Parametric representation of $\sqrt{x^2+y^2}\le z \le 2$

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just wondering how to parametrize this. Question is:

Let $C$ denote the conical region $\sqrt{x^2+y^2}\le z \le 2$. Find a parametric representation $\mathbf{x}(u,v)$ for $S$, the surface of $C$. (Hint: you'll need to split it into two parts.)

[Other parts to the question (that I can probably do after the parameterisation): Use simple geometry to write down the outwards pointing unit vector at each point on $S$. Find the surface area of the cone.]

I thought that maybe the parametrization could be $$\mathbf{x}(u,v)=(v\cos{u},v\sin{u},v)$$ with $0\le u\le2\pi$ and $0\le v\le 2$ but this would give me the volume of the cone if I integrated over it... Please help! Thank you :)

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Using your notation, the surface integral is

$$S=\int_{D(u,v)}\|x_u \times x_v\|dudv, $$

where $D(u,v)=\{(u,v)\in\mathbb R^{2} | 0\leq u\leq 2\pi, 0\leq v\leq 2\}$ and

$$x_u=(-v\sin u, v\cos u, 0), $$

$$x_v=(\cos u,\sin u, 1). $$

A quick computation gives $\|x_u\times x_v\|=\sqrt{2v^2}=\sqrt{2}v$. We arrive at

$$S=\int_{D(u,v)}\|x_u \times x_v\|dudv=\int_{0}^{2\pi}\int_{0}^{2}\sqrt{2}vdudv= 2\pi\sqrt{2}\frac{1}{2}2^2=4\sqrt{2}\pi.$$