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Let $R_t$ be the horizontal rotation matrix through angle $t$. Define the screw motionIts linear part is an orthogonal matrix with determinant one, and its remaining part is a translation. Thus it is a proper Euclidean motion of Euclidean three-space, with ${\Phi }_s{\Phi }_t={\Phi }_{s+t}$.

In horizontal-vector notation the given parametric surface satisfies $\mathbf{r}(u,v)=(R_v(1+\cos u,\sin u),v)$. ThereforeThe defining circle in each horizontal plane is entirely parametrized by $u$, so $\mathbf{r}$ covers $X$. The displayed equality and ${\Phi }_{-t}={\Phi }_t^{-1}$ prove ${\Phi }_t(X)=X$. This one-parameter subgroup preserves the dot products and cross products defining $I$, $II$ and the chosen unit normal. Consequently their coordinate coefficients, and the Gaussian curvature, depend only on $u$.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

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 readswhere the ${\epsilon }_j$ 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 $-\sin u/\sqrt{1+{\sin }^{2}u}$ 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 ${\mathbf{r}}_v$, soStarting 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 leaveswhich 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).