Finding the Radius of a Circle in 3D Using Stokes Theorem Let
$$\vec{F} (x, y, z) = xy\hat{i} +(4x - yz)\hat{j} + (xy - z^{1/2}) \hat{k},$$
and let $C$ be a circle of radius $R$ lying in the plane $x + y + z = 5$.
If
$$\int_C \vec{F} \cdot d\vec{r} = \pi \sqrt{3},$$
where $C$ is oriented in the counterclockwise direction when viewed from above the plane,
what is the value of $r$?
This question has some context and the version of Stokes Theorem to use.
 A: We know that $$\oint_C \vec{F} \cdot d\vec{r}=\pi\sqrt{3}$$
By the Stokes theorem, we can write that 
$$
\oint_C \vec{F} \cdot d\vec{r}=\iint_S \nabla \times \vec{F}\cdot d\vec{S} = \iint_S \nabla \times \vec{F}\cdot \vec{n},
\; dS$$
where $S$ is the surface inside the circle $C$ (the disc with radius $r$), lying in the plane $x+y+z=5$, and $\vec{n}$ is a unitary normal vector to this surface adequately orientated, i.e. $\vec{n}=\frac{1}{\sqrt{3}}(1,1,1)$. $\nabla \times \vec{F}$ represents the curl of $\vec{F}$, and is equal to $(x+y,-y,4-x)$.
It follows that 
$$
\iint_S \nabla \times \vec{F}\cdot \vec{n},
\; dS = \frac{1}{\sqrt{3}} \iint_S x+y-y+4-x \; dS = \frac{4}{\sqrt{3}} \;Area(S)=\frac{4}{\sqrt{3}} \pi r^2.
$$
Solving for $r$ yields:
$$
r=\frac{\sqrt{3}}{2}.
$$
A: According to your last comment, I think the main challenge of this question is to write the parametric equation of the circle in 3D. To do this, we first note that the parametric equation of the plane can be written as
$$ {\bf{r}}(x,y) = x {\bf{i}} + y {\bf{j}} + (5-x-y) {\bf{k}} \tag{1}$$
where $x$ and $y$ are the parameters. Next, suppose that the center of the circle is at
$$ {\bf{c}} = 5 {\bf{k}} \tag{2}$$
and hence the vector ${\bf{r-c}}$ always lie in our plane. In order to find a vector such that it lies on a circle with radius $R$ in the plane, we require that
$$\begin{array}{}
{\bf{r-c}} = x {\bf{i}} + y {\bf{j}} - (x+y) {\bf{k}}\\
({\bf{r-c}}) \cdot ({\bf{r-c}}) = R^2 \\
x^2+y^2+(x+y)^2=R^2 \\
2x^2+2y^2+2xy=R^2 \\
x^2+y^2+xy=\frac{R^2}{2}
\end{array} \tag{3}$$
Now, this last equation represent an ellipse rotated with angle $\alpha=-\frac{\pi}{4}$ around the origin. 
$\qquad \qquad \qquad \qquad$
This suggest us to do the following change of coordinates
$$\begin{array}{}
x = \cos\alpha \, u + \sin\alpha \,  v \\
y = -\sin\alpha \, u + \cos\alpha \, v
\end{array} \tag{4}$$
which represent a rotation. This will turn the last equation in $(3)$ into
$$(\sqrt{3}u)^2+v^2=R^2 \tag{5}$$
Also, we can go further and do another change of coordinates
$$\begin{array}{}
u = \frac{1}{\sqrt{3}} r \, \cos\theta \\
v = r \, \sin\theta
\end{array} \tag{6}$$
and this will turn $(5)$ into the simple equation
$$r=R \tag{7}$$
To conclude, we introduce the new curvelinear coordinates
$$\begin{array}{}
x = \frac{\cos\alpha}{\sqrt{3}}  \, r \, \cos\theta + \sin\alpha \,  r \, \sin\theta \\
y = - \frac{\sin\alpha}{\sqrt{3}} r \, \cos\theta + \cos\alpha \, r \, \sin\theta \\
z=z
\end{array} \tag{8}$$
and do the further computations in this new coordinates which simplifies the whole arithmetic.
