= Solution
Align the polar axis with the <magnetic dipole moment>. The <magnetic dipole field> has
$$
B_r=\frac{2m\cos\theta}{r^3},\qquad B_\theta=\frac{m\sin\theta}{r^3}.
$$
The <magnetic-field-line equation> gives $dr/(r\,d\theta)=2\cot\theta$, hence the <dipole magnetic-field line> is $r=C\sin^2\theta$. Normalize the loaded bundle by its small surface angular radius $\theta_\star$ at $r=R_\star$. Its boundary obeys
$$
\frac{\sin^2\theta(r)}{\sin^2\theta_\star}=\frac r{R_\star},
\qquad
\theta(r)^2\simeq\theta_\star^2\frac r{R_\star}.
$$
The <dipolar flux-tube area> for one polar cap is consequently
$$
\boxed{A(r)\simeq\pi\theta_\star^2\frac{r^3}{R_\star}.}
$$
Equivalently, $BA$ is constant by conservation of <magnetic flux> and $B\propto r^{-3}$ near the axis. Two equal loaded caps double the total area. The $r^3$ scaling is independent of the normalization; if the loading angle is specified at another radius $r_L$, use $A(r)=\pi\theta_L^2r^3/r_L$ instead.
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