VC4.9

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Let C1 denote the curve made up of the straight lines AB,BC and CA. Since the integration is performed in the xy-plane,we have r=xi+yj,dr=dxi+dyj+dzk\,. Then the required integral around the triangle ABC is given by

\int _{{C_{1}}}F\cdot \,dr=\int _{{C_{1}}}[(2x^{2}+y^{2})i+(3y-4x)j]\cdot (dxi+dyj)\,

=\int _{{C_{1}}}[(2x^{2}+y^{2})dx+(3y-4x)dy]\,

=\int _{{AB}}[(2x^{2}+y^{2})dx+(3y-4x)dy]+\int _{{BC}}[(2x^{2}+y^{2})dx+(3y-4x)dy]+\int _{{CA}}[(2x^{2}+y^{2})dx+(3y-4x)dy]\,

Here on AB,y=0,dy=0 and x varies from 0 to 2. On BC,x=2,dx=0 and y varies from 0 to 1. Now,equation of the line CA is y-0={\frac  {1-0}{2-0}}x,y={\frac  {x}{2}},

Therefore,on CA,dx=2dy\, and y varies from 1 to 0. Using the above,

\int _{{C_{1}}}F\cdot \,dr=\int _{{0}}^{{2}}2x^{2}dx+\int _{{0}}^{{1}}(3y-8)dy+\int _{{0}}^{{1}}[2(2y)^{2}+y^{2}]2dy+(3y-8ydy)\,

=2[{\frac  {x^{3}}{3}}]_{{0}}^{{2}}+[{\frac  {3y^{2}}{2}}-8y]_{{0}}^{{1}}+\int _{{1}}^{{0}}(18y^{2}-5y)dy\,

={\frac  {16}{3}}+{\frac  {3}{2}}-8+[6y^{3}-{\frac  {5y^{2}}{2}}]_{{1}}^{{0}}={\frac  {16}{3}}+{\frac  {3}{2}}-8-6+{\frac  {5}{2}}=-{\frac  {14}{3}}\,

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