VC4.2

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Since the integration is performed in the xy-plane,we have r=xi+yj\, and dr=dx i+dy j\,.

The path of integration C consists of straight lines OA,AB,BC and CO.Here the coordinates of A,B,C are (0,a),(a,b),(0,b)respectively.

Now,we have \int_{C} F\cdot\,dr=\int_{C} [(x^2+y^2)i+2xyj]\cdot(dx i+dy j)=\int_{C} [(x^2+y^2)dx-2xy dy]\,

Here, on OA,y=0,dy=0 and x varies from 0 to a. on AB,x=a,dx=0 and y varies from 0 to b. on BC,y=b,dy=0 and x varies from a to 0. on CO,x=dx=0 and y varies from b to 0.

Using the above integration,we have

\int_{C} F\cdot\,dr=\int_{0}^{a} x^2\,dx+\int_{0}^{b} (-2ay)\,dy+\int_{a}^{0} (x^2+b^2)\,dx+\int_{b}^{0} 0\,dy\,

=[\frac{x^3}{3}]_{0}^{a}-2a[\frac{y^2}{2}]_{0}^{b}+[\frac{x^3}{3}+b^2x]_{a}^{0}+0=\frac{a^3}{3}-ab^2-\frac{a^3}{3}-b^2a=-2ab^2\,

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