Compute the Surface area of $S=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2,1\le z\le2\right\}$

Question

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{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?

Answer 1

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.