Vector Calculus and Green's Theorem

Line integrals and Green's theorem

Green's theorem connects a line integral around a closed curve with a double integral over the plane region that the curve encloses. Let C be a piecewise smooth, simple closed curve and let R be the region it bounds. If C is positively oriented, meaning the region lies to the left as one walks along C (counterclockwise for an outer boundary), and P(x,y) and Q(x,y) have continuous first partial derivatives on an open set containing R, then

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

The integrand \partial Q/\partial x - \partial P/\partial y is the scalar curl of the field P\,\hat i + Q\,\hat j, so the theorem states that the circulation around C equals the total curl summed over R. Reversing the direction of traversal changes the sign of the left side, so checking the orientation is the first step in every problem.

In most exam questions the hard part is the double integral, and it often becomes simple. The most common case is a constant integrand: if \partial Q/\partial x - \partial P/\partial y = k, the line integral equals k times the area of R.

When the integrand is exactly 1, the line integral is just the area enclosed by C. For a circle (x-a)^2 + y^2 = r^2 traversed counterclockwise, this gives \pi r^2.

Area of a region under a graph y=f(x)\ge 0 over [a,b] is \int_a^b f(x)\,dx, and the area of a region between two curves is \int (\text{upper} - \text{lower})\,dx over the interval where they bracket the region. For example, the area between y=x and y=x^2 on [0,1] is \int_0^1 (x-x^2)\,dx = \tfrac16.

• positively oriented — the boundary direction that keeps the enclosed region on the left. Example: for the unit square traversed counterclockwise, the orientation is positive.
• simple closed curve — a closed curve that does not cross itself. Green's theorem applies directly to such curves.
• circulation — the value of \oint_C \vec F\cdot d\vec r around a closed curve. Green's theorem converts it to a double integral.

Small worked example: for \oint_C (y\,dx + 2x\,dy) around the rectangle [0,2]\times[0,1] traversed counterclockwise, \partial Q/\partial x - \partial P/\partial y = 2 - 1 = 1, so the value equals the rectangle's area, which is 2.