= Solution
Use the material-first <nominal stress tensor> convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}\dot A_{i\alpha}$. At a regular constraint point, $F_{,A}\ne0$, all admissible deformation rates satisfy $F_{,A_{i\alpha}}\dot A_{i\alpha}=0$. Equality of mechanical power and the derivative of the <strain energy density> gives
$$
\left(N_{\alpha i}-W_{,A_{i\alpha}}\right)\dot A_{i\alpha}=0
$$
for every <vector> in this eight-dimensional tangent hyperplane. Its <orthogonal complement> is the span of $F_{,A}$. Consequently the <constraint reaction in hyperelastic stress> is
$$
\boxed{N_{\alpha i}=W_{,A_{i\alpha}}+qF_{,A_{i\alpha}}.}
$$
The scalar <Lagrange multiplier> $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular <incompressibility> constraint by $(\det A-1)^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.
For <incompressibility> use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ gives
$$
\frac{\partial J}{\partial A_{i\alpha}}
=\frac12\epsilon_{ijk}\epsilon_{\alpha\beta\gamma}A_{j\beta}A_{k\gamma}
=J(A^{-1})_{\alpha i}.
$$
The first expression is the cofactor entry; contracting it with $A_{j\alpha}$ gives $J\delta_{ij}$, proving the inverse formula. On $J=1$ this yields
$$
\boxed{N_{\alpha i}=W_{,A_{i\alpha}}+q(A^{-1})_{\alpha i}.}
$$
With this plus-sign convention $q$ is the negative of the usual pressure multiplier.
For an equibiaxially stretched sheet, <incompressibility> requires the principal stretches $(\lambda,\lambda,\lambda^{-2})$. The printed sheet expression's $\lambda^{-1/2}$ is inconsistent with this constraint and must be $\lambda^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to <principal stretch> $\lambda_j$, evaluated on this constrained path. With no applied thickness-direction <traction>,
$$
0=N_{33}=W_3+q\lambda^2,
\qquad q=-\lambda^{-2}W_3.
$$
Isotropy gives $W_1=W_2$ at equal in-plane stretches. Hence
$$
N_{11}=N_{22}=W_1-\lambda^{-3}W_3,
\qquad
\frac d{d\lambda}W(\lambda,\lambda,\lambda^{-2})
=W_1+W_2-2\lambda^{-3}W_3.
$$
The <equibiaxial nominal tension of an incompressible sheet> is therefore
$$
\boxed{N_{11}=N_{22}=\frac12\frac d{d\lambda}W(\lambda,\lambda,\lambda^{-2}).}
$$
The two equal in-plane nominal tensions do work $2N_{11}\,d\lambda$ per reference volume; the thickness-direction <nominal stress> does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a <neo-Hookean solid> gives $N=\mu(\lambda-\lambda^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu/4$ at that state.
For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2h_0$. Its two tangential stretches are $\lambda=a/a_0$, while its thickness stretch is $\lambda^{-2}$ and its current thickness is $h=h_0\lambda^{-2}$. Define $\widehat W(\lambda)=W(\lambda,\lambda,\lambda^{-2})$. Its total <elastic energy> at leading thin-shell order is
$$
E=4\pi a_0^2h_0\widehat W(\lambda).
$$
During a quasistatic increment $da$, the enclosed volume changes by $d\mathcal V=4\pi a^2da$, and $d\lambda=da/a_0$. Equating energy change to work of the internal excess pressure gives
$$
p\,4\pi a^2da
=4\pi a_0h_0\widehat W'(\lambda)da,
\qquad
\boxed{p(a)=\frac{h_0}{a_0\lambda^2}\widehat W'(\lambda).}
$$
This is the <inflation pressure of a thin incompressible elastic balloon>. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll1$. It also agrees with membrane <force> balance: $p=2h\sigma_t/a$, $\sigma_t=\lambda N$, and $N=\widehat W'/2$. Here pressure is <force> per current area, while $N$ is <force> per reference area; confusing them would lose a factor of $\lambda$.
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