Solution
= Solution
Parametrize the <circle> by $\mathbf r(t)=(\cos t,\sin t)$, $0\le t\le2\pi$. Its tangent is $\mathbf r'=(-\sin t,\cos t)=\mathbf F(\mathbf r(t))$, so the <dot product> $\mathbf F\cdot\mathbf r'$ is one. For the other field,
$$
\mathbf G\cdot\mathbf r'=2\sin t\cos t(\cos^2t-\sin^2t)=\frac12\sin4t.
$$
Integrating around the full <circle> gives
$$
\boxed{\oint_{\rm circle}\mathbf F\cdot d\mathbf r=2\pi,\qquad
\oint_{\rm circle}\mathbf G\cdot d\mathbf r=0.}
$$