Compute the surface integral of the function $x+z$ over the first octant of the plane $x+y+z=1$
My attempt:
$$ \vec{n} \cdot \vec{k} = \sqrt{1 + f_x^2 + f_y^2} = \sqrt{3}$$
Then,
$$ \int_{\partial S} (x+z) dS= \int_{R} (x+z) \frac{dx dy}{\vec{n} \cdot \hat{k}} = \frac{1}{\sqrt{3}} \int_{R} (x+z)dx dy$$
Now, I plug in equation of plane for $z$ and rearrange the order ( I've ommited the root three outside)
$$ \int_{R} (1-y) dy dx = \int_?^? \int_?^? (1-y) dy dx$$
Now, for the inner integral I am running a double integral in the $x-y$ plane so I put $z=0$ in the equation of plane which gives $x+y=1$ or $ y=1-x$ , and the outer integral is from $ 0 \to 1$,
$$ \int_0^1 \int_{0}^{1-x} (1-y) dy dx = \int_0^1 [x^2 -1 ] dx = -\frac23$$
But the true answer of the original integral $ \frac{1}{\sqrt{3}}$ / the new integral should evaluate to one. What have I done wrong?
$z = f(x,y) = 1 - x - y$ is the parametrization of your surface.
$f_x = -1, f_y = - 1, \sqrt{1 + f_x^2 + f_y^2} = \sqrt3$
$\int_{S} (x+z) dS = \int_{A} (x+z) \sqrt{1 + f_x^2 + f_y^2} \, dA = \sqrt3 \displaystyle \int_0^1\int_{0}^{1-x} (1-y) \, dy \, dx = \frac{1}{\sqrt3}$
Also note $(x+z)$ is a scalar function and not a vector field so it is not necessary to find $\, \hat{n}\,$ for your integral.
EDIT:
Specifically to your question on the mistake, when you have a vector field your integral translates to
$\displaystyle \iint_R (\vec{F} \cdot \hat{n}) \frac{dx \, dy}{\hat{n}.\vec{k}}$
Here your surface is $x + y + z = 1$ which has a normal vector $(1, 1, 1)$.
Unit normal $\hat{n} = (\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3})$
$\hat{n} \cdot \vec{k} = \frac{1}{\sqrt3}$
So your integral should be
$\sqrt3 \displaystyle \iint_R (x + z) \, dx \, dy$