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 $S=\iint dS$.
 A: 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., $dz=\frac{-dr}{r^2}=-z^2dr$):
$$
\int_{1/3}^1\sqrt{r^{-2}+r^2}\; dr = \int_1^3\sqrt{z^{-2}+z^{-6}}\;dz
$$
A: Kuifje wonders in the comments how to calculate the surface area using the cylindrical parametrization $r(s,t)=t^{-1}$, $\theta(s,t)=s$, $z(s,t)=t$ for $(s,t)\in(0,2\pi)\times(1,3)$.
I admit this approach is hard to find in most multivariable calculus texts (feel free to speculate why it is more likely to be found in a physics book), but the procedure is the same: we cross tangent vectors to get the normal surface element, and then integrate its magnitude over the specified region in $(s,t)$ space. There are two catches. First, the position vector in cylindrical coordinates is not $r\mathbf{\hat{r}}+\theta\boldsymbol{\hat{\theta}}+z\mathbf{\hat{z}}$ but rather $r\mathbf{\hat{r}}+z\mathbf{\hat{z}}$. (Similarly, the position vector in polar coordinates is not $r\mathbf{\hat{r}}+\theta\boldsymbol{\hat{\theta}}$ but simply $r\mathbf{\hat{r}}$.) Second, writing down tangent vectors is not as simple as in Cartesian coordinates, because the basis vectors in cylindrical coordinates change from point to point. But these are important features of curvilinear coordinate systems, so it is important to be fluent with them.
I will give the argument for a general parametrization $\mathbf{u}(s,t)=r\mathbf{\hat{r}}+z\mathbf{\hat{z}}$, where $r$, $\theta$, and $z$ depend on $s$ and $t$, and then apply it to the special case above.
Partial differentiating the position vector with respect to $s$ and $t$ gives our tangent vectors
$$\mathbf{u}_s=r_s\mathbf{\hat{r}}+r\color{red}{\frac{\partial\mathbf{\hat{r}}}{\partial s}}+z_s\mathbf{\hat{z}}+z\color{red}{\frac{\partial\mathbf{\hat{z}}}{\partial s}}$$
and
$$\mathbf{u}_t=r_t\mathbf{\hat{r}}+r\color{blue}{\frac{\partial\mathbf{\hat{r}}}{\partial t}}+z_t\mathbf{\hat{z}}+z\color{blue}{\frac{\partial\mathbf{\hat{z}}}{\partial t}}$$
To calculate the partial derivatives of the basis vectors with respect to $s$ and $t$ (in red and blue), we must use the chain rule. Without loss of generality, I illustrate just for $s$:
$$\color{red}{\frac{\partial\mathbf{\hat{r}}}{\partial s}}=\frac{\partial\mathbf{\hat{r}}}{\partial r}r_s+\frac{\partial\mathbf{\hat{r}}}{\partial\theta}\theta_s+\frac{\partial\mathbf{\hat{r}}}{\partial z}z_s$$
and
$$\color{red}{\frac{\partial\mathbf{\hat{z}}}{\partial s}}=\frac{\partial\mathbf{\hat{z}}}{\partial r}r_s+\frac{\partial\mathbf{\hat{z}}}{\partial\theta}\theta_s+\frac{\partial\mathbf{\hat{z}}}{\partial z}z_s$$
But in cylindrical coordinates we have $\frac{\partial\mathbf{\hat{z}}}{\partial r}=\frac{\partial\mathbf{\hat{z}}}{\partial\theta}=\frac{\partial\mathbf{\hat{z}}}{\partial z}=\mathbf{0}$ and $\frac{\partial\mathbf{\hat{r}}}{\partial r}=\frac{\partial\mathbf{\hat{r}}}{\partial z}=\mathbf{0}$, while $\frac{\partial\mathbf{\hat{r}}}{\partial\theta}=\boldsymbol{\hat{\theta}}$, so
$$\frac{\partial\mathbf{\hat{r}}}{\partial s}=\theta_s\boldsymbol{\hat{\theta}},\,\,\frac{\partial\mathbf{\hat{z}}}{\partial s}=\mathbf{0}$$
Substituting into our expressions for the tangent vectors, we have
$$\boxed{\mathbf{u}_s=r_s\mathbf{\hat{r}}+r\theta_s\boldsymbol{\hat{\theta}}+z_s\mathbf{\hat{z}}}$$
and
$$\boxed{\mathbf{u}_t=r_t\mathbf{\hat{r}}+r\theta_t\boldsymbol{\hat{\theta}}+z_t\mathbf{\hat{z}}}$$
This may look like a lot of work, but for someone well-versed in vector analysis, this is common knowledge and can easily be looked up. 
For the special case $r(s,t)=t^{-1}$, $\theta(s,t)=s$, and $z=t$, we get
$$\mathbf{u}_s=t^{-1}\boldsymbol{\hat{\theta}},\,\,\mathbf{u}_t=-t^{-2}\mathbf{\hat{r}}+\mathbf{\hat{z}}$$
Crossing these gives our surface element
$$\mathbf{u}_s\times\mathbf{u}_t=t^{-1}\boldsymbol{\hat{\theta}}\times\big(-t^{-2}\mathbf{\hat{r}}+\mathbf{\hat{z}}\big)=t^{-1}(\underbrace{\boldsymbol{\hat{\theta}}\times\mathbf{\hat{z}}}_{=\mathbf{\hat{r}}})-t^{-3}(\underbrace{\boldsymbol{\hat{\theta}}\times\mathbf{\hat{r}}}_{=-\mathbf{\hat{z}}})=\boxed{t^{-1}\mathbf{\hat{r}}+t^{-3}\mathbf{\hat{z}}}$$
The magnitude of this vector is $\sqrt{t^{-2}+t^{-6}}$, and integrating this function over the region $(s,t)\in(0,2\pi)\times(1,3)$ gives our surface area, the same result as Kuifje's second approach:
$$\boxed{\int_0^{2\pi}\int_1^3\sqrt{t^{-2}+t^{-6}}\,dt\,ds\approx 7.6030}$$
