Let be $h:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}$ a continuous function,
$f:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}^n$ a vector field where $f(x):=h(x)x$,
$C$ a continuously differentiable curve with $C\subset\{x\in\mathbb{R}^n\mid \Vert x\Vert_2=R\}$, where $R>0$.
Show that the line integral of the vector field $f$ along $C$ satisfies $$ \int_C f(x)\cdot dx = 0. $$
My approach:
Let be $\varphi:[\alpha,\beta]\to\mathbb{R}^n$ a parametrization of $C$ then we use the definition of the line integral and get $$ \int_C f(x)\cdot dx= \int\limits_{\alpha}^{\beta}f(\varphi(t))\cdot \varphi'(t)dt=\int\limits_{\alpha}^{\beta}h(\varphi(t))\varphi(t)\cdot \varphi'(t)dt=\int\limits_{\alpha}^{\beta}h(\varphi(t))\left(\varphi_1(t) \varphi_1'(t)+\varphi_2(t) \varphi_2'(t)+\cdots+\varphi_n(t) \varphi_n'(t)\right)dt. $$ But this doesn't help a lot.
If I take for example $\varphi(t)=\begin{pmatrix}R\cos(t)\\R\sin(t)\end{pmatrix}$ then $\varphi(t)\cdot \varphi'(t)$ vanishes and $\int_C f(x)\cdot dx = 0$. But why is that so? Do you have any hints which way to go?
As the curve is contained in the sphere, the squared length $||\varphi(t)||^2 = \varphi(t)\cdot \varphi(t)$ is constantly equal to $R^2$. Differentiating with respect to $t$ gives $2\varphi(t)\cdot\dot{\varphi}(t) = 0$.