Assuming there's a typo and it's the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1, and you do a complete revolution, I believe the answer should be zero. You can compute it directly from the line integral using x = a cos(t), y = b sin(t), but it's easier to use Green's theorem and integrate 2x + 2y over the interior of the ellipse, which obviously must yield zero by symmetry of the function over the region of integration.