Question: Using Stokes's Theorem for $A=(y+z)\hat i-(xz) \hat j+(y^2)\hat k$ where $S$ is the surface of the region in the first octant bounded by $2x+z=6$ and $y=2$ which is not included in the (a) $xy$ plane, (b) the plane $2x+z=6$.
I can't seem to get the answer in the textbook.
First I evaluated $\nabla$ x $A$
$$ \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ y+z & -xz & y^2 \\ \end{vmatrix} $$
Which gave me: $\nabla$ x $A$= $(2y+x) \hat i+ \hat j + (1-z) \hat k$
Now in my mind the surface I'm supposed to integrate over looks something like this, which can be divided into 5 surfaces:
I'm quite new to Stokes' Theorem, so do correct me if I have any conceptual errors.
For the 2 triangle surfaces, I started calculating for one of the surfaces first.
$$\hat n= \begin{pmatrix} 0\\ 1\\ 0 \\ \end{pmatrix}$$
$$\int \int_R \begin{pmatrix} 0\\ 1\\ 0 \\ \end{pmatrix}. \begin{pmatrix} 2y+x\\ 1\\ 1-z \\ \end{pmatrix}dx dz=\int \int_R 1 dx dz = 9 $$
Since the surface integral of the curl of A is independent of $x,y,z$, both triangle surfaces give the same value. (ie. 9)
For the surface lying on the $yz$ plane
$$\hat n= \begin{pmatrix} 1\\ 0\\ 0 \\ \end{pmatrix}$$
$$\int \int_R \begin{pmatrix} 1\\ 0\\ 0 \\ \end{pmatrix}. \begin{pmatrix} 2y+x\\ 1\\ 1-z \\ \end{pmatrix}dy dz=\int_0 ^6 \int_0 ^2 (2y+x) dy dz = 24 (x=0) $$
For the surface on the $xy$ plane:
$$\hat n= \begin{pmatrix} 0\\ 0\\ 1 \\ \end{pmatrix}$$
$$\int \int_R \begin{pmatrix} 0\\ 0\\ 1 \\ \end{pmatrix}. \begin{pmatrix} 2y+x\\ 1\\ 1-z \\ \end{pmatrix}dx dy=\int \int_R (1-z) dx dy = 6 (z=0) $$
For the plane surface: Let $f=2x+z$
$\nabla f= \begin{pmatrix} 2\\ 0\\ 1 \\ \end{pmatrix}$
For this surface, I didn't bother finding $\hat n dS$ as from my understanding, $\hat n dS=\nabla f dxdy$
$$\int \int_R \begin{pmatrix} 2\\ 0\\ 1 \\ \end{pmatrix}. \begin{pmatrix} 2y+x\\ 1\\ 1-z \\ \end{pmatrix}dx dy=\int \int_R (4y+2x-z+1) dx dy$$
Subbing in $-z=2x-6$ $$\int_0 ^2 \int_0 ^3 (4y+4x-5) dx dy=-42$$
Adding all of them up, $9.2+24+6-42=6$
a) Subtracting $xy$ plane $6-6=0$
b) Subtracting the plane surface $2x+z=6, 6+42=48$.
The answer in my textbook is $-6 and -18$ respectively. I'm not really concerned about the answer here, but I think I have several misconceptions of Stokes' theorem even after studying when its applicability. I'd be grateful if someone can point out any misconceptions I have here (hence the typing of this long working). Thanks.