Let \(\gamma\) be a path in the plane \(\mathbb{R}^2\). Define what is meant by \[\int_\gamma f(x,y)\,dx + g(x,y)\,dy.\]
Compute this line integral in the case where \(f(x,y) = 2xy, g(x,y) = x^2 + y^2\) and \(\gamma\) is the straight-line path from \(P = (1,0)\) to \(Q = (0,1)\).
Is there another path \(\beta\) from \(P\) to \(Q\) such that the corresponding line integral takes a different value?
Solution
Part (a).
1. If \(\gamma: [a,b] \to \RR^2\) is a piecewise-smooth path with \(\gamma(t) = (x(t), y(t))\), then\[\int_\gamma f\,dx + g\,dy = \int_a^b \left[ f(x(t), y(t))\, x'(t) + g(x(t), y(t))\, y'(t) \right] dt.\]
definition of the line integral of the differential form \(f\,dx + g\,dy\) along \(\gamma\).
Part (b).
1. Parametrize \(\gamma(t) = (1-t, t)\) for \(t \in [0,1]\).
this is the straight line from \((1,0)\) to \((0,1)\).
2. \(x'(t) = -1\) and \(y'(t) = 1\).
differentiate.
3. The integrand is \(f\,x' + g\,y' = 2xy(-1) + (x^2 + y^2)(1) = -2(1-t)t + (1-t)^2 + t^2\).
substitute \(x = 1-t\), \(y = t\).
4. \(\int_\gamma f\,dx + g\,dy = \int_0^1 \left[ -2t(1-t) + (1-t)^2 + t^2 \right] dt = \int_0^1 (1 - 4t + 4t^2)\, dt\).
expand \(-2t + 2t^2 + 1 - 2t + t^2 + t^2 = 1 - 4t + 4t^2\).
5. \(\int_0^1 (1 - 4t + 4t^2)\, dt = \left[ t - 2t^2 + \tfrac{4}{3}t^3 \right]_0^1 = 1 - 2 + \tfrac{4}{3} = \tfrac{1}{3}\).
evaluate the antiderivative.
6. Hence \(\int_\gamma f\,dx + g\,dy = \tfrac{1}{3}\).
5.
Part (c).
1. No such path \(\beta\) exists.
1.1. The form \(f\,dx + g\,dy\) is exact.
\(\frac{\partial f}{\partial y} = 2x = \frac{\partial g}{\partial x}\), so the form is closed; on the simply connected domain \(\RR^2\) every closed form is exact.
1.2. The line integral of an exact form depends only on the endpoints.
if \(f\,dx + g\,dy = dF\), then \(\int_\gamma f\,dx + g\,dy = F(Q) - F(P)\) by the fundamental theorem of line integrals.
1.3. Hence every path from \(P\) to \(Q\) gives the same value \(\tfrac{1}{3}\).
1 and 2.
2. Q.E.D.
1.