Calculate $\oint_\gamma f(X) dX$ where $f(x,y,z) = (yz,xz,xy)$, $\gamma = \{X(t)| 0 \le t \le 2\pi\}$, $X(t) = (cos(t)cos(\frac{\pi}{8} + \frac{t(2\pi-t)}{4\pi}),sin(t)cos(\frac{\pi}{8} + \frac{t(2\pi-t)}{4\pi}),sin(\frac{\pi}{8} + \frac{t(2\pi-t)}{4\pi}))$
I know that I need to use the Divergence theorem(that's the subject of the exercise). I want to calculate the normal $N(x)$ of the manifold $\gamma$, so I could try and get anything close to $\oint_\gamma f(X)dX$ in the form of $\int<F,N> dx$ where $F$ is some vector field (and then use the Divergence theorem).
I know that usually the normal of manifold which is given by parameterization is $N(x)=\frac{r_{u_1}\times\dots\times r_{u_{n-1}}}{||r_{u_1}\times\dots\times r_{u_{n-1}}||}$ (Where $r(U)$ is the parameterization) but here I'm given a parameterization that operate over single variable, $t$.
How does that formula for the normal $N(x)$ works in that case? Or perhaps is there any other way to calculate the normal?
Or is there any other trick that I'm missing to use the Divergence theorem with the current integral?
The given vector field $f$ is nothing else but $f=\nabla F$ for the scalar function $F(x,y,z):=xyz$. We are therefore told to integrate $$\int_\gamma \nabla F(X)\cdot dX$$ along the curve $$\gamma: \quad t\mapsto X(t):=\bigl(\cos\psi(t)\cos t,\cos\psi(t)\sin t,\sin\psi(t)\bigr)\qquad(0\leq t\leq2\pi)\ ,$$ with $$\psi(t):={\pi\over8}+{t(2\pi -t)\over4\pi}\ ,$$ hence $\psi(0)=\psi(2\pi)={\pi\over8}$. It follows that $\gamma$ begins and ends at the point ${\bf p}:=\gamma(0)=\bigl(\cos{\pi\over8},0,\sin{\pi\over8}\bigr)$. Since $F$ is $C^1$ on all of ${\mathbb R}^3$ this allows to conclude that $$\int_\gamma \nabla F(X)\cdot dX=F({\bf p})-F({\bf p})=0\ .$$