surface area of $\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 =\frac{1}{z^2}, 1I want to calculate the surface area of the surface that bounds the solid $$K=\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 \leq\frac{1}{z^2},   1<z<3\right\}$$ 
I'm stuck with the differential surface area that I shall consider so that I can solve $S=\iint dS$.

Here is how I would do it:
The surface can be parametrized as follows 
\begin{cases}
x=x\\
y=y\quad\quad\quad\quad\quad\quad\quad(x,y)\in D=\{(x,y)\;|\;\frac{1}{9}\le x^2+y^2\le 1\}\\
z=\frac{1}{\sqrt{x^2+y^2}}\\
\end{cases}
You can plot this surface and its domain $D$ with WolframAlpha:

Now, the surface area is given by
$$
A=\iint_D ||r_x\times r_y ||\; dA = \iint_{\{(x,y)\;|\;\frac{1}{9}\le x^2+y^2\le 1\}} \sqrt{ \frac{x^2+y^2+(x^2+y^2)^3}{(x^2+y^2)^3}}\; dA 
$$
Switching to polar coordinates yields:
$$
\boxed{
A=\int_0^{2\pi}\int_{1/3}^1\sqrt{r^{-2}+r^2}\; drd\theta \approx  7.6030
}
$$
Alternatively you could proceed as follows:
\begin{cases}
x=\frac{\cos\theta}{z}\\
y=\frac{\sin\theta}{z}\quad\quad\quad\quad\quad\quad\quad0\le \theta\le 2\pi, \; 1\le z\le 3\\
z=z\\
\end{cases}
$$
A=\iint_{\{(\theta,z)|0\le\theta\le 2\pi, \; 1\le z\le 3\}} ||r_{\theta}\times r_z ||\; dA 
$$
Computing the integral yields
$$
\boxed{A=\int_0^{2\pi}\int_1^3\sqrt{z^{-2}+z^{-6}}\;dzd\theta \approx  7.6030}
$$
Also note that using the change of variables $z=\frac{1}{r}$ (i.e., $dz=\frac{-dr}{r^2}=-z^2dr$):
$$
\int_{1/3}^1\sqrt{r^{-2}+r^2}\; dr = \int_1^3\sqrt{z^{-2}+z^{-6}}\;dz
$$
