Problem on divergence, rotation, flux In $(x, y, z)$ space is considered the vector field 
$V(x,y,z)=(y^2 z, yx^2, ye^y)$ solid spatial region $Ω$ is given by the parameterization:
$\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =r(u,v,w)=\left[ \begin{matrix} wu \\ wv \\ 2-2w \end{matrix} \right]$
Where $u\in \left[ -2,2 \right] ,\quad v\in \left[ -2,2 \right] ,\quad w\in \left[ 0,1 \right]$
a) Determine the divergence and the rotation of V.
b) Determine$\int _{ \partial \Omega  }^{  }{ V\cdot n }\ d\mu$, $n$ is an outwardly directed on unit-normal vector field $\partial \Omega$.
For any number represents the intersection of points $\Omega$ and the plan has equation $y = c$, a surface termed $F_c$. The surface likely equipped with a unit normal vector field $n$ pointing away from the $z-axis$. The basket rim $\partial F_c$ likely equipped with a unit tangent vector field $e$ pointing counter-clockwise when viewed from the $y-axis$ positive end.
c) Determine the value for $c$ which determines,
$\int _{ F_{ c } }^{  }{ V\cdot n } d\mu =\int _{ \partial F_{ c } }^{  }{ V\cdot e } d\mu $
Approache/Solution
So I have actually solved the first two parts (a) and (b). For (a) I have the found the following results,
$Div(V)=x^2$
$Rot(V)=(y^2e^y, y^2,y2x-2yz)$
For b) I found the answer to be $\frac{128}{15}$. 
But I cannot seem to solve part (c) and I am totally lost about what the question is asking!!? Any help would be great and highly appreciated. 
Thank You
 A: It does seem a bit on the ugly side in that the region seems to be now down to one square pyramid with vertex at $(0,0,2)$. I think the $x$-component of $\vec\nabla\times\vec V$ should be $(y+1)e^y$: check it. For part b), I get your answer now that the domain of $w$ has been corrected. For the Jacobian, I get
$$J=\left|\det\begin{bmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\
\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial u}\\
\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial u}\end{bmatrix}\right|
=\left|\det\begin{bmatrix}w&0&u\\0&w&v\\0&0&-2\end{bmatrix}\right|=\left|-2w^2\right|=2w^2$$
Then
$$\begin{align}\int\int_{\partial\Omega}\vec V\cdot d^2\vec A&=\int\int\int_{\Omega}\vec\nabla\cdot\vec Vd^3\text{volume}\\
&=\int_0^1\int_{-2}^2\int_{-2}^2w^2u^2\cdot2w^2\,du\,dv\,dw\\
&=2\cdot\frac232^3\cdot2\cdot2\cdot\frac151^5=\frac{128}{15}\end{align}$$
For part c) I think you are supposed to use Stokes' theorem to get
$$\oint_{\partial F_c}\vec V\cdot d\vec r=\int\int_{F_c}\vec\nabla\times\vec V\cdot d^2\vec A$$
$w=0$ is above the pyramid, $c=y=vw\le2w$ so $w=\frac c2$ is the upper surface of the pyramid, and $w=1$ is the base of the pyramid. We need $\vec r$ along the surface:
$$\vec r=\langle x,y,x\rangle=\langle uw,vu,2-2w\rangle=\langle uw,c,2-2w\rangle$$
$$d\vec r=\langle w,0,0\rangle\,du+\langle u,0,-2\rangle\,dw$$
Then the vector areal element is
$$\begin{align}d^2\vec A&=\langle w,0,0\rangle\,du\times\langle u,0,-2\rangle\,dw\\
&=\pm\langle0,2w,0\rangle\,du\,dw=\langle0,2w,0\rangle\,du\,dw\end{align}$$
To be pointing away from the $z$-axis, assuming $c>0$. So there are $2$ integrals to evaluate:
$$\int_{\frac c2}^1\int_{-2}^2\langle c^2(2-2w),cu^2w^2,ce^c\rangle\cdot\langle0,2w,0\rangle\,du\,dw=2c\cdot\frac14\left(1-\frac{c^4}{16}\right)\cdot\frac232^3$$
$$\int_{\frac c2}^1\int_{-2}^2\langle (c+1)e^c,c^2,2cuw-2c(2-2w)\rangle\cdot\langle0,2w,0\rangle\,du\,dw=2c^2\cdot\frac12\left(1-\frac{c^2}{4}\right)\cdot4$$
The condition we are trying to satisfy is
$$\frac16c(16-c^4)=c^2(4-c^2)$$
So either $c=0$, $c^2-4=0$, or $\frac16(4+c^2)=c$. The last has the solutions $c=3\pm\sqrt5$ of which only $c=3-\sqrt5$ is valid, and $c=\pm2$ is degenerate, so it looks like the only thing left is $c=0$, which seems rather odd because how can you point away from the $z$-axis when $y=0$? When $c=0$ neither vector field has a nonzero $y$-component, so I guess that's a reasonable solution. I suppose it can be seen that $c\ge0$ because otherwise the two vector fields have $y$-components that disagree in sign.
