Consider the vector field F=$\langle a x + b y , c x + d y \rangle$. Show that F has zero circulation on any oriented circle centered at the origin, for any a, b, c, and d, provides b=c.

The force field is

$$F(x,y)=(ax+by,cx+dy).$$

The curve is

$$r(t)=(\alpha \cos t,\alpha \sin t),\ \forall t \in [0,2\pi ]\Rightarrow r'(t)=(-\alpha \sin t,\alpha \cos t).$$

Thus, the integral is

$$\begin{align*}\int _C F\cdot Tds&=\int _0^{2\pi }F(r(t))r'(t)dt\\&=\int _0^{2\pi }(a\alpha \cos t+b\alpha \sin t, c\alpha \cos t+d\alpha \sin t)\cdot (-\alpha \sin t,\alpha \cos t)dt\\&=\alpha ^2\int _0^{2\pi}(-b\sin ^2t+(-a+d)\sin t\cos t+c\cos ^2t)dt\\&=\alpha ^2\left (\dfrac{-bt+b\sin t\cos t}{2}+\dfrac{a-d}{4}\cdot \cos (2t)+\dfrac{ct+c\sin t\cos t}{2}\right )\Big |_0^{2\pi }\\&=\alpha ^2\pi (-b+c).\end{align*}$$

Provided $b=c$, this result equals $0$.