Alternative approach for a line integral?

Question

Let $F$ be the vector field in ${\Bbb R}^3\setminus\{0\}$ defined by $$ F(x,y,z):=\left(0,\frac{xz}{(y^2+z^2)\sqrt{x^2+y^2+z^2}},\frac{-xy}{(y^2+z^2)\sqrt{x^2+y^2+z^2}}\right). $$ Compute the line integral $\int_CF\cdot ds$, where $C$ is the unit circle centered at the point $(1,1,1)$ that lies on the plane $x+y+z=3$ and has the orientation from the point $\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}}\right)$ to $\left(1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}},1-\frac{1}{\sqrt{6}}\right)$ to $\left(1+\frac{2}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}}\right)$ and back to $\left(1-\frac{1}{\sqrt{6}},1-\frac{1}{\sqrt{6}},1+\frac{2}{\sqrt{6}}\right)$.

A key step is to parameterize the curve: $$ \begin{cases} (x-1)^2+(y-1)^2+(z-1)^2=1,\\ x+y+z=3. \end{cases} $$

---

Let $$ \begin{cases} x=1+\sin\theta\cos\phi\\ y=1+\sin\theta\sin\phi\\ z=1+\cos\theta \end{cases} $$ Substituting it into $x+y+z=3$, we have $\theta=g(\phi)$ for some function $g$, which can be written down explicitly. Then we have $$ (x,y,z)=(x(\phi),y(\phi),z(\phi)). $$

Here is my question:

Is there an alternative approach which might be more efficient than using the parameterization above to calculate the line integral?

---

[Added] I'm not sure if Stokes' theorem would help. Some calculations render that $$ \nabla\times F(x,y,z)=\left(\frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}}, \frac{z}{(x^2+y^2+z^2)^{3/2}}\right). $$ It seems that this would not simplify the calculation when we do integration on the "upper" sphere (since the center of the circle is not at the origin), of which the unit circle in the problem is the boundary.

Answer 1

The unit circle $\gamma$ in ${\mathbb R}^2$ is parametrized most simply as $$ \gamma:\quad t\mapsto(\cos t,\sin t)\qquad(0\leq t\leq 2\pi)\ .$$We now transport this parametrization to your situation. Your circle $C$ has $(1,1,1)$ as its center and $n:={1\over\sqrt{3}}(1,1,1)$ as its axis. First we need two unit vectors which are orthogonal to $n$ and orthogonal to each other. A first such vector would be $e:={1\over\sqrt{2}}(1,-1,0)$, and the second is then given by $f=n\times e={1\over\sqrt{6}}(1,1,-2)$. The resulting parametrization of $C$ is then $$C:\quad t\mapsto(1,1,1)+\cos t\>e+\sin t\>f\qquad(0\leq t\leq 2\pi)\ .\tag{1}$$ You still have to check whether the $C$ given by $(1)$ is oriented as described in the problem. If it isn't replace $f$ by $-f$ in $(1)$.