Calculate the area of ​the helicoid defined by the image of $\phi:D\subset \mathbb{R}\to \mathbb{R}^3$; $\phi (u, v) = (u (\cos v), u (\sin v), v)$ **Calculate the area of ​​the helicoid defined by the image of $\phi:D\subset \mathbb{R}\to \mathbb{R}^3$; $\phi (u, v) = (u (\cos v), u (\sin v), v)$, with $D=\\{(u, v) ∈ \mathbb{R}^2 : 0 ≤ u ≤ 1; 0 ≤ v ≤ 2π\\}$.** I have thought about doing the following but I am not sure: I have to calculate P where $\int\int_{\phi(D)}dA=\int\int_D|\det(D_{\phi})|dudv$, but $D_{\phi}$ is not square, am I making an error? How can I do this? Thank you.

Your surface area element $\mathrm d A$ is wrong. The area of a paramerized surface surface $\phi:D\to \mathbb{R}^3$ is defined as $$ A(\phi) = \int_{\phi(D)} \, \mathrm dA = \iint_D |\phi_u \times \phi_v| \, \mathrm du\, \mathrm dv. $$ We have $$ \mathrm d\phi = \begin{pmatrix}\phi_u & \phi_v\end{pmatrix} = \begin{pmatrix} \cos v & -u\sin v\\\ \sin v & u\cos v\\\ 0 & 1 \end{pmatrix},$$ so that \begin{align} |\phi_u\times \phi_v| = |(\sin v, -\cos v, u)| = \sqrt{1+u^2}. \end{align} Thus $$ A(\phi) = \iint_D \sqrt{1+u^2}\, \mathrm du\,\mathrm dv = 2\pi \int_0^1 \sqrt{1+u^2}\, \mathrm du. $$ Can you compute this integral on your own?