Table of Contents

Last modified on August 3rd, 2023

An elliptic cone is a cone with an elliptical cross-section. It has a directrix, which is an ellipse. Such a cone is different from the standard circular cone (with a circular cross-section) in shape.

It is thus one of the quadric surfaces with traces composed of conic sections.

An elliptic cone is also called a degree two cone as it has two angles at the vertex.

The parametric equations for an elliptic cone of height h, semi-major axis a, and semi-minor axis b are:

- ${x=a\dfrac{h-u}{h}\cos v}$
- ${x=b\dfrac{h-u}{h}\sin v}$
- z = u

where v âˆˆ [0, 2Ï€] and u âˆˆ [0, h]

When the coordinates line are the curvature lines of the cone, such that â€˜aâ€™ represents the x- axis, â€˜bâ€™ represents the y-axis, â€˜câ€™ represents the z-axis, and if a â‰¥ b, then:

- ${x=\dfrac{h}{\sqrt{a^{2}+h^{2}}\sqrt{b^{2}+h^{2}}}uv}$
- ${y=\dfrac{a}{\sqrt{a^{2}-b^{2}}\sqrt{a^{2}+h^{2}}}u\sqrt{a^{2}+h^{2}}-v^{2}}$
- ${z=\dfrac{a}{\sqrt{a^{2}-b^{2}}\sqrt{b^{2}+h^{2}}}u\sqrt{v^{2}-b^{2}-h^{2}}}$

An elliptic cone is a quadratic ruled surface, and has volume calculated as:

Volume ${\left( V\right) =\dfrac{1}{3}\pi abh}$, here a = major axis between x = 0 and x = a, b = minor axis between y = 0 and y = b, h = height between z = 0 and z = h, Ï€ = 3.141

**Find the volume of an elliptic cone with a major axis of 6 units, a minor axis of 4 units, and a height of 8 units.**

Solution:

As we know,

Volume ${\left( V\right) =\dfrac{1}{3}\pi abh}$, here a = 6, b = 4, h = 8, Ï€ = 3.141

${=\dfrac{1}{3}\times 3\cdot 141\times 6\times 4\times 8}$

= 201.06 cubic units

To find the lateral surface area (S) of an elliptic cone we first need the coefficients of the first and second fundamental form.

The coefficients of the first fundamental form are:

- ${E=\dfrac{h^{2}+a^{2}\cos ^{2}v+b^{2}\sin ^{2}v}{h^{2}}}$
- ${F=\dfrac{\left( a^{2}-b^{2}\right) \left( h-u\right) \cos v\sin v}{h^{2}}}$
- ${G=\dfrac{\left( h-u\right) ^{2}\left( a^{2}\sin ^{2}v+b^{2}\cos ^{2}v\right) }{h^{2}}}$

The coefficients of the second fundamental form are:

- e = 0
- f = 0
- ${g=\dfrac{\sqrt{2}ab\left( h-u\right) }{\sqrt{2a^{2}b^{2}+\left( a^{2}+b^{2}\right) h^{2}+\left( b^{2}-a^{2}\right) h^{2}\cos 2v}}}$

Thus, the lateral surface area (LSA) of an elliptical cone is calculated as:

${S=\int ^{2\pi }_{0}\int ^{h}_{0}\sqrt{EG-F^{2}}dudv}$

When a > b

- ${2a\sqrt{b^{2}+h^{2}}E\sqrt{\dfrac{1-\dfrac{b^{2}}{a^{2}}}{1+\dfrac{b^{2}}{h^{2}}}}}$

When a < b

- ${2b\sqrt{a^{2}+h^{2}}E\sqrt{\dfrac{1-\dfrac{a^{2}}{b^{2}}}{1+\dfrac{a^{2}}{h^{2}}}}}$

Assuming 0 < b < a, where E(k) is a complete elliptic integral of the second kind

The total surface area (TSA) of an elliptical cone is given as:

- TSA = LSA + Ï€ab

**Calculate the lateral and total surface area of an elliptic cone having height 5 units, semi-major axis 3 units, and semi-minor axis 2 units. Round your answers to the nearest decimal places.**

Solution:

Here, a = 3 unit > b = 2 units

Thus,

LSA = ${2a\sqrt{b^{2}+h^{2}}E\sqrt{\dfrac{1-\dfrac{b^{2}}{a^{2}}}{1+\dfrac{b^{2}}{h^{2}}}}}$, here h = 5 units, a = 3 units, b = 2 units

Â = 44 units

Again,

TSA = LSA + Ï€ab

= 63 units

Last modified on August 3rd, 2023