Evaluate the integral $ \ \int_C F \cdot dr \ $ 
Evaluate the integral $ \ \int_C F \cdot dr \ $. 
Answer:
$ \ F=\left\langle -x^2y,x,0 \right\rangle \ $
$ curl \ F= \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial {x}} & \frac{\partial}{\partial {y}} & \frac{\partial}{\partial {z}} \\ -x^2y & x &0 \end{vmatrix} \ =(1+x^2) \hat k $ 
Now equation the line $ \ BC \ $ is 
$ \frac{y-3}{x-6}=\frac{3-3}{6-0} \\ i.e. \ y=3 $ 
I think we have to find the equation of the plane $ \ ABCD \ $
But I can not finish the problem.
 A: The equation of the plane that goes through the curve is going to be something of the form $z = ay$ since we can see that it is not dependent on $x$. So, we can just look at the slice through the $yz$-plane, and determine that the equation of that line, and hence the plane itself, is $z = -\frac{4}{3}y + 4$. Without fractions, that's $4y+3z=12$.
Since we are told to use Stokes', we need to evaluate the corresponding surface integral
$$\oint_{C} \textbf{F} \cdot d\textbf{r} = \iint_{S} \text{curl}(\textbf{F}) \cdot d\textbf{S},$$
where $S$ is the section of the plane bounded by $ABCD$.
To evaluate surface integrals, we need to create a parameterization of the surface. Naturally, we can do $G(x, y) = \langle x, y, -\frac{4}{3}y + 4 \rangle = \langle X, Y, Z \rangle$ with it's domain $D = [0,6]\times[0,3],$ which is a fancy way of saying the rectangle directly "below" the surface in the $xy$-plane. If $N = G_x \times G_y$, then $N = \langle 0, \frac{4}{3}, 1 \rangle$.
So it follows that
\begin{align*}
\oint_{C} \textbf{F} \cdot d\textbf{r} & = \iint_{S} \text{curl}(\textbf{F}) \cdot d\textbf{S}\\
& = \iint_{D} \left(\text{curl}(\textbf{F})\bigg\vert_{G(x, y)}\right) \cdot N \;\,dA \\
& = \int_{0}^{6}\!\!\!\int_{0}^{3} \langle 0, 0, 1+X^2 \rangle \cdot \langle 0, \frac{4}{3}, 1 \rangle \;\,dy\,dx \\
& = \int_{0}^{6}\!\!\!\int_{0}^{3} \langle 0, 0, 1+x^2 \rangle \cdot \langle 0, \frac{4}{3}, 1 \rangle \;\,dy\,dx \\
& = \int_{0}^{6}\!\!\!\int_{0}^{3} 1 + x^2 \;\,dy\,dx \\
& = 3 \int_{0}^{6} 1 + x^2 \;\,dx \\
& = 3 \left[ \left(6 + \frac{6^3}{3}\right) - \left(0 + \frac{0^3}{3}\right) \right] \\
& = 234
\end{align*}
For the evaluation over the open-topped box, you'll need to parameterize each side, take the surface integral of curl($\textbf{F}$) over each side and add up all those surface integrals. (Hint: your final answer should still be 234.)
