Question

Consider the vector field F=ax+by,cx+dy\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.

Solution

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The force field is

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

The curve is

r(t)=(αcost,αsint), t[0,2π]r(t)=(αsint,αcost).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

CFTds=02πF(r(t))r(t)dt=02π(aαcost+bαsint,cαcost+dαsint)(αsint,αcost)dt=α202π(bsin2t+(a+d)sintcost+ccos2t)dt=α2(bt+bsintcost2+ad4cos(2t)+ct+csintcost2)02π=α2π(b+c).\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=cb=c, this result equals 00.

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