Calculate the surface integral of a scalar function $f=z$
$I=\int {f\cdot ds}$
the surface is the part of hyperbolic paraboloid $z=xy$ which is cut by the cylinder $x^2+y^2=4$ and sketch the graph
How am i supposed to start? i am confused with all these shapes. How can i find the limits i have to use?
Note that you are integrating on the surface $z=xy$ where $x^2+y^2\leq4$, in other words on the hyperbolic paraboloid where $(x,y)$ has length $\leq2$. First of all, you need a parameterization of your surface (let's call it $S$), for example: $$ \phi: [0,2\pi]\times (0,2]\longrightarrow S\\ (\theta,\rho)\mapsto \left(\rho\cos(\theta),\rho\sin(\theta),\frac{\rho^2}{2}\sin(2\theta)\right). $$
Now you should know that any surface integral in "regular" condition is defined by $$ \int_Sfd\sigma=\iint_Df(\phi(u,v))\,|d_u\phi\wedge d_v\phi|du\,dv, $$ where $\phi:D\to S$ is a parameterization of the surface $S$, "$\wedge$" is the cross-product in $\mathbb{R}^3$ and "$|\cdot|$" is the usual euclidean norm in $\mathbb{R}^3$ (this is a classical definition and you can find it in most basic analysis books). Hence in this case we have $$ \int_Sfd\sigma=\int_0^{2\pi}d\theta\int_0^2d\rho \frac{\rho^2}{2}\sin(2\theta)\rho\sqrt{\rho^2+1}. $$ Try to calculate $|d_u\phi\wedge d_v\phi|$ by yourself, is not hard, just a little bit time expensive.
here the plot of your surface... a Pringles, as you can see.