Discussion
The work done by the force \(F = (x + y)\hat{i} - (x^{2} + y^{2})\hat{j}\), where \(\hat{i}\) and \(\hat{j}\) are unit vectors in \(\overrightarrow{OX}\) and \(\overrightarrow{OY}\) directions, respectively, along the upper half of the circle \(x^{2} + y^{2} = 1\) from (1, 0) to (-1, 0) in the xy-plane is
Answer: B
The work done is \(-\frac{\pi}{2}\).
Parametrise the upper semicircle by x = cos t, y = sin t, t from 0 to \(\pi\). On it \(x^{2} + y^{2} = 1\), so
\[ \begin{aligned} F \cdot dr &= (x+y)\,dx - (x^{2}+y^{2})\,dy \\ &= -(\cos t + \sin t)\sin t\,dt - \cos t\,dt \end{aligned} \]
\[ \begin{aligned} W &= \int_{0}^{\pi} (-\sin t\cos t - \sin^{2} t - \cos t)\,dt \\ &= 0 - \frac{\pi}{2} - 0 = -\frac{\pi}{2} \end{aligned} \]
The terms \(\sin t\cos t\) and \(\cos t\) integrate to 0 over [0, \(\pi\)], and \(\sin^{2} t\) integrates to \(\pi/2\). \(\pi/2\) has the sign reversed (the path traversed the other way), and \(\pm\pi\) would come from integrating \(\sin^{2} t\) as if it were 1.
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