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.

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:
$$
r=\frac{\sqrt{3}}{2}.
$$
