Surface area of a torus

Another way to proceed would be by writing out the surface integral using differential forms. To this end we need to set up a chart and a coordinate system on the Torus. Fortunately we need only one chart and the coordinate system can be made global (using the coordinates already introduced in the question $u,v$), the surface integral will be

$$ S=\int_{M} ab~ du \wedge dv $$ The integration runs over the coordinate range of the open cover $M=T^2$, that is $u \in [0, 2\pi], ~v \in [0, 2\pi]$ leading to

$$ S= ab \int_0^{2\pi}du \int_0^{2\pi} dv = 4\pi^2 ab $$

Use Pappus theorem. If the radius of the transversal section of the torus is $r$ then its perimeter is $2\pi r$ and Pappus theorem states that the surface of the torus (it is a revolution surface) equals $A=2\pi r \cdot 2 \pi R$ where $R$ is the radius of rotation that generates the torus. In your case this is

$$ A = 4\pi^2 ab $$

$$\frac{\partial\vec{g}}{\partial u}=(-a\sin u\cos v, -a\sin u\sin v, a \cos u),$$

$$\frac{\partial\vec{g}}{\partial v}=(-(b+a\cos u)\sin v, (b+a\cos u)\cos v, 0).$$

Without surprise, these two vectors are orthogonal, and the norm of their cross product is the product of their norms, $a(b+a\cos u)$.

Integrating on $u,v$ both in range $[0,2\pi]$, noticing that the average value of the cosine is zero, will yield $$(2\pi)^2ab.$$ Nothing really tedious.