The area of the outer part of the torus surface

To find the area of the torus, we can use the parametrization $$f(u, v) = ((a+b\cos u)\cos v, (a+b\cos u)\sin v, b\sin u))^{T}, (u,v)\in D_{uv}=[0,2\pi]^2$$ and evaluate the surface integral $S=\iint_{D_{uv}} {\lVert \frac{∂f}{∂u} \times \frac{∂f}{∂v} \rVert \, du \, dv}$.

But how can I find a area of the outer part of the torus surface that is outside the cylinder $x^2+y^2=a^2$?

The surface of the given torus outside the cylinder $x^2+y^2=a^2$ is generated by the semicircle $$S:=\{(a+b\cos(u),0,b\sin(u)): u\in [-\pi/2,\pi/2]\}$$ revolved around the $z$-axis (the whole circle generates the surface of the whole torus!).

We may apply Pappus's centroid theorem for areas: since the centroid of $S$ is $(a+\frac{2b}{\pi},0,0)$ (check Arc of circle in this list), we have that the area of the outer part of the torus surface is $$\text{Area}=2\pi\left(a+\frac{2b}{\pi}\right)\cdot \pi b=2\pi^2ab+4\pi b^2.$$

P.S. You can verify that above formula is correct by computing: $$\iint_{D_{uv}} {\left\lVert \frac{∂f}{∂u} \times \frac{∂f}{∂v} \right\rVert \, du \, dv}$$ with $D_{uv}=\{(u,v): u\in[-\pi/2,\pi/2], v\in[0,2\pi]\}$.