the spring balance.
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