VC5.12

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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.

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