Question Evaluate$\int_{S}\vec{F.d\vec{S}}$ where$\vec{F}$ = y$\hat{i}$ +$2x$$\hat{j}$-z$\hat{k}$ and S is the surface of the plane $2x+y=4$ in the first octant cut off by the plane $z=4$
My Approach $\hat{n}=\frac{\nabla s}{\mid\nabla s\mid}=\frac{1}{\sqrt{5}}\left(2\hat{i}+\hat{j}\right)$ and $dS =\frac{dxdy}{\mid\hat{n.}\hat{j\mid}}=\sqrt{5} dxdy$
$\vec{F}\cdot\hat{n}=\frac{2}{\sqrt{5}}\left(x+y\right)$
$\vec{F}\cdot\hat{ndS}=\frac{2}{\sqrt{5}}\left(x+y\right)$$\sqrt{5} dxdy=2\left(x+y\right)dxdy$
Now i can't decide the region of integral$\int_{S}2\left(x+y\right)dxdy$
Book's Approach my book uses this formula for $dS$
$$dS =\frac{dxdz}{\mid\hat{n.}\hat{j\mid}}=\sqrt{5} dxdz$$
Is my formula wrong? Or are both formulas correct?
If both formulas are correct then why my method is not giving the right answer?
The standard way is to parametrise the surface. The surface can be parametrised in the following way: $\mathbf{r}(u,v)=\langle u, 4-2u, v \rangle$, where $u\in (0,2)$ and $v\in(0,4)$. Then, take partial derivatives $\mathbf{r}_u=\langle 1,-2,0\rangle$ and $\mathbf{r}_v=\langle 0, 0, 1\rangle$; consequently, $\mathbf{r}_u\times\mathbf{r}_v=\langle -2, -1, 0\rangle$. This will give you one choice of normal vector $\mathbf{n}$, but whether this is the correct orientation, it is dependent on the question(the other being $-\mathbf{n}$). That is why I ask for the correct orientation.
We compute the surface integral in the following way: $\int_S\mathbf{F} \mathrm{d}S=\iint_D\mathbf{F}\cdot(\mathbf{r}_u\times\mathbf{r}_v)\mathrm{d}A$, where $D$ is comprised of all the $(u,v)$ in domain. Here, the equation gives $\iint_D\langle 4-2u,2u,-v\rangle\cdot\langle -2, -1, 0\rangle\mathrm{d}A=\int_0^4\int_0^2(-8+2u)\mathrm{d}u\mathrm{d}v=-48$