Stokes theorem equates the circulation of a continuously differentiable vector field around an oriented boundary to the flux of its curl across a spanning oriented surface:

\oint_{\partial S} A \cdot d r = \int_{S} (\nabla \times A) \cdot \hat{n} d S .

The boundary direction follows the right-hand rule relative to the normal vector. For the polynomial vector field, only the

z

component of the curl can be nonzero, and

\nabla \times A = (\frac{\partial ( 2 x y )}{\partial x} - \frac{\partial ( x ^{2} - y ^{2} )}{\partial y}) \hat{z} = 4 y \hat{z} .

b

Solution

+ \hat{z}

. The surface integral of the curl is an ordinary planar integral:

I = \int_{0}^{a} \int_{0}^{b} 4 y d y d x = 2 a b^{2} .

If the opposite orientation is chosen, both this flux and the corresponding boundary line integral change sign.

For positive

a, b

and upward orientation, traverse the four edges counterclockwise. The bottom edge contributes

\int_{0}^{a} x^{2} d x = a^{3} /3

. The right edge contributes

\int_{0}^{b} 2 a y d y = a b^{2}

. On the top edge

x

decreases, giving

\int_{a}^{0} (x^{2} - b^{2}) d x = - a^{3} /3 + a b^{2}

. The left edge has

x = 0

\oint_{\partial S} A \cdot d r = \frac{a ^{3}}{3} + a b^{2} - \frac{a ^{3}}{3} + a b^{2} = 2 a b^{2} .

This agrees with the surface integral and explicitly verifies Stokes theorem, including the direction of the top-edge traversal.

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