Problem in calculating the surface integral Problem
Let S denote the plane surface whose boundary is the triangle with vertices at (1,0,0),(0,1,0), and (0,0,1) , and let F(x,y,z) = xi+yj+zk. Let n denote the unit normal to S having a nonnegative z-component . Evaluate the surface integral using the vector representation 
r(u,v)= (u+v)i+(u-v)j+(1-2u)k.
Attempt
$\frac{\partial r}{\partial u} = (1,1,-2)$
$\frac{\partial r}{\partial v}=(1,-1,0)$
$\frac{\partial r}{\partial u}×\frac{\partial r}{\partial v}=-2(1,1,1)$
$\iint_T F.n dS= -2\iint_T du dv $
I am not able to calculate the range of u and v
i think u will vary from 0 to 1/2 . 
Attempt 2
$0 \leq y \leq x$ which implies $0 \leq u-v \leq u+v$ which in turn implies $v \leq u$ and $0 \leq v$.So 
-2$\int_{0}^{1} \int_{0}^{u} du dv $ = 1/4( change of sign takes place)
 A: For $\textbf{r} = (u+v) \textbf{i} + (u-v)\textbf{j} + (1-2u)\textbf{k}$ to fill the triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ the following restrictions must be placed on the components of $\textbf{r}$
\begin{equation} \tag{1}
0 \leq u+v \leq 1
\end{equation}
\begin{align} \tag{2}
0 \leq u-v \leq 1
\end{align}
\begin{align} \tag{3}
0 \leq 1-2u \leq 1
\end{align}
Subtracting $(3)$ from $(1)$
$ \implies 0\leq u+v - (1-2u) \leq 0 \implies 3u-1-v =0 $
$$ v=3u-1 \tag{4}$$ 
Substitution of $(4)$ into either $(1)$ or $(2)$ shows that $$0 \leq u \leq \frac{1}{2} \tag{5}$$ 
Notice that $(2)$ implies $v \leq u$. Therefore, for all $u$ that satisfy $(5)$, we have
$$ 3u-1\leq v \leq u  \tag{6} $$ 
Hence $(5)$ and $(6)$ are the bounds for $u$ and $v$

Note: You want $\textbf{n}$ to be in the opposite direction of $-2(1,1,1)$ since this has a negative $z$-component
Edit
The range for $v$ I originally provided was wrong, but the above ranges should be correct. This can be shown by using the fact that surface integrals are independent under a change of parametric representation. Consider a second parameterization 
$$\textbf{r}(x,y) = x \textbf{i} + y \textbf{i} + (1-x-y) \textbf{i} $$
this gives a normal 
$\textbf{n} = \frac{\partial \textbf{r} }{\partial x} \times \frac{\partial \textbf{r} }{\partial y}  = \textbf{i} + \textbf{j} + \textbf{k} $. 
Then 
$$ \iint_S \textbf{F} \cdot \textbf{n} \, dS = \iint_T (x,y,1-x-y) \cdot (1,1,1) dx\, dy = \iint_T dx \, dy = \int_{0}^{1-x} \int_{0}^{1} dx\, dy = \frac{1}{2} $$
Now the parameterization 
$\textbf{r} = (u+v) \textbf{i} + (u-v)\textbf{j} + (1-2u)\textbf{k}$ together with $\textbf{n} = 2\textbf{i} + 2\textbf{j} + 2\textbf{k}$ and the bounds for $u$ and $v$ must give the same answer
$$\iint_S \textbf{F} \cdot \textbf{n} \, dS  = \iint_T (u+v,u-v,1-2u) \cdot (2,2,2) du\, dv = 2 \iint_T du\, dv = 2 \int_{0}^{\frac{1}{2}} \int_{3u-1}^{u} dv\, du = \frac{1}{2}  $$
A: Let $\mathbf{a}=\langle 1,1,-2\rangle$ and $\mathbf{b}=\langle 1,-1,0 \rangle.$  Then
$$\mathbf{r}(u,v) = u\mathbf{a}+v\mathbf{b}+\mathbf{k}. $$
If you solve $\mathbf{r}(u,v) = (1,0,0)$ you get $u=1/2$ and $v=1/2$.
If you solve $\mathbf{r}(u,v) = (0,1,0)$ you get $u=1/2$ and $v=-1/2$.
If you solve $\mathbf{r}(u,v) = (0,0,1)$ you get $u=0$ and $v=0$.
So the ranges are $0\leq u \leq 1/2$ and $-1/2\leq v\leq 1/2.$
