= Solution
{c}
For the <parametrized surface> in <3/b/i>,
$$
\begin{aligned}
\mathbf R_t&=((c-a)\cos\theta,(c-a)\sin\theta,d-b),\\
\mathbf R_\theta&=(-\rho\sin\theta,\rho\cos\theta,0).
\end{aligned}
$$
These <tangent vectors> are <orthogonal>, and their <Euclidean norms> are $\sqrt{(c-a)^2+(d-b)^2}$ and $\rho$, respectively. Their <cross product> therefore has <Euclidean norm> $\rho\sqrt{(c-a)^2+(d-b)^2}$. Integration gives
$$
\begin{aligned}
\operatorname{Area}(\Sigma_1)
&=2\pi\sqrt{(c-a)^2+(d-b)^2}\int_0^1(a+(c-a)t)\,dt\\
&=\boxed{\pi(a+c)\sqrt{(a-c)^2+(b-d)^2}.}
\end{aligned}
$$
This is the lateral <surface area> of a <conical frustum>, with no end disks included.
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