Let  \(I = \mathop \oint \limits_C \left( {{x^2}y\;dy - {y^2}x\;dx} \right)\), where C is the boundary of square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Then I equals

Concept:

By using Greens theorem:

 \(\mathop \oint \limits_C^{} Mdx + Ndy\) = \(\;\mathop \int\!\!\!\int \limits_R^{} \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

Calculation:

Given:

\(I = \mathop \oint \limits_C \left( {{x^2}y\;dy - {y^2}x\;dx} \right)\)

 where C is the boundary of square  0 ≤ x ≤ 1, 0 ≤ y ≤ 1

\(\mathop \oint \limits_C \left( {{x^2}y\;dy - {y^2}x\;dx} \right)\) = \(\;\mathop \int\!\!\!\int \limits_R^{} \left( {\frac{{∂ N}}{{∂ x}} - \frac{{∂ M}}{{∂ y}}} \right)dxdy\)

∂N/∂x = 2xy, ∂M/∂y = -2xy

using the values in the Greens theorem :

\(\;\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {2xy + 2xy} \right)dxdy\) = \( \smallint \limits_{x = 0}^1\smallint \limits_{y = 0}^1 \left( {4xy} \right)dxdy\)

\(\;\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {2xy + 2xy} \right)dxdy\) =\(\;4\mathop \smallint \limits_{x = 0}^1 x\left[ {\frac{{{y^2}}}{2}} \right]_0^1dx\)

=\(\;2\mathop \smallint \limits_{x = 0}^1 xdx\)

=\(\;2\left[ {\frac{{{x^2}}}{2}} \right]_0^1\)

= 1