Green's Theorem

Green's Theorem

Green's theorem connects a line integral around a closed curve to a double integral over the enclosed region:

$\oint_{\partial R} P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

This is one of the central results of vector calculus — it shows that circulation around the boundary equals the total curl inside.

Area Formula

A beautiful consequence: if we choose $P = -y/2$, $Q = x/2$, then:

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} + \frac{1}{2} = 1$

So:

$\text{Area}(R) = \frac{1}{2} \oint_{\partial R} (-y\,dx + x\,dy)$

Numerical Verification

We can verify Green's theorem numerically by computing both sides independently:

1. Double integral side: integrate the curl $\partial Q/\partial x - \partial P/\partial y$ over $R$
2. Line integral side: integrate $P\,dx + Q\,dy$ around the boundary

For $P = -y$, $Q = x$ over the unit square $[0,1]\times[0,1]$:

• Curl: $\partial Q/\partial x - \partial P/\partial y = 1 - (-1) = 2$
• Double integral: $2 \times \text{Area} = 2 \times 1 = 2$

Implementing the Double Integral Side

double greens_double_integral(
    double (*curl)(double, double),
    double ax, double bx, double ay, double by, int n) {
    double hx = (bx - ax) / n, hy = (by - ay) / n, sum = 0.0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            sum += curl(ax + i*hx, ay + j*hy) * hx * hy;
    return sum;
}

Your Task

Implement double greens_double_integral(curl, ax, bx, ay, by, n) that numerically integrates a curl function over the rectangle $[ax, bx] \times [ay, by]$.