= Solution
Compute the <flux> directly using the outward <vector area element>, as required. Restrict the <vector field> to the surface <parametric equation> and take the <dot product>:
$$
\mathbf F(\mathbf R)\cdot d\mathbf S
=\left[\frac ba r^2(\cos^2\theta+\sin^2\theta)+br\left(1-\frac ra\right)\right]\,dr\,d\theta
=br\,dr\,d\theta.
$$
The terms quadratic in $r$ cancel. Therefore the <surface integral> is
$$
\boxed{\int_C\mathbf F\cdot d\mathbf S=\int_0^{2\pi}\int_0^a br\,dr\,d\theta=\pi a^2b.}
$$
This is a direct evaluation of the <flux of the position vector through a cone>.
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