= Solution
For $0<b-a<2\pi$, let $D=\mathbf r([a,b]\times[c,d])$ and use the <orientation of a surface> determined by $N=\mathbf r_u\times\mathbf r_v/|\mathbf r_u\times\mathbf r_v|$. Its positive <boundary orientation> follows the bottom side in increasing $u$, the right side in increasing $v$, the top side in decreasing $u$, and the left side in decreasing $v$. The <Gauss-Bonnet theorem> for this disc with four corners reads
$$
\int_D\kappa\,dA+\int_{\partial D}\kappa_g\,ds+\sum_{j=1}^4\varepsilon_j=2\pi,
$$
where the $\varepsilon_j$ are the signed exterior <angles>.
The bottom and top sides have cancelling <geodesic curvature> integrals: a <screw motion> takes one to the other with opposite <boundary orientation>. Explicitly, along increasing $u$ at fixed $v$, the <speed> is one and the signed <geodesic curvature> is $-\sin u/\sqrt{1+\sin^2u}$ on both sides. The right and left sides contribute, respectively, the integral along $\gamma_b$ and minus the integral along $\gamma_a$.
Although the coordinate sides are generally not perpendicular, the corner contributions still sum to $2\pi$. Indeed, let $\theta(u)$ be the <angle> between $\mathbf r_u$ and $\mathbf r_v$, so
$$
\cos\theta(u)=\frac{1+\cos u}{\sqrt{3+2\cos u}}.
$$
Starting at the bottom-left corner, the exterior <angles> are $\pi-\theta(a)$, $\theta(b)$, $\pi-\theta(b)$ and $\theta(a)$. Their sum is $2\pi$. Cancelling this from the <Gauss-Bonnet theorem> leaves
$$
\int_D\kappa\,dA=\int_{\gamma_a}\kappa_g\,ds-\int_{\gamma_b}\kappa_g\,ds,
$$
which is the identity in part (e). For a rectangle covering more than one turn, apply the same argument on the parameter rectangle with its induced <Riemannian metric>, or partition it into narrower rectangles and cancel the artificial internal <boundary> contributions. This gives the version counting multiplicity, as required by the correction in part (e).
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