How do I find the lateral surface area of an elliptical cone? (not a frustum)

Question

The formula for the lateral surface area of a right circular cone is $\pi*r*l$ where $r$ is the circle's radius.

But ellipse has a semi-major axis as well as a semi-minor one.

Which one of them should be implemented in the formula and how?

Answer 1

If your cone is a section of a circular right cone, then a simple answer is possible.

Let $\alpha$ be the semi-aperture of the cone, $\beta$ the angle between the axis of the cone and the normal to the intersecting plane, $q$ the distance of the vertex from the intersection between the plane and the axis of the cone.

A parametric equation of the conic surface is then: $$ x=t\sin\alpha\cos\phi,\quad y=t\sin\alpha\sin\phi,\quad z=t\cos\alpha, $$ where $t$ is the distance from vertex $(0,0,0)$ and $\phi$ is the azimuthal angle about $z$-axis (which is the axis of the cone).

A suitable equation for the plane is simply $z=(\tan\beta) x+q$, and substituting here $x$ and $z$ from above one gets: $$ t={q\over \cos\alpha-\tan\beta\sin\alpha\cos\phi}, $$ which is the distance from the vertex to a generic point $P$ on the ellipse at the base of the cone.

The area of the lateral surface is then: $$ A=\int_0^{2\pi}\int_0^t r\sin\alpha\,dr\,d\phi= {q^2\sin\alpha\over2\cos^2\alpha} \int_0^{2\pi}{d\phi\over (1-\tan\alpha\tan\beta\cos\phi)^2}= {\pi q^2\sin\alpha\over\cos^2\alpha} {1\over(1-\tan^2\alpha\tan^2\beta)^{3/2}}. $$