Hello Kuba,
To find the surface area of a torus whose major radius is $R$ and minor radius is $r$ (where $r\le R$), we may revolve the function:
$$f(x)=\sqrt{r^2-(x-R)^2}$$
about the $y$-axis and double the resulting surface of revolution $S$.
Hence, we may state:
$$S=4\pi\int_{R-r}^{R+r}x\sqrt{1+\left[f'(x) \right]^2}\,dx$$
Computing the required derivative, we find:
$$f'(x)=\frac{-(x-R)}{\sqrt{r^2-(x-R)^2}}$$
And so we obtain:
$$S=4\pi\int_{R-r}^{R+r}x\sqrt{1+\left[\frac{x-R}{\sqrt{r^2-(x-R)^2}} \right]^2}\,dx$$
Now, let's employ the substitution:
$$u=x-R\,\therefore\,du=dx$$
and now we may write:
$$S=4\pi\int_{-r}^{r}(u+R)\sqrt{1+\frac{u^2}{r^2-u^2}}\,du$$
Simplifying the radicand, we obtain:
$$S=4\pi r\int_{-r}^{r}\frac{u+R}{\sqrt{r^2-u^2}}\,du=4\pi r\left(\int_{-r}^{r}\frac{u}{\sqrt{r^2-u^2}}\,dx+\int_{-r}^{r}\frac{R}{\sqrt{r^2-u^2}}\,du \right)$$
On the far right the first integrand is odd and the second is even, and so we are left with:
$$S=8\pi rR\int_{0}^{r}\frac{1}{\sqrt{r^2-u^2}}\,du$$
Now, using the substitution:
$$u=r\sin(\theta)\,\therefore\,du=r\cos(\theta)$$
we obtain:
$$S=8\pi rR\int_{0}^{\frac{\pi}{2}}\frac{\cos(\theta)}{ \sqrt{1-\sin^2(\theta)}}\,d\theta= 8\pi rR\int_{0}^{\frac{\pi}{2}}\,d\theta$$
Applying the FTOC, we obtain:
$$S=8\pi rR\left(\frac{\pi}{2}-0 \right)$$
$$S=4\pi^2rR$$
This demonstrates that you can indeed treat the torus as a cylinder with respect to its surface area:
$$S=(2\pi r)(2\pi R)=4\pi^2 rR$$
Similarly, see http://mathhelpboards.com/questions-other-sites-52/roisins-question-yahoo-answers-regarding-volume-torus-7992.html?highlight=torus for the derivation using calculus of the volume of a torus, given as:
$$V=2\pi^2r^2R$$
which we may obtain by treating the torus as a cylinder as follows:
$$V=\pi r^2(2\pi R)=2\pi^2r^2R$$
Now, using the given data:
$$r=4\text{ in}$$
$$R=13\text{ in}$$
we find the surface area of the given torus is:
$$S=4\pi^2(4\text{ in})(13\text{ in})=208\pi^2\text{ in}^2\approx2052.87771543\text{ in}^2$$