I'm stuck on the following question:
Find $\iint\limits_S {ydS}$ where $S$ is the part of the plane $z=1+y$ that lies inside the cone $z = \sqrt {2({x^2} + {y^2})} $
I tried to parametrize $x,y$ as following:
$$\left\{ \begin{gathered} x = u \hfill \\ y = v \hfill \\ z = 1 + v \hfill \\ \end{gathered} \right. \Rightarrow dS = \sqrt 2 dudv$$
So, the double integral turns out to be:
$$\iint\limits_D {v\sqrt 2 dudv}$$ It seems to be right, but there does not seem to be any way to set up the proper boundaries for the domain $D$. I experimented with a polar substitution, but unfortunately, since the domain is an infinite cone, I am unable to quantify it.
The answer should be $2\pi$. Does anyone know a good strategy to solve problems with these kinds of domains?