Para calcularmos a função usando o teorema de Green, realizaremos os cálculos abaixo:

\(\begin{align} & f\left( x,y \right)=\left[ x\text{ }+2y{}^\text{2}\text{ }x{}^\text{2}\text{ }+3y \right] \\ & \\ & \int_{{}}^{{}}{f(x,y)dx+g(x,y)dy=\int_{{}}^{{}}{\int_{R}^{{}}{\left( \frac{dg}{dx}-\frac{df}{dy} \right)}}} \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\oint\limits_{C}{x\text{ }+2y{}^\text{2},\text{ }x{}^\text{2}\text{ }+3y} \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\int_{0}^{1}{\int_{0}^{1}{\left( x\text{ }+2y{}^\text{2}-\text{ }x{}^\text{2}-3y \right)}}dxdy \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\int_{0}^{1}{\frac{{{x}^{2}}}{2}+}2x{{y}^{2}}-\frac{{{x}^{3}}}{3}-3yx \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\int_{0}^{1}{\frac{1}{2}+}{{y}^{2}}-\frac{1}{3}-3y \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\left( \frac{1}{2}y+\frac{{{y}^{3}}}{3}-\frac{1}{3}y \right)_{0}^{1} \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\frac{1}{2}+\frac{1}{3}-\frac{1}{3} \\ & \int_{C}^{{}}{f(x,y)dx+g(x,y)dy=}\frac{1}{2} \\ \end{align}\ \)