$\newcommand{\curl}{\operatorname{curl}}$I have the box that lies "between the planes $x=\pm1$ (with $-1\le y, \ z \le 1$), $\ y=\pm1$ (with $-1\le x, \ z \le 1$), and $z=-1$ with ($-1\le x, \ y \le 1)$ with his normal outward oriented". And i have to find the circulation of $\vec G=curl\ \vec F$ over the surface $S$ of the mentioned region, where $\vec F(x,y,z)=(x, y-2x,x-z)$.
If I have not been confused by the complicated instructions, this is a box centered in the origin, without the top, and there should be multiple ways to solve the problem. I tried two but they do not give me the same result:
- By Stokes Theorem I know: $\iint_S\curl \ \vec F\ dS=\int_C\vec F \cdot d\vec r=\iint_{S_1}\curl \ \vec F\ dS$, where $S_1$ is a surface with boundary $C$, in this case the square:
$$S_1=\{(x,y,1);-1\le x\le 1, -1 \le y\le1\}$$
$$\vec G= \curl\ \vec F=(0,1,-2)$$
$$\int^1_{-1}\int_{-1}^1(0,1,-2)\cdot(0,0,1)\ dxdy=-8$$
- With the Divergence or Gauss theorem, I evaluate the entire cube to later subtract the flux through the top.
$$\operatorname{div}\ G=0$$ $$\int_{-1}^1 \int_{-1}^1 \int_{-1}^1 0 \ dV-\iint_{S_1}\curl \ \vec F\ dS=8$$
I have also calculated the flow on each surface and it resulted in 8.
What i am doing wrong?
Also, if i dont steal too much time: how could i solve this using only the curve $C$? I tried to divide it in four pieces and calculate every line integral but i end with $0$ always.
The orientation on $C$ induced by $S$ is the the one which keeps $S$ on your left as you walk around $C$ with head pointing in the direction of the normal vector. Picture doing this, and at the same time watching yourself from a vantage point above the cube. You'll see you are walking around the square $C$ in a clockwise direction.
This is the opposite of the normal (as in, usual) orientation given to the boundary of a horizontal surface. So your first integral is the one that's off by a factor of $-1$.