= Solution
For a regular <parametrized surface> $\sigma(u,v)$, the <first fundamental form> is the induced metric
$$
I=E\,du^2+2F\,du\,dv+G\,dv^2,\qquad E=\sigma_u\cdot\sigma_u,\quad F=\sigma_u\cdot\sigma_v,\quad G=\sigma_v\cdot\sigma_v.
$$
Choose the <unit normal> $N=(\sigma_u\times\sigma_v)/|\sigma_u\times\sigma_v|$. With the convention $II_{ij}=N\cdot\sigma_{ij}$, the <second fundamental form> is
$$
II=e\,du^2+2f\,du\,dv+g\,dv^2,\qquad e=N\cdot\sigma_{uu},\quad f=N\cdot\sigma_{uv},\quad g=N\cdot\sigma_{vv}.
$$
The <Gaussian curvature>, the product of the <principal curvatures>, is
$$
\boxed{K=\frac{eg-f^2}{EG-F^2}.}
$$
For a compact <smooth embedded surface> without boundary, the <Gauss-Bonnet theorem> gives
$$
\boxed{\int_S K\,dA=2\pi e(S),}
$$
where $e(S)$ is the <Euler characteristic>. The no-boundary condition is needed for this form; a <smooth embedded surface> with boundary has the additional boundary <geodesic curvature> integral.
For the <surface of revolution>, use
$$
\sigma(u,v)=\bigl((r+a\sin u)\cos v,(r+a\sin u)\sin v,b\cos u\bigr).
$$
The parameters are periodic, with local open parameter rectangles providing the <coordinate charts>. Put $R=r+a\sin u$ and $D=\sqrt{a^2\cos^2u+b^2\sin^2u}$. Since $r>a>0$ and $b>0$, both $R$ and $D$ are positive. The tangent vectors are
$$
\sigma_u=(a\cos u\cos v,a\cos u\sin v,-b\sin u),\qquad\sigma_v=(-R\sin v,R\cos v,0).
$$
Their <inner products> are $E=D^2$, $F=0$, and $G=R^2$. Thus
$$
\boxed{I=(a^2\cos^2u+b^2\sin^2u)\,du^2+(r+a\sin u)^2\,dv^2.}
$$
Their <cross product> is $R(b\sin u\cos v,b\sin u\sin v,a\cos u)$, so its normalization gives the outward <unit normal>
$$
N=\frac1D(b\sin u\cos v,b\sin u\sin v,a\cos u).
$$
Next,
$$
\begin{aligned}
\sigma_{uu}&=(-a\sin u\cos v,-a\sin u\sin v,-b\cos u),\\
\sigma_{uv}&=(-a\cos u\sin v,a\cos u\cos v,0),\\
\sigma_{vv}&=(-R\cos v,-R\sin v,0).
\end{aligned}
$$
Taking their <inner products> with $N$ yields $e=-ab/D$, $f=0$, and $g=-bR\sin u/D$. Hence the <fundamental forms of an elliptic ring torus> give
$$
\boxed{II=-\frac{ab}{\sqrt{a^2\cos^2u+b^2\sin^2u}}\,du^2-\frac{b(r+a\sin u)\sin u}{\sqrt{a^2\cos^2u+b^2\sin^2u}}\,dv^2.}
$$
Choosing the inward <unit normal> changes both displayed coefficients of $II$ to their negatives and leaves $I$ and $K$ unchanged. As a consistency check, $K=ab^2\sin u/(RD^4)$; its integral over the <torus> is zero, since $K\,dA=ab^2\sin u\,du\,dv/D^3$ cancels under $u\mapsto u+\pi$. This agrees with $e(S)=0$ in the <Gauss-Bonnet theorem>.
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