Conservative field

The central result is the fundamental theorem for line integrals: if \(\mathbf F=\nabla f\) and \(C\) goes from \(A\) to \(B\), then \(\int_C\mathbf F\cdot d\mathbf r=f(B)-f(A)\). It is the multivariable version of the fundamental theorem of calculus. Two consequences follow: the integral does not depend on the path and, on a closed curve, equals \(0\). These three properties (being a gradient, path independence and zero circulation) are equivalent on a connected region.

To decide whether a plane field \(\mathbf F=\langle P,Q\rangle\) is conservative, use the cross-partials test: if it is, then \(P_y=Q_x\), because \(f_{xy}=f_{yx}\). Conversely, if \(P_y=Q_x\) on a region without holes (simply connected), the field is conservative. In space the three analogous equalities among the partials of \(P,Q,R\) are required. For \(\langle y,x\rangle\) we get \(1=1\) and the potential is \(xy\); for \(\langle -y,x\rangle\) it fails (\(-1\ne1\)).

To find the potential, integrate \(f_x=P\) with respect to \(x\), giving \(f=\int P\,dx+g(y)\); then differentiate in \(y\), set it equal to \(Q\) and solve for \(g\). The potential is defined up to a constant, which does not affect \(f(B)-f(A)\). With it, the integral is computed without parametrizing anything.

Typical mistake: applying \(P_y=Q_x\) on a region with a hole. The field \(\langle -y,x\rangle/(x^2+y^2)\) satisfies the equality on \(\mathbb R^2\setminus\{0\}\), but its circulation around the origin is \(2\pi\): it is not conservative there. This concept uses the gradient and the vector line integral, and is the case where work does not depend on the path.