Questions about the line integral

Question

Here's how we get to the formula for the line integral:

$$\overrightarrow{R}(t)=x(t) \hat{\imath}+y(t) \hat{\jmath}+z(t) \hat{k}, \ \ \ \ \ \ a \leq t \leq b$$

We subdivide the curve into the elements $\Delta S_1, \Delta S_2, \dots , \Delta S_n$

$$\sum_{k=1}^n w(x_k, y_k, z_k) \Delta S_k$$ $$n \rightarrow \infty$$ $$\lim_{n \rightarrow \infty} \sum_{k=1}^n w(x_k, y_k, z_k) \Delta S_k= \int_C w(x,y,z)ds \ \ \ \ \ \ (*)$$

$ds=|d \overrightarrow{R}| \ \ \ \ (**)$

$|\overrightarrow{R}|=|\frac{\overrightarrow{R}}{dt}dt|=|\frac{\overrightarrow{R}}{dt}|dt=\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt$

$$\int_C w ds=\int_a^b w(x(t), y(t), z(t))\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt$$

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Could you explain me why the relations $(*)$ and $(**)$ stand?

Answer 1

if $r(t)$ is your position vector, the total movement in time,

$$\Delta S=|r(t_1)-r(t_0)|=|\Delta r(t)|$$

$$\dfrac{\Delta S}{\Delta t}=\dfrac{|\Delta r(t)|}{\Delta t}$$

if you take limit when $\Delta t \to 0$, you will get $$\dfrac{ds}{dt}=|r'(t)|$$ or simply $$ds=|r'(t)|dt$$

which explain the "**", I think "*" is natural result of Rieaman sum.