How to evaluate double integrals of a surface over a specific region? I found this exercise while exercising for the exam:
Let $T$ $\subset$ $R^2$ be the triangle with these vertices  $(0,0), (2,0), (0,1)$ and let $\Omega$ be the surface defined like this:
$\Omega$ = {$(x,y,z) \in R^3 : z^2 - x^2 - y^2 = 0, z > 0, (x,y) \in T$}
Evaluate $\iint_{\Omega}x^2 y dS $
I'm having a hard time solving it because it confounds me... I can't seem to "visualize" the situation. What's exactly the role of the triangular region, where am I going to have to use that region when solving the integral?
Could you help me visualize the problem in some way?
 A: Intuition : We can think this surface integral as surface $ \Omega $ above the triangular region. You can see 3D graph on GeoGebra. 
We know that $ z^2 = x^2 + y^2 $ is cone and the part of cone above the triangular region $ (0,0) \, (2,0) \, (0,1) $ So from here we see that
$ 0 \le x \le2 $ And $0 \le y \le 1$.
so we can parametrize the curve 
$$\vec{r} = (x)\hat{i} + (y)\hat{j} + (\sqrt{x^2+y^2})\hat{k}$$

So from the definition of surface integral,

$$ \int \int_S \vec{F}.\mathrm{d}S = \int \int_S \vec{F}\hspace{2mm}\lvert\vec{r}_x \times \vec{r_y} \rvert dA ....(1)$$
So now ,
$$ \vec{r}_x \times \vec{r_y} = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & \frac{x}{\sqrt{x^2+y^2}} \\
0 & 1 & \frac{y}{\sqrt{x^2+y^2}} \\
\end{vmatrix}
 $$
$$ \vec{r}_x \times \vec{r_y} = \frac{-x}{\sqrt{x^2+y^2}}\hat{i} + \frac{-y}{\sqrt{x^2+y^2}}\hat{j} + (1)\hat{k} $$
$$ \lvert \vec{r_x} \times \vec{r_y} \rvert = \sqrt{2} $$
So putting this in our $(1)$ and in the place of $\vec{F}$ we are given $x^2y$ so we get,
$$ 
\begin{align}
S.I. &= \int_0^2 \int_0^1 \sqrt{2}\,x^2 \, y \, \mathrm{d}y \, \mathrm{d}x \\
&= \sqrt{2} \int_0^2 x^2 \, \left(\frac{1}{2}\right) \, \mathrm{d}x \\
&= \frac{1}{\sqrt{2}} \int_0^2 x^2\, \mathrm{d}x \\
&= \frac{8}{3\sqrt{2}}
\end{align}
 $$
