Given the following question and temptative answers:
I am not sure what the answer is for the $V(x,y)$ box.
Normally I would have converted the problem to spherical coordinates and solved it that way.
My guess is that when one is converting $\iint_M(\nabla\times F)\cdot dS$ to $\iint_DV(x,y)dydx$, there is a term that appears from the change in the differential that needs to be multiplied to the curl in order to obtain the correct term, but I am not sure what it should be.
We have that $\nabla \times \mathbf{F}=\langle2yzx^3,-3y^2zx^2,-2\rangle$ and we have $z=-\sqrt{\frac{1-x^2-y^2}{6}}$ since $z\leq0$. Then we have that $$\begin{align} \iint_M(\nabla\times F)\cdot dS & = \iint_M \langle2yx^3z,-3y^2x^2z,-2\rangle \cdot\langle-z_{x},-z_{y},1\rangle\,dA\\ & = \iint_M\langle-2yx^3\sqrt{\frac{1-x^2-y^2}{6}}, 3y^2x^2\sqrt{\frac{1-x^2-y^2}{6}}, -2\rangle \cdot\langle-z_{x},-z_{y},1\rangle\,dA\\ & = \frac{1}{6}\iint_M -2yx^4+3y^3x^2-12\,dA. \end{align}$$ Then $$V(x,y)= \frac{-2yx^4+3y^3x^2-12}{6}.$$ Now we will convert to polar coordinates. Since $x^2+y^2=1$, we have that $r$ will range from $0$ to $1$ and since our figure is an ellipsoid we have that $\theta$ will range from $0$ to $2\pi$. Then our equation becomes $$\begin{align} \iint_DV(x,y)\,dy\,dx & = \frac{1}{6}\int_{0}^{1}\int_0^{2\pi}(-2r^5\cos^4(\theta)\sin(\theta)+3r^5\cos^2(\theta)sin^3(\theta)-12)r\,d\theta\,dr \\ & = \frac{1}{6}\int_{0}^{1}(0+0-12(2\pi))r\,dr \\ & = -2\pi\,r^2 \big|_{0}^{1} \\ & = -2\pi. \end{align}$$