Compute the integral:
$$\iint_\sigma (x+y+z)\,\mathrm dS $$
on the cube $0\le x\le1, 0\le y\le1, 0\le z \le 1$.
I am really confused how should I compute it as I used to just take the projection of $\ z $ on the $\ xy $ plane but here the projection is just a square from $\ 0 $ to $\ 1 $. Trying to compute $\ \int_0^1 \int_0^1 (x + y) dy dx $ doesn't lead me to the right answer.
$\displaystyle\int_0^1 \int_0^1 (x + y + 0)\ dy\ dx =\int_0^1 x + \frac12\ dx =1$ >> the value on the plane $z=0$.
$\displaystyle\int_0^1 \int_0^1 (x + y + 1)\ dy\ dx =1 + \int_0^1 \int_0^1 (x + y)\ dy\ dx =2$ >> the value on the plane $z=1$.
The correct answer is $3\cdot(1 + 2) = 9$. The $3$ means that there are three dimensions. (same value by symmetry)