= Solution
The sloping height relation on the <cone> is $z=b(1-r/a)$. A <parametric equation> of its curved surface is therefore
$$
\mathbf R(r,\theta)=(r\cos\theta,r\sin\theta,b(1-r/a)),\qquad0\le r\le a,\quad0\le\theta<2\pi.
$$
Its two <tangent vectors> are $\mathbf R_r=(\cos\theta,\sin\theta,-b/a)$ and $\mathbf R_\theta=r(-\sin\theta,\cos\theta,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their <cross product> is the <vector area element>:
$$
\boxed{d\mathbf S=(\mathbf R_r\times\mathbf R_\theta)\,dr\,d\theta
=r\left(\frac ba\cos\theta,\frac ba\sin\theta,1\right)\,dr\,d\theta.}
$$
Its radial component is positive, so this <normal vector> points outwards from the solid <cone>. Taking its <norm> gives the scalar <area element>
$$
\boxed{dS=r\sqrt{1+\frac{b^2}{a^2}}\,dr\,d\theta.}
$$
The <parametric equation> degenerates at the apex, but that single point contributes no <area>.
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