VC4.14

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Given curve is given by r=xi+yj+zk=a\cos \theta i+a\sin \theta j+a\theta k\, --(1)

Its parametric equations are x=a\cos \theta ,y=a\sin \theta ,z=a\theta \, --(2)

Here F=xyi+yzj+zxk=s^{2}\sin \theta \cos \theta i+a^{2}\theta \sin \theta j+a^{2}\theta \cos \theta k\, --(3)

From (1),{\frac  {dr}{d\theta }}=-a\sin \theta i+a\cos \theta j+ak\, --(4)

Therefore,\int _{{C}}F\cdot \,dr=\int _{{0}}^{{{\frac  {\pi }{2}}}}(-a^{3}\sin ^{2}\theta \cos \theta +a^{3}\theta \sin \theta \cos \theta +a^{3}\theta \cos \theta )\,d\theta \,

=-a^{3}\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin ^{2}\theta \cos \theta \,d\theta +{\frac  {a^{3}}{2}}\int _{{0}}^{{{\frac  {\pi }{2}}}}\theta \sin 2\theta \,d\theta +a^{3}\int _{{0}}^{{{\frac  {\pi }{2}}}}\theta \cos \theta \,d\theta \, by using (3)and(4)

=-a^{3}[{\frac  {\sin ^{3}\theta }{3}}]_{{0}}^{{{\frac  {\pi }{2}}}}+{\frac  {a^{3}}{2}}[-{\frac  {\theta \cos 2\theta }{2}}]_{{0}}^{{{\frac  {\pi }{2}}}}-\int _{{0}}^{{{\frac  {\pi }{2}}}}[-{\frac  {\cos 2\theta }{2}}\,d\theta +a^{3}{[\theta \sin \theta ]_{{0}}^{{{\frac  {\pi }{2}}}}-\int _{{0}}^{{{\frac  {\pi }{2}}}}\sin \theta \,d\theta }\,

=-{\frac  {a^{3}}{3}}+{\frac  {a^{3}}{2}}[{\frac  {\pi }{4}}+{\frac  {1}{2}}{{\frac  {1}{2}}\sin 2\theta }_{{0}}^{{{\frac  {\pi }{2}}}}]+a^{3}[{\frac  {\pi }{2}}+(\cos \theta )_{{0}}^{{{\frac  {\pi }{2}}}})]\,

=-{\frac  {a^{3}}{3}}+{\frac  {a^{3}}{2}}{\frac  {\pi }{4}}+a^{3}({\frac  {\pi }{2}}-1)=a^{3}[{\frac  {5\pi }{8}}-{\frac  {4}{3}}]\,


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