 Solution

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe 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. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe 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.