Let $\varphi: (0, +\infty) \times (0, 2\pi) \to \mathbb{R}^3$ with $\varphi(r, \theta) = (r \cos(\theta), r \sin(\theta), \theta)$, and let $S$ be the image of $\varphi$.
(a) Show that $S$ is a surface, and $\varphi$ is a parametrization.
(b) Calculate the first and second fundamental forms associated with the given parametrization.
(c) Calculate the Gaussian curvature of $S$ at each of its points.
Here is my solution:
(a) Let's show that $\varphi$ is a parametrization: $\varphi$ is a homeomorphism with its image. It has a continuous inverse (and $\varphi$ itself is continuous): $$ \varphi^{-1}(x, y, z) = \biggl(\sqrt{x^2 + y^2},\, z\biggr). $$ Additionally, $\varphi$ is differentiable, and its differential is $$ d\varphi(r, \theta) = \begin{pmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \\ 0 & 1 \end{pmatrix}. $$ The determinant of the $2 \times 2$ leading minor is $r$, which is nonzero. Therefore, $\varphi$ is a parametrization, and it is global as $S = \operatorname{Im}, \varphi$. Thus, $S$ is a surface.
(b) We have $\dfrac{\partial\varphi}{\partial r} = \bigl(\cos(\theta),\, \sin(\theta),\, 0 \bigr)$ and $\dfrac{\partial\varphi}{\partial\theta} = \bigl(-r\sin(\theta),\, r\cos(\theta),\, 1\bigr)$. Hence, $$ E = \Bigl\langle\frac{\partial\varphi}{\partial r}, \frac{\partial\varphi}{\partial r}\Bigr\rangle = 1, \quad F = \Bigl\langle\frac{\partial\varphi}{\partial\theta}, \frac{\partial\varphi}{\partial r}\Bigr\rangle = 0, \quad\text{and}\quad G = \Bigl\langle\frac{\partial\varphi}{\partial\theta}, \frac{\partial\varphi}{\partial\theta}\Bigr\rangle = r^2 + 1. $$ Then, $$ N = \frac{\frac{\partial\varphi}{\partial r} \wedge \frac{\partial\varphi}{\partial\theta}}{\left|\frac{\partial\varphi}{\partial r} \wedge \frac{\partial\varphi}{\partial\theta}\right|} = \left(\frac{\sin(\theta)}{\sqrt{r^2+1}},\, \frac{\cos(\theta)}{\sqrt{r^2+1}},\, \frac{r}{\sqrt{r^2+1}}\right). $$ Therefore: $$ \begin{align} e &= \Bigl\langle N, \frac{\partial^2\varphi}{\partial\theta^2} \Bigr\rangle = 0 \\ f &= \Bigl\langle N, \frac{\partial^2\varphi}{\partial\theta\,\partial r} \Bigr\rangle = -\frac{1}{\sqrt{r^2+1}} \\ g &= \Bigl\langle N, \frac{\partial^2\varphi}{\partial r^2} \Bigr\rangle = 0 \end{align} $$
(c) Gaussian curvature: $$ K\bigl(\varphi(r,\theta)\bigr) = \frac{eg - f^2}{EG - F^2} =-\frac{1}{(r^2 + 1)^2} $$