verify stoke's theorem for $F=xz \vec i - y \vec j +x^2y \vec k $ where S is surface of region bounded by $x =0, y=0, 2x+y+2z=8$ which is not included in xz plane I am confused over magnitude(moving along curve so that surface is in left hand side)
My attempt
$\iint_{S_1+S_2} Curl \vec F \cdot \vec n = \iiint div Curl \vec F dV = 0 \implies \iint_{S_1} =- \iint_{S_2}$
$S_2 $ --> i am taking in direction so that normal $\vec n = -\vec k$
In this case i am getting $\iint_{S_2} = \frac{-32}{3} \implies \iint_{S_1} = - (\frac{-32}{3})=(\frac{32}{3}) $
Now calculating single integral $\int_C \vec F.d\vec r$
Now how should i take direction? I took in following way but i am getting sign change. can you pls have a look?
Single integrals along $\vec {OA}, \vec {BO}$ i am getting zero.
For single integral along $\vec {AB}$:
$y=0, x+z = 4 \implies dx = -dz \implies \int_C \vec F \cdot d \vec r = \int_C F.(-dz \vec i+0 \vec j+dz \vec k)$ where $F = xz \vec i - y \vec j +x^2y \vec k$
So, $\int_C \vec F.d\vec r = \int^4_{z=0} (-xz+x^2y)dz , y =0 \implies $ integral = $\int^4_{z=0} (-(4-z)z)dz = \frac{-32}{3} $
i am gettting negative sign. how should i identify direction of C quickly without confusion
Using the right-hand rule convention, the contour $C$ that bounds the open surface of interest is comprised of the line segment from (i) $(4,0,0)$ to $(0,0,0)$, (ii) $(0,0,0)$ to $(0,0,4)$, and (iii) the line segment from $(0,0,4)$ to $(4,0,0)$$
The line integral of $\vec F$ on $C$ can be written as
$$\begin{align} \oint_C \vec F\cdot d\vec \ell&=\oint_C (\hat xxz-\hat y y+\hat z x^2y)\cdot d\vec\ell\\\\ &=\color{blue}{\int_0^4\left.\left(\hat xxz-\hat y y+\hat z x^2y\right)\right|_{y=z=0}\cdot (-\hat x)\,dx}\\\\ &+\color{red}{\int_0^4 \left.\left(\hat xxz-\hat y y+\hat z x^2y\right)\right|_{x=y=0}\cdot (\hat z)\,dz}\\\\ &+\int_0^4 \left.\left(\hat xxz-\hat y y+\hat z x^2y\right)\right|_{y=0, z=4-x}\cdot \left(\hat x-\hat z\right)\,dx\\\\ &=\color{blue}{0}+\color{red}{0}+\int_0^4 (4x-x^2)\,dx\\\\ &=\frac{32}{3} \end{align}$$