I want to parametrise the surface $x^{2}+y^{2}=36$ to then calculate a surface integral however I'm not really sure how to parametrise this. Can we use $$\mathbf{r}(t) = (6\cos{t}, 6\sin{t}, 0).$$
I want to use the parametrisation to calculate the surface integral of $\phi = \sqrt{x^{2}+y^{2}}$ but i'm not really sure how to find the surface normal for $S$.
Based on your second comment, the surface is then a circle in the $xy$ plane, whose radius is 6.
Then a parametrization is $$\psi(r,\varphi)=(r\cos\varphi,r\sin\varphi,0) $$ where $0\le r<6$ and $0\le \varphi\le2\pi$. The surface integral is $$ \int_S\phi\circ\psi dA=\int_0^{2\pi}\int_0^6rdA $$ where $dA$ is $rdrd\varphi$, so $$\int_0^{2\pi}\int_0^6r^2drd\varphi=\left.\frac{r^3}{3}\right|_0^6\cdot\left.\varphi\right|_0^{2\pi}=72\cdot2\pi=144\pi $$