= Solution
For the anticlockwise arc use $x=\cos\theta$, $y=\sin\theta$, $0\le\theta\le\pi/2$. Since $dx=-\sin\theta\,d\theta$ and $dy=\cos\theta\,d\theta$, the <line integral> becomes
$$
I_3=\int_0^{\pi/2}\left[-\alpha(\cos^2\theta\sin\theta+
\cos\theta\sin^2\theta)+\beta\cos\theta\right]d\theta.
$$
The two integrals multiplying $-\alpha$ each equal $1/3$, by substituting $\cos\theta$ and $\sin\theta$ respectively. The remaining integral equals one. Thus
$$
\boxed{I_3=\beta-\frac{2\alpha}{3}}.
$$
Back to article page
