VC4.1

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If F=3xyi-y^2j\,,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=i dx+j dy=(i+4t j)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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