VC4.19

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Since the integration is performed in xy-plane r=xi+yj,dr=dxi+dyj\,

In the xy-plane, we have z=0. So, in xy-plane,given force becomes F=(2x-y)i+(x+y)j\,

The parametric equation of the given circle x^{2}+y^{2}=3^{2}\, are x=3\cos \theta ,y=3\sin \theta ,0\leq \theta \leq 2\pi \,

Hence the required work done \int _{C}F\cdot \,dr=\int _{C}[(2x-y){\frac  {dx}{d\theta }}+(x+y){\frac  {dy}{d\theta }}]\,d\theta \,

=\int _{{\theta =0}}^{{2\pi }}[(6\cos \theta -3\sin \theta )(-3\sin \theta )+(3\cos \theta +3\sin \theta )(3\cos \theta )]\,d\theta \, =\int _{{0}}^{{2\pi }}(9-9\sin \theta \cos \theta )\,d\theta =[9\theta ]_{{0}}^{{2\pi }}-{\frac  {9}{2}}\int _{{0}}^{{2\pi }}\sin 2\theta \,d\theta \, =18\pi -{\frac  {9}{2}}[{\frac  {\cos 2\theta }{2}}]_{{0}}^{{2\pi }}=18\pi -{\frac  {9}{4}}(1-1)\, =18\pi \,


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