 Solution

For positive $a,b$ and upward orientation, traverse the four edges counterclockwise. The bottom edge contributes ${\int }_0^a{x}^{2}dx=a^3/3$. The right edge contributes ${\int }_0^{b}2aydy=a{b}^2$. On the top edge $x$ decreases, giving ${\int }_a^0(x^2-b^2)dx=-a^3/3+a{b}^2$. The left edge has $x=0$ and zero tangential component. HenceThis agrees with the surface integral and explicitly verifies Stokes theorem, including the direction of the top-edge traversal.