Calculate $\int_{T} \vec{A} \cdot \hat{n} \ dS$ for given $\vec{A}$ and $T$

Question

Calculate $$\int_{T} \vec{A} \cdot \hat{n} \ dS$$ where $\vec{A}=4 \ \hat{i}$, and $T$ is the intersection between $x+y+2z=1$ and $\Gamma=\{(x,y,z): x\ge 0, y\ge0,z\ge 0\}$

My work. Normal to the plane $x+y+2z=1$ :

$$\hat{n}=\frac{1}{\sqrt{1+1+4}} \ (\hat{i}+\hat{j}+2 \hat{k})=\frac{1}{\sqrt{6}} \ (\hat{i}+\hat{j}+2 \hat{k})$$

$$\vec{A} \cdot \hat{n}=\frac{4}{\sqrt{6}}$$

Integration bounds, considering $z=0$:

$$0 \le y \le 1-x$$ $$0 \le x \le 1$$

$$\int_0^1 \int_0^{1-x} \Big(\frac{4}{\sqrt{6}} \ dy\Big) \ dx=\int_0^1 \Big( \frac{4}{\sqrt{6}} (1-x)\Big) dx=\Big[ \frac{4}{\sqrt{6}} \Big(x-\frac{1}{2} x^2\Big) \Big]_0^1=\frac{2}{\sqrt{6}}$$

Is it correct?

Thanks!

Answer 1

Note that $z=1-(x+y)/2$, and (see link), $$dS=\sqrt{1+z_x^2+z_y^2}\,dxdy=\frac{\sqrt{6}}{2}\,dx dy.$$ Hence, it should be $$\int_{T} \vec{A} \cdot \hat{n} \ dS=\int_{x=0}^1 \int_{y=0}^{1-x} \frac{4}{\sqrt{6}} \cdot \frac{\sqrt{6}}{2}\ dy \ dx=1.$$ The same result can be obtained as the product of $4$ multiplied by the area of the projection of triangle $T$ in the $yz$ plane (which is orthogonal to $\vec{A}$) whose area is $\frac{1\cdot 1/2}{2}=1/4$.