= Solution
The original PDF has angle $\theta(t)=2\pi t^n$, with $n$ an exponent on $t$. It is not $2\pi nt$: each permitted value of $n$ changes the speed but traverses the same circle once counterclockwise. Along the path, $r^2=2$ and
$$
\boldsymbol r'(t)=\theta'(t)(-\sin\theta,\cos\theta,0),\qquad
\theta'=2\pi n t^{n-1}.
$$
The radial contribution to the <line integral> vanishes, since $(\cos\theta,\sin\theta,1)\cdot(-\sin\theta,\cos\theta,0)=0$. The rotational contribution gives
$$
\boldsymbol F(\boldsymbol r(t))\cdot\boldsymbol r'(t)=a\theta'(t).
$$
Therefore
$$
\boxed{\int_\Gamma\boldsymbol F\cdot d\boldsymbol r
=a[\theta(1)-\theta(0)]=2\pi a,\qquad n=1,2,\ldots.}
$$
The answer is independent of the monotone parametrization, as an oriented <line integral> should be.
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