= Solution
{c}
Use the <cylindrical coordinate system> basis $\mathbf e_r,\mathbf e_\varphi,\mathbf e_z$. The meridional unit <tangent vector> and outward unit <normal vector> consistent with the given geometry are
$$
\mathbf e_s=\cos\theta\,\mathbf e_r-\sin\theta\,\mathbf e_z,
\qquad
\mathbf n=\sin\theta\,\mathbf e_r+\cos\theta\,\mathbf e_z.
$$
An <surface of revolution> has $\theta$ independent of $\varphi$. Consequently
$$
\partial_s\mathbf e_s=-\theta_s\mathbf n,\qquad
\partial_\varphi\mathbf e_\varphi=-\mathbf e_r,
\qquad
\mathbf e_r=\cos\theta\,\mathbf e_s+\sin\theta\,\mathbf n.
$$
Apply the <product rule> to the vector <force balance>:
$$
\begin{aligned}
\partial_s(rt_s\mathbf e_s)
&=(r_s t_s+r\partial_s t_s)\mathbf e_s-rt_s\theta_s\mathbf n,\\
\partial_\varphi(t_\varphi\mathbf e_\varphi)
&=(\partial_\varphi t_\varphi)\mathbf e_\varphi
-t_\varphi\cos\theta\,\mathbf e_s-t_\varphi\sin\theta\,\mathbf n.
\end{aligned}
$$
Equating the three independent components to zero, and using $r_s=\cos\theta$, gives
$$
\begin{aligned}
\mathbf e_s &: \quad r\partial_s t_s+(t_s-t_\varphi)\cos\theta=0,\\
\mathbf e_\varphi &: \quad \partial_\varphi t_\varphi=0,\\
\mathbf n &: \quad rP-rt_s\theta_s-t_\varphi\sin\theta=0.
\end{aligned}
$$
Thus, away from the rotation axis where $r>0$, the scalar <mechanical equilibrium> equations are
$$
\boxed{P=t_s\theta_s+t_\varphi\frac{\sin\theta}{r},\qquad
\partial_s t_s=\frac{\cos\theta}{r}(t_\varphi-t_s),\qquad
\partial_\varphi t_\varphi=0.}
$$
The two geometric factors $\theta_s$ and $\sin\theta/r$ are the signed meridional and azimuthal <principal curvatures> in this convention. The normal equation is the <Young–Laplace equation> generalized to unequal principal <membrane tensions>.
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