Line Integral of a Scalar Field

Line Integrals

A line integral (or path integral) integrates a scalar field $f(x,y)$ along a curve $C$:

$\int_C f(x,y)\, ds$

where $ds$ is the arc-length element — each tiny piece of the curve weighted by the function value at that point.

Parameterization

If the curve is given by $\mathbf{r}(t) = (x(t), y(t))$ for $t \in [a, b]$, then:

$ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt$

So:

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

Examples

Arc length (set $f = 1$): $\int_C 1\, ds = \text{length of } C$

For a unit circle ($x = \cos t$, $y = \sin t$, $t \in [0, 2\pi]$): $\int_C 1\, ds = \int_0^{2\pi} 1 \cdot 1\, dt = 2\pi \approx 6.2832$

Integral of $f = x$ along $x$-axis from 0 to 2 ($x=t, y=0, ds=dt$): $\int_0^2 t\, dt = 2.0000$

Numerical Computation

double line_integral(double (*f)(double, double),
                     double (*xt)(double), double (*yt)(double),
                     double (*dxt)(double), double (*dyt)(double),
                     double a, double b, int n) {
    double h = (b - a) / n, sum = 0.0;
    for (int i = 0; i < n; i++) {
        double t = a + i * h;
        double dx = dxt(t), dy = dyt(t);
        sum += f(xt(t), yt(t)) * sqrt(dx*dx + dy*dy) * h;
    }
    return sum;
}

Your Task

Implement double line_integral(f, xt, yt, dxt, dyt, a, b, n) using a left Riemann sum over the parameterization.