Scalar Line Integrals

Suppose you have a function \(f(x,y)\), and you want to integrate with respect to the \(x\) on some interval \([a,b]\):

You would have to do an integration like so:

$$ \int_a^b f(x,0) \ dx $$

You should already know how to do this. Similarly, if you want to integrate with respect to \(y\) on some interval \([c,d]\):

You would have to do an integration like so:

$$ \int_c^d f(0,y) \ dy $$

What if you don't want to integrate over just the axis alone? What if you want to integrate over some curve \(C\) on the \(xy\)-plane:

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Scalar line integrals are integrals of a scalar function over a curve in a plane or in space.

Let \(C\) be a smooth curve in the \(xy\) plane given by the parameterization \(\textbf{r}(t)=⟨x(t),y(t)⟩\), \(a≤t≤b\). To define the line integral of the function \(f\) over \(C\), we start by chopping the curve into small pieces.

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 \(\textbf{r}(t_i)\).

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Now, we evaluate the function \(f\) at point \(P^*_i\) for \(1≤i≤n\). Multiply \(f(P^*_i)\) by the length \(\Delta s_i\) of \(C_i\), which gives the area of the "sheet" with base \(C_i\), and height \(f(P^*_i)\).

The scalar line integral of \(f\) along \(C\) is:

$$\int_C f(x,y) \ ds = \lim_{n \to \infty} \sum^n_{i=1} f(P^*_i) \Delta s_i$$

We can use the arc length formula for \(\Delta s_i\). Since \(\Delta s_i\) is the length between \(t_{i-1}\) and \(t_{i}\):

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

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 \(t_{i}\):

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

Remember that \(\textbf{r}(t_i^*)\) is some value within \([t_{i-1},t_{i}]\), and that \(\Delta t\) is constant. From the above formula, we can say:

$$\int_C f(x,y) \ ds = \lim_{n \to \infty} \sum^n_{i=1} f(\textbf{r}(t_i^*)) \ \Vert \textbf{r}'(t_i^*) \Vert \Delta t = \int_a^b f(\textbf{r}(t)) \ \Vert \textbf{r}'(t) \Vert dt $$

Since \(\Vert \textbf{r}'(t) \Vert = \sqrt{x'(t) + y'(t)}\):

$$\int_C f(x,y) \ ds = \int_a^b f(\textbf{r}(t)) \ \sqrt{x'(t) + y'(t)} dt $$

If \(f\) and \(\textbf{r}\) were functions that use three dimensions, then we can use a similar argument as above to state:

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

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