Let \(\tau\) be the arc of the unit circle in the first quadrant, from \((1,0)\) to \((3/5, 4/5)\). Compute \[\int_\tau x\,dy + y\,dx.\]
Solution
1. Exact differential formulation:
1.1. Consider the differential 1-form \(\omega = y\,dx + x\,dy\).By the product rule for differentials:\[ \omega = d(xy). \]
1.2. Thus \(\omega\) is an exact 1-form on \(\mathbb{R}^2\) with potential function \(f(x, y) = xy\).
2. Application of the Fundamental Theorem of Line Integrals:
2.1. Let \(\tau\) be any piecewise smooth path starting at \(A = (1, 0)\) and ending at \(B = (3/5, 4/5)\).By the Fundamental Theorem for Line Integrals:\[ \int_\tau (x\,dy + y\,dx) = \int_\tau d(xy) = f(B) - f(A). \]
2.2. Evaluate the potential function at the endpoints:- \(f(A) = f(1, 0) = 1 \cdot 0 = 0\),- \(f(B) = f\left(\frac{3}{5}, \frac{4}{5}\right) = \frac{3}{5} \cdot \frac{4}{5} = \frac{12}{25}\).
2.3. Therefore:\[ \int_\tau (x\,dy + y\,dx) = \frac{12}{25} - 0 = \frac{12}{25}. \]
3. Alternative verification via trigonometric parametrization:
3.1. Parametrize the arc \(\tau\) by \(x(t) = \cos t, y(t) = \sin t\) for \(t \in [0, t_0]\), where \(\cos t_0 = 3/5\) and \(\sin t_0 = 4/5\).Then \(dx = -\sin t\,dt\) and \(dy = \cos t\,dt\).
3.2. The integrand becomes:\[ x\,dy + y\,dx = (\cos^2 t - \sin^2 t)\,dt = \cos(2t)\,dt. \]
3.3. Integrating:\[ \int_0^{t_0} \cos(2t)\,dt = \left[ \frac{\sin(2t)}{2} \right]_0^{t_0} = \sin(t_0)\cos(t_0) = \frac{4}{5} \cdot \frac{3}{5} = \frac{12}{25}. \]
4. Conclusion:\(\int_\tau x\,dy + y\,dx = \frac{12}{25}\). Q.E.D.