VC4.17

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

Let C denote the arc of the parabola y^{2}=x\, from the point (0,0) to the point (1,1).The parametric equations of y^{2}=x\, are

x=t^{2},y=t\,

Hence at the point (0,0),t=0 and at the point (1,1),t=1.

Therefore,the required work done are

\int _{C}F\cdot \,dr=\int _{C}[(x^{2}-y^{2}+x)i-(2xy+y)j]\cdot (dxi+dyj)\,

=\int _{C}[(x^{2}+y^{2}+x)dx-(2xy+y)dy]=\int _{{t=0}}^{{1}}[(x^{2}-y^{2}+x){\frac  {dx}{dt}}-(2xy+y){\frac  {dy}{dt}}]\,dt\,

=\int _{0}^{1}[(t^{4}-t^{2}+t^{2})(2t)-(2t^{3}+t)(1)]\,dt\,

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


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