VC4.1

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If F=3xyi-y^{2}j\,,Evaluate \int _{{C}}F\cdot dr\,\,,where C is the curve y=2x^{2}\, in the xy-plane from (0,0)to (1,2).

Since the path of integration C lies in the xy-plane (z=0),so we take r=xi+yj\,.

Take x=t,so that y=2t^{2}\, and hence the parametric equations of curve C are x=t,y=2t^{2}\,. Again the given points (0,0) and (0,1) correspond to t=0 and t=1 respectively.

Sice r=xi+yj\,

dr=idx+jdy=(i+4tj)dt\,

Therefore, \int _{{C}}F\cdot \,dr=\int _{{0}}^{{1}}(6t^{3}i-4t^{4}j)\cdot (i+4tj)\,dt\,

=\int _{{0}}^{{1}}(6t^{3}-16t^{5})\,dt=\left[{\frac  {3}{2}}t^{4}-{\frac  {8}{3}}t^{6}\right]_{{0}}^{{1}}=-{\frac  {7}{6}}\,

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