# VC5.12

Solution Verify Green's theorm in plane for $\oint _{C}[(3x^{2}-8y^{2})dx+(4y-6xy)dy]\,$ where C is the region bounded by the parabolas $y^{2}=x,y=x^{2}\,$

By Green's theorm,$\oint _{C}[(3x^{2}-8y^{2})dx+(4y-6xy)dy]=\oint _{R}[{\frac {\partial }{\partial x}}(4y-6xy)-{\frac {\partial }{\partial y}}(3x^{2}-8y^{2})]dxdy\,$

Where R is the region bounded by $y^{2}=x,y=x^{2}\,$.The bounding curve C is closed curve OAO where O is the origin and A(1,1).Now,on $y=x^{2}\,$,dy=2xdx and x varies from 0 to 1 and on $x=y^{2}\,$ dx=2ydy and y varies from 1 to 0.

LHS=$\int _{{x=0}}^{{1}}[(3x^{2}-8y^{2})dx+(4y-6xy)dy]+\int _{{y=1}}^{{0}}[(3x^{2}-8y^{2})dx+(4y-6xy)dy]\,$

=$\int _{0}^{1}[(3x^{2}-8x^{4})dx+(4x^{2}-6x^{3})(2xdx)]+\int _{1}^{0}[3y^{4}-8y^{2})(2ydy)+(4y-6y^{3})dy]\,$

=${\frac {3}{2}}\,$ on simplification of the above.

Again,RHS=$\oint _{R}(-6y+16y)dxdy=10\int _{0}^{1}\int _{{y=x^{2}}}^{{y={\sqrt {x}}}}ydxdy\,$

=$10\int _{0}^{1}[{\frac {y^{2}}{2}}]_{{x^{2}}}^{{{\sqrt {x}}}}dx=10\int _{0}^{1}{\frac {1}{2}}(x-x^{4})dx\,$

=${\frac {3}{2}}\,$ on simplification.

Therefore,LHS=RHS,hence the theorm is verified.