Use the Stokes's theorem to evaluate the line integral $\oint_c y dx + x dy + (x^2 + y^2 + z^2) dz$ I am using the Stokes's theorem to evaluate the following line integral, 
$$
\oint_c y dx + x dy + (x^2 + y^2 + z^2) dz
$$
where $C$ is the curve $x^2 + y^2 = 1, z = xy$ directed clockwise as viewed from the point $(0,0,1)$. I have
$$\oint_c y dx + x dy + (x^2 + y^2 + z^2) dz = \int \int_{S} \operatorname{curl}(y, x, x^2 + y^2 + z^2) \cdot \bar{n} dS$$

I am not looking to solve the full question right now, but I'm just wondering how to get $\bar{n}$ and $\operatorname{curl}$.
For $\bar{n}$ the solution is 
$$
\bar{n} = \frac{(y, x, -1)}{\sqrt{x^2 + y^2 + 1}}.
$$ 
How do we get this result?
For the $\operatorname{curl}$ we have
$$
\begin{split}
\operatorname{curl}(y, x, x^2 + y^2 + z^2)
&=
  \begin{bmatrix}
    i & j & k \\
    \frac{d}{dx} & \frac{d}{dy} & \frac{d}{dz} \\
    y & x & x^2 + y^2 + z^2
  \end{bmatrix}\\
\\
&=   \begin{bmatrix}
    i & j & k \\
    0 & 0 & 2z \\
    y & x & x^2 + y^2 + z^2
  \end{bmatrix} \\
\\
& = i(-2zx) + j(2zy) + 0k \equiv (-2zx,  2zy, 0)
\end{split}
$$
while the stated answer of the exercise is $(2y, -2x, 0).$ What did I do wrong? 
 A: For any surface, eg $F(x,y,z)=f(x,y)-z=0$, the vector $\nabla F= \left(\dfrac{\partial F}{\partial x},\dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\right)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},-1\right)$ is perpendicular to this surface, so, a vector normal and unitary is 
$\bar n=\nabla F/|\nabla F|=\bar{n} = \dfrac{(y, x, -1)}{\sqrt{x^2 + y^2 + 1}}$
With $F(x,y,z)=xy-z$
I am not sure how you understood the determinant to calculate the curl but this is how the shorthand works (it's a purely formal device to facilitate calculations: we aren't in fact writing vectors into the determinant).
$
\operatorname{curl}(y, x, x^2 + y^2 + z^2)
=
  \begin{vmatrix}
    i & j & k \\
    \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
    y & x & x^2 + y^2 + z^2
  \end{vmatrix}=$
$=\left(\dfrac{\partial (x^2+y^2+z^2)}{\partial y}-\dfrac{\partial x}{\partial z}\right)i+\left(\dfrac{\partial y}{\partial z}-\dfrac{\partial (x^2+y^2+z^2)}{\partial x}\right)j+\left(\dfrac{\partial x}{\partial x}-\dfrac{\partial y}{\partial y}\right)j=2yi-2xj$
As expected.
