= Solution
{c}
Use the <parametric curve>
$$
X=2\sqrt2\cos t,\quad Y=\sqrt2\sin t,\qquad x(t)=2\cos t+\sin t,\quad y(t)=-2\cos t+\sin t,\quad 0\leq t\leq2\pi.
$$
Here $|y(t)|\leq\sqrt5<3$, so the <ellipse> stays strictly above the <axis of rotation> $y=-3$. Rotating in three dimensions gives a <surface of revolution> with <parametrization>
$$
\mathbf s(t,\varphi)=\bigl(x(t),-3+(y(t)+3)\cos\varphi,(y(t)+3)\sin\varphi\bigr),\qquad 0\leq\varphi\leq2\pi.
$$
The two parameter <tangent vectors> are <orthogonal>, with <Euclidean norms> $\sqrt{x'(t)^2+y'(t)^2}$ and $y(t)+3$. Hence the <area element of a surface> is
$$
|\mathbf s_t\times\mathbf s_\varphi|=(y(t)+3)\sqrt{x'(t)^2+y'(t)^2}.
$$
The generating <ellipse> is simple and has positive distance from the <axis of rotation>; its <rotation> sweeps the <surface> once, apart from the identified parameter endpoints. Also
$$
x'(t)^2+y'(t)^2=(-2\sin t+\cos t)^2+(2\sin t+\cos t)^2=8\sin^2t+2\cos^2t.
$$
Writing $w(t)=\sqrt{8\sin^2t+2\cos^2t}$, the <surface area> is therefore $2\pi\int_0^{2\pi}(3+y(t))w(t)\,dt$. Since $y(t+\pi)=-y(t)$ and $w(t+\pi)=w(t)$,
$$
\int_0^{2\pi}y(t)w(t)\,dt=\int_0^\pi\bigl(y(t)w(t)+y(t+\pi)w(t+\pi)\bigr)\,dt=0.
$$
Consequently
$$
\boxed{\operatorname{Area}(S)=6\pi\int_0^{2\pi}\sqrt{8\sin^2t+2\cos^2t}\,dt.}
$$
The cancellation is an instance of the <surface area of revolution of a centrally symmetric curve>: the <surface area> is the generating curve’s <arc length> times the distance $6\pi$ traveled by its center.
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