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When computing the scalar line integral, we find the integral of f dot ds over some path by integrating the product of f evaluated at the path function with the magnitude of the derivative of the path function. And in the vector line integral, one does something similar except with a dot product of f evaluated at the path function and the derivative of the path function. However, in the differential form: $$\int_{\bf X} {\bf F} \,{\bf\cdot}\, d{\bf s} = \int_{\bf X} M(x,y,z)\,dx+N(x,y,z)\,dy+P(x,y,z)\,dz$$ I don't see that sort of multiplying by the derivative happening anywhere. Why is that?

In other words, why is it that in the differential form of the line integral, we don't have an integrand that is the product of a function evaluated along a parameterized path and the derivative of that parameterized path?


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Let ${\bf X}$ be a path and let ${\bf F}(x,y,z)=\langle M(x,y,z),N(x,y,z),P(x,y,z) \rangle$.

In addition let ${\bf r}(t)=\langle x(t),y(t),z(t) \rangle$, $a \leq t \leq b$ be a parametrization of ${\bf X}$.

Then by definition: $\int_{\bf X} {\bf F} \,{\bf\cdot}\, d{\bf s} = \int_a^b {\bf F}({\bf r}(t)) \,{\bf\cdot}\, {\bf r}'(t)\,dt$ which is $\int_a^b \left(M({\bf r}(t))x'(t)+N({\bf r}(t))y'(t)+P({\bf r}(t))z'(t)\right) \,dt=\int_a^b M\,dx+N\,dy+P\,dz$.

So the derivative of your parametrization does appear in the Mdx+Ndy+Pdz form. In fact: $d{\bf s} = {\bf r}'(t)\,dt = \langle x'(t),y'(t),z'(t) \rangle\,dt = \langle dx,dy,dz \rangle$.

Notice this all involves "${\bf s}$" (boldface s = vector s). I think your confusing this somewhat with "$s$" (regular s = arc length parameter). In that case $ds=|{\bf r}(t)|\,dt$ (this involves the magnitude of the derivative of the parametrization).

Of course these are related. If ${\bf T}(t) = \frac{{\bf r}'(t)}{|{\bf r}'(t)|}$ is the unit tangent, then $d{\bf s} = {\bf r}'(t)\,dt = \frac{{\bf r}'(t)}{|{\bf r}'(t)|}|{\bf r}'(t)|\,dt = {\bf T}(t)ds$. Thus $\int_{\bf X} {\bf F}\,{\bf\cdot}\, d{\bf s} =\int_{\bf X} {\bf F}\,{\bf\cdot}\, {\bf T}\,ds$

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