Vector Line Integrals

If an object feels a constant force \(\textbf{F}\) while moving along a displacement \(\textbf{s}\), then the work done, \(W\), by \(\textbf{F}\) is:

$$W = \textbf{F} \cdot \textbf{s}$$

What about if the force and the path is not constant? Suppose a particle moves in a three dimensional space along a curve \(C\) defined by \(\textbf{r}(t)\) from \(t=a\) to \(t=b\):

Suppose the particle feels a force while moving along \(C\), where the force is defined by the vector field:

$$\textbf{F}(x,y,z) = P(x,y,z)\textbf{i} +Q(x,y,z)\textbf{j} +R(x,y,z)\textbf{k}$$

Partition the parameter interval \([a,b]\) into \(n\) subintervals \([t_{ i-1 },t_i]\), where \(t_0=a\) and \(t_n=b\). Let \(t^*_i\) be a value in the \(i\)th interval \([t_{i-1}, t_i]\).

Denote the endpoints of \(\textbf{r}(t_0), \textbf{r}(t_1), \ldots , \textbf{r}(t_n)\) by \(P_0, \ldots,P_n\). Points \(P_i\) divide curve \(C\) into \(n\) pieces \(C_1,C_2, \ldots,C_n\), with lengths \(\Delta s_1, \Delta s_2, \ldots, \Delta s_n\), respectively. Let \(P^*_i\) denote the endpoint of \(\textbf{r}(t^*_i)\) for \(1≤i≤n\). The endpoint of \(\textbf{r}(t^*_i)\) lies between \(\textbf{r}(t_{i-1})\) and \(\textbf{r}(t_i)\).

Suppose \(n\) is very large such that the individual pieces of the curve look almost like straight lines, and such that \(\textbf{F}\) doesn't change much. The approximate work done by \(\textbf{F}\) between \(t_{i-1}\) and \(t_i\) is:

$$ \textbf{F}(P^*_i) \cdot (\textbf{T}(P^*_i) \ \Delta s_i) $$

\(\textbf{T}(P^*_i)\) gives the unit tangent vector at point \(P^*_i\), and the force \(\textbf{F}(P^*_i)\) can be treated as constant within the interval \([t_{i-1},t_i]\) since the curve piece is very small. The value of \((\textbf{T}(P^*_i) \ \Delta s_i)\) gives the displacement from \(t_{i-1}\) to \(t_i\), since \(\textbf{T}\) gives the direction and \(\Delta s_i\) scales the unit tangent vector to the distance between \(P_{i-1}^*\) and \(P_i^*\).

To find total work done from \(t=a\) to \(t=b\), we need to sum the individual approximation of the work done on every piece:

$$ \begin{align} W &= \sum_{i=1}^n \textbf{F}(P^*_i) \cdot (\textbf{T}(P^*_i) \ \Delta s_i) \\ &= \sum_{i=1}^n \textbf{F}(P^*_i) \cdot \textbf{T}(P^*_i) \ \Delta s_i \end{align} $$

If \(n\) goes to infinity, then the above is represented like so:

$$ \int_C \textbf{F} \cdot \textbf{T} ds = \lim_{n \to \infty} \sum_{i=1}^n \textbf{F}(P^*_i) \cdot \textbf{T}(P^*_i) \ \Delta s_i$$

We are summing infinitely many \(\textbf{F} \cdot \textbf{T} \Delta s_i\). Since \(\textbf{T}(P^*_i) = \frac{\textbf{r}'(t^*_i)}{\Vert \textbf{r}'(t^*_i) \Vert}\):

$$ \textbf{F}(P^*_i) \cdot \textbf{T}(P^*_i) \Delta s_i = \textbf{F}(P^*_i) \cdot \left(\frac{\textbf{r}'(t^*_i)}{\Vert \textbf{r}'(t^*_i) \Vert}\right) \Delta s_i $$

Suppose there are many intervals (i.e. \(n\) is very large), and that \(\Delta s_i\) is so small such that \(\textbf{r}'(t)\) barely changes between \(t_{i-1}\) and \(t_{i}\):

$$\Delta s_i = \int_{t_{i-1}}^{t_i} \Vert \textbf{r}'(t) \Vert dt \approx \Vert \textbf{r}'(t_i^*) \Vert \Delta t$$

This means:

$$\begin{align} \textbf{F}(P^*_i) \cdot \textbf{T}(P^*_i) \Delta s_i &\approx \textbf{F}(P^*_i) \cdot \left(\frac{\textbf{r}'(t^*_i)}{\Vert \textbf{r}'(t^*_i) \Vert}\right) (\Vert \textbf{r}'(t_i^*) \Vert \Delta t) \\ &\approx \textbf{F}(P^*_i) \cdot \textbf{r}'(t^*_i) \Delta t\end{align} $$

Summing infinite of these:

$$W =\int_a^b \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t) d t$$

We can use these notations to represent the above:

$$\int_a^b \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t) d t = \int_C \textbf{F} \cdot \textbf{T} ds = \int_C \textbf{F} \cdot d\textbf{r}$$

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