= Solution
{c}
The rotation axis is $L=\{(x,1,0):x\in\mathbb R\}$. The distance of the generating <function graph> from this axis is $\sqrt x$, so a <parametrization> is
$$
\mathbf R(x,\phi)=(x,1+\sqrt x\cos\phi,\sqrt x\sin\phi),\qquad0<x<1.
$$
The <surface of revolution> therefore has equation $(y-1)^2+z^2=x$: it is a circular <elliptic paraboloid> opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the <surface area> formula, so
$$
\begin{aligned}
A&=2\pi\int_0^1\sqrt x\sqrt{1+\frac1{4x}}\,dx
=\pi\int_0^1\sqrt{4x+1}\,dx\\
&=\boxed{\frac\pi6(5\sqrt5-1).}
\end{aligned}
$$
The cancellation makes the <improper integral> finite at the vertex. Neither the missing vertex nor the missing boundary <circle> changes the <surface area>, and no cap is included.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iv-q3-revolution.png]
{title=The <elliptic paraboloid> obtained by rotating $y=1+\sqrt x$ around $L$}
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