Answer to Question #235251 in Calculus for Fozia Sayda

Question

Verify the green,s theorem in the plane for $\oint$ (2xy-x²)dx+(x+y²)dy where c is the closed curve of the region bounded by y=x² and y²=x .

Answer 1

Plot the curve:

$\oint {(2xy - {x^2})dx + (x + {y^2}} )dy = \int\limits_{{L_2}} {(2xy - {x^2})dx + (x + {y^2})dy + } \int\limits_{{L_1}} {(2xy - {x^2})dx + (x + {y^2})dy}$

${L_1}:x = {y^2} \Rightarrow dx = 2ydy \Rightarrow \int\limits_{{L_1}} {(2xy - {x^2})dx + (x + {y^2})dy} = \int\limits_1^0 {\left( {2{y^2}y - {{\left( {{y^2}} \right)}^2}} \right)} 2ydy + \left( {{y^2} + {y^2}} \right)dy = \int\limits_1^0 {\left( {4{y^4} - 2{y^5} + 2{y^2}} \right)} dy = \frac{4}{5}\left. {{y^5}} \right|_1^0 - \frac{1}{3}\left. {{y^6}} \right|_1^0 + \frac{2}{3}\left. {{y^3}} \right|_1^0 = - \frac{4}{5} + \frac{1}{3} - \frac{2}{3} = - \frac{{17}}{{15}}$

${L_2}:y = {x^2} \Rightarrow dy = 2xdx \Rightarrow \int\limits_{{L_2}} {(2xy - {x^2})dx + (x + {y^2})dy} = \int\limits_0^1 {\left( {2{x^3} - {x^2}} \right)} dx + (x + {x^4})2xdx = \int\limits_0^1 {\left( {2{x^3} - {x^2} + 2{x^2} + 2{x^5}} \right)dx} = \frac{1}{2}\left. {{x^4}} \right|_0^1 + \frac{1}{3}\left. {{x^3}} \right|_0^1 + \frac{1}{3}\left. {{x^6}} \right|_0^1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{3} = \frac{7}{6}$

Then

$\oint {(2xy - {x^2})dx + (x + {y^2}} )dy = \frac{7}{6} - \frac{{17}}{{15}} = \frac{1}{{30}}$

$\frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {2xy - {x^2}} \right) = 2x;\,\,\,\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {x + {y^2}} \right) = 1$

Then

$\int {\int\limits_D {\left( {\frac{\partial }{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy} } = \int\limits_0^1 {dx} \int\limits_{{x^2}}^{\sqrt x } {\left( {1 - 2x} \right)dy} = \int\limits_0^1 {\left( {1 - 2x} \right)\left. y \right|_{{x^2}}^{\sqrt x }dx} = \int\limits_0^1 {\left( {1 - 2x} \right)\left( {\sqrt x - {x^2}} \right)dx} = \int\limits_0^1 {\left( {\sqrt x - 2{x^{\frac{3}{2}}} - {x^2} + 2{x^3}} \right)} dx = \frac{2}{3}\left. {{x^{\frac{3}{2}}}} \right|_0^1 - \frac{4}{5}\left. {{x^{\frac{5}{2}}}} \right|_0^1 - \frac{1}{3}\left. {{x^3}} \right|_0^1 + \frac{1}{2}\left. {{x^4}} \right|_0^1 = \frac{2}{3} - \frac{4}{5} - \frac{1}{3} + \frac{1}{2} = \frac{1}{{30}}$

So, $\int {\int\limits_D {\left( {\frac{\partial }{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy} } = \oint {Pdx + Qdy}$ then Green's theorem is valid.