= Solution
The new <cone> can be parametrized by
$$
\mathbf R'(r,\theta)=(r\cos\theta,b(1-r/a),r\sin\theta).
$$
The outward <vector area element> is $r(\tfrac ba\cos\theta,1,\tfrac ba\sin\theta)\,dr\,d\theta$. This is $\mathbf R'_\theta\times\mathbf R'_r\,dr\,d\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its <dot product> with $\mathbf F(\mathbf R')$ is again $br\,dr\,d\theta$. Hence
$$
\boxed{\int_{C'}\mathbf F\cdot d\mathbf S=\pi a^2b.}
$$
This equality also follows because a rigid <rotation> takes the first <cone> into the second and leaves the radial <vector field> $\mathbf F(\mathbf x)=\mathbf x$ invariant.
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