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Here the parabola y^{2}=8x\, and the straight line x=2 intersect at the point say A(2,4) and B(2,-4). Thus the closed curve C consists of straight line BA and AOB of the parabola y^{2}=8x\,. Let R be the region bounded by C. Then,by Green's theorm in plane,we have

\oint _{C}[(x^{2}-2xy)dx+(x^{2}y+3)dy]=\iint _{R}[{\frac  {\partial }{\partial x}}(x^{2}y+3)-{\frac  {\partial }{\partial y}}(x^{2}-2xy)]dxdy\,

RHS of (1) =\iint _{R}(2xy+2x)dxdy=2\int _{{x=0}}^{{2}}\int _{{y=-{\sqrt  {8x}}}}^{{{\sqrt  {8x}}}}x(y+1)dxdy\,

=2\int _{0}^{2}x[{\frac  {y^{2}}{2}}+y]_{{y=-{\sqrt  {8x}}}}^{{{\sqrt  {8x}}}}dx\, on integrating w.r.t. y.

=2\int _{0}^{2}x[{\frac  {1}{2}}(8x-8x)+{\sqrt  {8x}}+{\sqrt  {8x}}]dx\,

=4\int _{0}^{2}{\sqrt  {8}}x^{{{\frac  {3}{2}}}}dx=8{\sqrt  {2}}[{\frac  {2x^{{{\frac  {5}{2}}}}}{5}}]_{0}^{2}={\frac  {16}{5}}{\sqrt  {2}}2^{{{\frac  {5}{2}}}}={\frac  {128}{5}}\, --(2)

Now,LHS=\int _{{AOB}}[(x^{2}-2xy)dx+(x^{2}y+3)dy]+\int _{{BA}}[(x^{2}-2xy)dx+(x^{2}y+3)dy]\,

=\int _{{y=4}}^{{-4}}[({\frac  {y^{4}}{64}}-2{\frac  {y^{2}}{8}}y){\frac  {1}{4}}ydy+({\frac  {y^{2}}{64}}y+3)dy]+\int _{{-4}}^{{4}}(4y+3)dy\,

=\int _{{-4}}^{{4}}[{\frac  {y^{5}}{256}}-{\frac  {y^{4}}{16}}+{\frac  {y^{3}}{64}}+3]dy+\int _{{-4}}^{{4}}(4y+3)dy\,

=-2\int _{0}^{4}[3-{\frac  {y^{4}}{16}}]dy+2(16-16)+3(4+4)\, [Note that {\frac  {y^{5}}{256}}+{\frac  {y^{3}}{64}}\, and (3-{\frac  {y^{4}}{16}})\, are respectively odd and even function of y]

=-2[3y-{\frac  {1}{16}}{\frac  {y^{5}}{5}}]_{0}^{4}+24=-2[12-{\frac  {4^{5}}{80}}]+24\,

={\frac  {128}{5}}\, on simplification. --(3)

From (2) and (3) LHS = RHS,hence Green's theorm is verified.

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