$\def\hl#1#2{\bbox[#1,1px]{#2}} \def\box#1#2#3#4#5{\color{#2}{\bbox[0px, border: 2px solid #2]{\hl{#3}{\color{white}{\color{#3}{\boxed{\underline{\large\color{#1}{\text{#4}}}\\\color{#1}{#5}\\}}}}}}} \def\verts#1{\left\vert#1\right\vert} \def\Verts#1{\left\Vert#1\right\Vert} \def\R{\mathbb{R}}$ $\box{black}{black}{} {Question} {\text{Compute the Surface area of $S=\left\{(x,y,z)\in\R^3:x^2+y^2=z^2,1\le z\le2\right\}$}}$
My Attempts
Since $z>0$, consider a injective fuction $G:T\subseteq\R^2\to S$ such that $G(x,y)=(x,y,\sqrt{x^2+y^2})$ have \begin{align} G_x=&\left(1,0,\frac{x}{\sqrt{x^2+y^2}}\right)\hspace{5ex} G_y=\left(0,1,\frac{y}{\sqrt{x^2+y^2}}\right)\\ G_x\times G_y=&\left(-\frac{x}{\sqrt{x^2+y^2}},-\frac{y}{\sqrt{x^2+y^2}},1\right)\\ \Verts{G_x\times G_y}=&\sqrt{2} \end{align} Therefore \begin{align} \text{area}(S)=&\iint_T\Verts{G_x\times G_y}dA\\ =&\int_0^{2\pi}\int_1^2\sqrt{2}\cdot r~drd\theta\\ =&3\sqrt{2}~\pi \end{align} Is my solution correct?
I checked it directly in cylindrical coordinates as parametric surface and obtain $$\begin{cases} E=\cos^2 \phi + \sin^2 \phi +1 = 2 \\ G=r^2 \sin^2 \phi + r^2 \cos^2 \phi +0 = r^2 \\ F = r\cos \phi (-\sin \phi) + r\sin \phi \cos \phi +0 =0 \end{cases} $$ So $\sqrt{EG-F} = r\sqrt{2}$ so your integral is correct.