Q:

Having trouble with a simple problem. Evaluate this Surface integral.

I feel like this problem is very simple, but no matter what I try, I just can't seem to get the correct answer: $\sqrt6$
Define the Parametrized surface $X:D->R^3$ $$X(s,t)=(2s,s+t,2t)$$
where D is the triangular region with verticies at (0,0), (0,1) and (1,1) 
$$\int\int_{x}ydS$$
The first thing I did was draw the triangle defined by the 3 points. I am unsure if that triangle is on the parameter "graph" (s,t) or the (x,y,z) coordinate graph. The line that defines the triangle is y=x; or s=t.
I'm not sure what I was supposed to do next. I took the normal vector to the triangle. 2 vectors on the triangle$<1,0,0>$ and $<0,1,0>$ cross product: $<0,0,1>$ the magnitude is $\sqrt1=1$
Normally when I do surface integrals, I'm allowed to define x,y and z myself. But in this problem, I'm already given a set of parameters and I have no idea when and where to implement them. So I decided to just take the integral without parametrizing and see what happens. 
$$\int_{x=0}^{x=1}\int_{y=0}^{y=x}y*dS=\int_{x=0}^{x=1}\frac{x^2}{2}=\frac{1^3}{(2*3)}=1/6$$  Which is not the right answer. 
How can I implement the parameters they gave me to solve this question? 

A:

$S$ is not the same surface as $D$.  $S$ is defined in terms of $X$ which is in turn defined in terms of $s$ and $t$.  You need to find out how much $x,y$ and $z$ move as $s$ and $t$ move.
$dS = ||\frac{\partial X}{\partial s} \times \frac{\partial X}{\partial t}||$
$(2,1,0)\times(0,1,2) = ||(2,-2,4)|| = \sqrt{24}$
$dS = \sqrt{24} ds\, dt$
So, if $s$ and $t$ had the freedom to define a 1x1 square, $S$ would be a $\sqrt{24}$ sq unit parallelogram.
$y = (s+t)$ 
$\int_0^1\int_0^t (s+t) \sqrt{24} \,ds\, dt$  
$\int_0^1 ((1/2)s^2+st) \sqrt{24} \, dt |_0^t$
$\int_0^1 ((3/2)t^2) \sqrt{24} \, dt$
$((1/2)t^3) \sqrt{24}|_0^1 $
$(1/2) \sqrt{24}$
$\sqrt{6}$

