= Solution
{c}
The endpoint <radii> are $5$ and $13$, but \b[substituting those two numbers in the finite <conical frustum> formula is invalid]. A rotation about the $z$-axis preserves the horizontal <angle> between the endpoint <position vectors>; here $(3,4)$ and $(12,5)$ are not parallel. The <radius> along the generating segment is not linear in height.
Parametrize the actual segment by
$$
(x,y,z)=(3+9t,4+t,1+6t),\qquad0\leq t\leq1.
$$
Its rotated <circle> at height $1+6t$ has <radius>
$$
\rho(t)=\sqrt{(3+9t)^2+(4+t)^2}=\sqrt{25+62t+82t^2},\qquad
\rho'(t)=\frac{31+82t}{\rho(t)}.
$$
Thus the rotated image is the <surface of revolution> $\mathbf R(t,\theta)=(\rho(t)\cos\theta,\rho(t)\sin\theta,1+6t)$. Applying the <conical frustum> formula to infinitesimal pieces of this radial profile, or directly using the <cross product> as in <3/b/ii>, gives the <surface area of a surface of revolution>
$$
\begin{aligned}
\mathcal A
&=2\pi\int_0^1\rho(t)\sqrt{\rho'(t)^2+36}\,dt\\
&=2\pi\int_0^1\sqrt{9676t^2+7316t+1861}\,dt.
\end{aligned}
$$
To evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. Define
$$
G(u)=u\sqrt{59u^2+19602}+\frac{19602}{\sqrt{59}}\log\bigl(\sqrt{59}\,u+\sqrt{59u^2+19602}\bigr).
$$
The <chain rule> gives $G'(u)=2\sqrt{59u^2+19602}$, so the answer is
$$
\boxed{\mathcal A=\frac{\pi}{82\sqrt{41}}\bigl(G(113)-G(31)\bigr)\simeq561.8721076764.}
$$
The positive <logarithm> arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical <plane> before joining them would instead create a different surface, with <surface area> $180\pi\simeq565.4867$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-q3-swept-segment.png]
{title=The rotated segment has a curved radial profile, although its three-dimensional generator is straight.}
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