= Solution
For an <injective> <holomorphic function>, the real <Jacobian determinant> is $|h'|^2$. The <area formula for a univalent power series>, obtained by <change of variables>, is therefore
$$
\operatorname{Area}(h(r\mathbb D))=\int_{|z|<r}|h'(z)|^2\,dA(z)=\boxed{\pi\sum_{n\geq1}n|c_n|^2r^{2n}}.
$$
Indeed, insert $h'(\rho e^{i\theta})=\sum_{n\geq1}nc_n\rho^{n-1}e^{i(n-1)\theta}$. <Orthogonality> eliminates the cross terms in the angular <integral>, and radial integration gives $2\pi n^2\int_0^r\rho^{2n-1}d\rho=\pi n r^{2n}$. The <power series> converges uniformly on each smaller <closed disk>, justifying the calculation for every $r<1$.
Since $|h(z)|^2=|\phi(z^2)|$, the <Koebe growth theorem> gives $|h(z)|\leq|z|/(1-|z|^2)$. Hence $h(r\mathbb D)$ lies in the <disk> of radius $r/(1-r^2)$, so
$$
\boxed{\operatorname{Area}(h(r\mathbb D))\leq\frac{\pi r^2}{(1-r^2)^2}}.
$$
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