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I'm finding the concept of line integrals with differentials including dx or dy hard to swallow intuitively. Specifically, I'm having trouble differentiating them from a Riemann integral. What are the precise differences intuitively, rigorously, and geometrically? I'm also perhaps asking implicitly, what makes a line integral a line integral if its not the differential ds?

As requested:

$$\int_c F(x,y,z) * dr$$

$$\int_c F_1(x,y,z) * dx + \int_c F_2(x,y,z) * dy + \int_c F_3(x,y,z)*dz$$

That is the general form. The specifics of the vector field F(x,y,z) aren't the issue. I can perform the evaluations, but I don't quite understand what I am doing. The meaning of the sub-integrals is what is troubling me.

SystemFun
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2 Answers2

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One dimension less $\dots$

Let $f$ be defined on a set of points of the $xy$-plane including the parameterized curve $C$.
Then the graph of $f$ over $C$ is a space curve. Project this curve on the $xz$-plane. Then the symbol $$\int_C f(x,y)\,dx$$ is called, by definition, the line integral of $f$ on $C$ with respect to $x$ and represents, by definition, a limit which can be thought as the area of the region on the $xz$-plane, between the above projection and the $x$-axis.

By due assumptions it can be shown it is given by the ordinary (Riemann) integral $$\int_{t_1}^{t_2} f(x(t),y(t))\,x'(t)\,dt$$

If $C$ is a segment $[a,b]$ on the $x$-axis, one has $$\int_C f(x,y)\,dx=\int_a^b f(x,0)\,dx$$ If you add one dimension, intuition runs out.

Tony Piccolo
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The line integral $$\int_\gamma {\bf F}\cdot d{\bf r}$$ in your first displayed line has an intuitive physical meaning; the three integrals in the second displayed line don't and should not be envisaged at all.

When you have to push a cart along the curve $\gamma$ against the force field ${\bf f}$ then the total work $W$ done is roughly $$\sum_{k=1}^N {\bf F}({\bf r}_k)\cdot ({\bf r}_k-{\bf r}_{k-1})\ ,$$ where $({\bf r}_0,{\bf r}_1,\ldots,{\bf r}_N)$ is a polygonal approximation of $$\gamma:\quad t\mapsto{\bf r}(t)\qquad(a\leq t\leq b)\ .$$ It follows that $$W\doteq\sum_{k=1}^N {\bf F}\bigl({\bf r}(t_k)\bigr)\cdot \dot{\bf r}(t_k)\ (t_k-t_{k-1})\doteq \int_a^b{\bf F}\bigl({\bf r}(t)\bigr)\cdot\dot{\bf r}(t)\ dt\ .$$ Note that the integral appearing on the right hand side is a bona fide Riemann integral over the interval $[a,b]$, albeit with a complicated integrand $\Psi(t)$. Introducing the components $(F_1,F_2,F_3)$, resp. $(x, y, z)$, of the involved vectors we arrive at $$\Psi(t)=F_1\bigl(x(t),y(t),z(t)\bigr)\dot x(t)+F_2\bigl(x(t),y(t),z(t)\bigr)\dot y(t)+F_3\bigl(x(t),y(t),z(t)\bigr)\dot z(t)\ .$$ The individual summands appearing here have no meaning by themselves. Arguing purely formally one of course could write $$\int_a^b F_1\bigl(x(t),y(t),z(t)\bigr)\dot x(t)\ dt=\int_\gamma F_1(x,y,z)\ dx\ ,$$ which leads to the integrals in your second displayed line.

Christian Blatter
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