= Oblique section of a circular cylinder
{title2=$a=R/\cos\theta,\quad b=R$}
A mathematical plane cutting an infinite <circular cylinder> at a nonzero angle to its axis produces an <ellipse>, unless it is parallel to that axis. More precisely, let the cylinder radius be $R>0$, and let $\theta$ be the acute angle between the cylinder axis and the plane's <normal vector>, with $0\leq\theta<\pi/2$. The <semi-major and semi-minor axes> and <area> are
$$
\boxed{a=\frac{R}{\cos\theta},\qquad b=R,\qquad A=\frac{\pi R^2}{\cos\theta}}.
$$
The angle is measured from the normal, not from the cutting plane itself. At $\theta=0$, the profile is a <circle>.
To prove the result, choose <Cartesian coordinates> so the cylinder is $x^2+y^2=R^2$, with axis along $z$. Translate the cut so it crosses the axis at $(0,0,h)$. After rotating about the axis if necessary, an <orthonormal basis> in the cutting plane is
$$
\mathbf e_1=(\cos\theta,0,\sin\theta),\qquad \mathbf e_2=(0,1,0).
$$
A point in the plane is $(0,0,h)+u\mathbf e_1+v\mathbf e_2$. Substitution into the cylinder equation gives $u^2\cos^2\theta+v^2=R^2$, or
$$
\frac{u^2}{(R/\cos\theta)^2}+\frac{v^2}{R^2}=1.
$$
Because the basis is orthonormal, $u$ and $v$ measure actual lengths within the plane. The <area of an ellipse> then gives the stated area.
For an original example, $R=2$ and $\theta=60^\circ$ give $a=4$, $b=2$ and $A=8\pi$, twice the perpendicular-profile area. As $\theta$ approaches $\pi/2$, the ellipse grows without bound. At the excluded parallel orientation, a plane through the axis instead meets the lateral cylinder in two generator lines; the limiting formula does not describe a finite ellipse there.
<Microtomy> applies this geometry to thin physical sections of a circular <material fibre>. Its worked example and figure show how an elongated profile can arise from <section orientation> without <deformation (mechanics)>. Applying the ideal lemma requires accounting for finite thickness, a finite specimen and departures from a straight circular cylinder.
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