Proof Line Integral is zero avoiding Stokes

Question

Give is: $C$ which is a closed curve which forms the surface $\Sigma$., $\vec{v} $ which is a constant vector.

I should prove the following expression without using Stokes' Theorem:

$$\oint_C \vec{v} \cdot d\vec{l} = 0$$

How do I go about doing it for an arbitrarily closed (even overlapping) curve ?

Answer 1

Let $C$ be parameterized by $\alpha(t)$, where $\alpha : \mathbb{R} \to \mathbb{R}^n$. Then using the definition of the line integral,

$$\int_{t_0}^{t_1}\mathbf{v} \cdot \alpha'(t) dt = \int_{t_0}^{t_1} (\mathbf{v} \cdot \alpha(t))' dt = (\mathbf{v} \cdot \alpha(t)) \Big|_{t_0}^{t_1} = 0$$

since $\alpha(t_0) = \alpha(t_1)$.

Answer 2

Let $\vec{v}=\langle c_1,c_2,\ldots, c_n\rangle$ be a constant vector in $\mathbb{R}^n$.

Since $\vec{v}=\nabla f$ for some differentiable function $f$, that is, we can take $f$ to be $$ f(x_1,x_2,\ldots, x_n)=c_1x_1 + c_2 x_2+\ldots + c_n x_n + C, \mbox{ where }C\in \mathbb{R}, $$ and $\nabla = \langle \partial_1,\partial_2,\ldots, \partial_n \rangle$, $\partial_i := \partial_{x_i}$, $\vec{v}$ is conservative (and so $\nabla f$ is a gradient field).

Thus a line integral along any smooth curve depends only on the endpoints of the path, i.e., given a parametrization $\vec{l}:[t_1,t_2]\rightarrow \mathbb{R}^n$ of a smooth closed curve $C$, say $$ \vec{l}(t_1) = \vec{l}(t_2)=(p_1,p_2,\ldots, p_n), $$ we have
$$ \begin{align*} \oint_C \vec{v}\cdot d\vec{l} &= \oint_C \nabla f \cdot d\vec{l} \\ &= f(\vec{l}(t_2)) - f(\vec{l}(t_1)) \\ &= f(p_1,p_2,\ldots, p_n) - f(p_1,p_2,\ldots, p_n) \\ &= 0. \end{align*} $$

Note that if $C$ is not smooth, you can generalize this argument for piecewise smooth curves.