# Line integral along curve

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?

• If I understood you correctly, im$(\varphi)\subseteq C$ and so $\|\varphi(t)\|^2=R^2$ for all $t$. Then $\langle \varphi(t),\varphi(t)\rangle=R^2$, which implies $\langle\varphi(t),\varphi'(t)\rangle = 0$. Commented Mar 18, 2022 at 22:00

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 $$2\varphi(t)\cdot\dot{\varphi}(t) = 0$$.