Using Stokes theorem to integrate $\vec{F}=5y \vec{\imath} −5x \vec{\jmath} +4(y−x) \vec{k}$ over a circle Find $\oint_C \vec{F} \cdot d \vec{r}$ where $C$ is a circle of radius $2$ in the plane $x+y+z=3$, centered at $(2,4,−3)$ and oriented clockwise when viewed from the origin, if $\vec{F}=5y \vec{\imath} −5x \vec{\jmath} +4(y−x) \vec{k}$
Relevant equations:

Stokes theorem: $$\int_S \operatorname{curl}{F} \cdot \mathbf{n} \, dS = \oint_{\partial S} F \cdot d\mathbf{r}$$

My attempt:

*

*For the curl I get $(4,4,-10)$.


*For $d\vec{S}$ I get $(1,1,1)$ from $z = 3-x-y$


*Dotted together its $-2$.


*So: $-2 \iint_S dA$.


*Area of circle is $4\pi$.
My answer would be $-8\pi$ but the online homework system says it's not correct. Please help!
 A: $x+y+z=3 \implies z = -x-y+3$
$\begin{align} \\ d\vec{S} =\hat{n}dS &= \langle -f_x,-f_y,1 \rangle dxdy\\~\\&= \langle 1,1,1 \rangle dA \end{align}$
you get : $-2\iint_S dA$
So far your work is correct ! there is a mistake in your next step :
$-2\iint_{\color{green}{\mathbb S}} dA \color{red}{\ne} 4\pi $  
why ? because you're assuming that the given circle itself is a shadow in the $xy$ plane, which is wrong. 
since $dA=dxdy$, for the range of $x$ and $y$, you need to take shoadow of $\color{green}{\mathbb S} $  in $xy$ plane :

If $\alpha$ is the angle between $xy$ plane and $\color{green}{\mathbb S}$, then clearly the area scales by a factor of $\cos \alpha $ :  
$\text{Area of ellipse in xy plane} =  (\text{Area of circle in } \color{green}{\mathbb S})\times \cos \alpha =  4\pi \cos \alpha = 4\pi \dfrac{ \langle 1,1,1 \rangle .  \langle 0,0,1 \rangle}{|| \langle 1,1,1 \rangle||} = \dfrac{4\pi}{\sqrt{3}}$
A: well, I think that it is rather a question of getting C in its parametric form.
Your plane is $\Pi :x+y+z=3$.
I will try here to find the orthogonal vectors to the $\Pi$:
let's say that $\underline e_1=(a_1,a_2,0)$ and $\underline{e_2}=(-a_1,a_2,0)$.
Dot product of those vectors with the normal vector of $\Pi$ gives you these 2 eqautions: 
$a_1+a_2+a_3=0, -a_1+a_2=0$.
So you can say that $\underline{e_1}= \left(\begin{array}{c}
  1  \\
  1  \\
  -2 
 \end{array} \right)$
and $\underline{e_2}=\left(\begin{array}{c}
  -1  \\
  1  \\
  0 
 \end{array}\right)$
So that your curve parametric form is:
$$C = \sqrt 3 +4 \cos t\cdot \left(\begin{array}{c}
  1  \\
  1  \\
  -2 
 \end{array}\right) -4\sin t\cdot \left(\begin{array}{c}
  -1  \\
  1  \\
  0 
 \end{array} \right)$$ where $t\in[0,2\pi]$
And from here it becomes much easy.
I hope I helped!
