Evaluate $ \int_{C} xy\,ds $ where C is the arc of the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ in the first quadrant.
Let $x = a\cos t$ and $ y= b\sin t$ and use a substitution of $ u = a^2 \sin^2t + b^2\cos^2t $ to simply the expression under the sqrt root when dealing with $ ds$. I eventually get to $\frac{ab}{2(a^2-b^2)} [\frac{2}{3} \sqrt{(a^2\sin^2t + b^2 \cos^2t)^3}]$evaluated between $\pi/2$ and $0$.
Should I take $\sqrt{a^6} = a^3$ here? (my thoughts being that $ a>0 $ in first quadrant)