The vector function F(r) = -x î + yĵ is defined over a circular are C shown in the figure.The line integral of ∫C F(r) ⋅ dr is

Concept:∫C F(r) dr can be solved by putting:x = r cos θy = r sin θdx = -r sin θ dθdy = r cos θ dθApplication:GIVEN r = 1θ = 0 to 45°The required integral can be written as:\(\mathop \smallint \nolimits_C \VEC F \CDOT dr = \mathop \smallint \nolimits_0^{45^\circ } \left[ {\left( { - r\cos \theta } \right)\left( { - r\sin \theta } \right)d\theta + \left( {r\sin \theta } \right)\left( {r\cos \theta } \right)d\theta } \right]\)\(\mathop \smallint \nolimits_0^{45^\circ } \left( {{r^2}\cos \theta \sin \theta + {r^2}\sin \theta \cos \theta } \right)d\theta \)With r = 1, the above integral becomes:\( = \FRAC{1}{2}\mathop \smallint \nolimits_0^{45^\circ } \left( {\sin 2\theta + \sin 2\theta } \right)d\theta \)r = 1\( = \mathop \smallint \nolimits_0^{45^\circ } \sin 2\theta \;d\theta \)\( = \frac{{\left( { - \cos 2\theta } \right)_0^{45^\circ }}}{2}\)\( = \frac{{ - 0 - \left( { - 1} \right)}}{2} = \frac{1}{2}\)