I 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., $dz=\frac{-dr}{r^2}=-z^2dr$): $$ \int_{1/3}^1\sqrt{r^{-2}+r^2}\; dr = \int_1^3\sqrt{z^{-2}+z^{-6}}\;dz $$