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Volume of an Elliptical Cone

```
Date: 10/15/98 at 12:33:51
From: francesca
Subject: Volume for elliptical cone

Can you enlighten me on how to find the volume for an elliptical cone?

I've tried many approximations. I'm looking for an equation or analysis
by integration.

Thanks,
Francesca
```

```
Date: 10/16/98 at 08:22:03
From: Doctor Rob
Subject: Re: Volume for elliptical cone

If it is a right elliptical cone, i.e., the elliptical base is
perpendicular to the axis, then the formula is:

V = Pi*a*b*h/3

where V is the volume, h is the height, and a and b are the semi-major
and semi-minor axes of the elliptical base.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/16/98 at 15:39:45
From: francesca weaver
Subject: Re: Volume for elliptical cone

Could you show or explain, how you would begin to solve for the volume
through integration (triple integral)? Just the initial setup - need
not solve.

Yesterday, I ended up solving for it through geometric properties,
using an assumption that was valid in my case. I considered your exact
equation by modifying that of the cone but had no proof/resource to
check it against.

Thanks much,
Francesca
```

```
Date: 10/16/98 at 16:11:50
From: Doctor Rob
Subject: Re: Volume for elliptical cone

The equation of the cone is x^2/a^2 + y^2/b^2 - z^2/h^2 = 0, for
0 <= z <= h, where the origin is at the vertex, the z-axis is the axis
of the cone, the plane z = h is the plane of the base, the semi-major
axis of  the elliptical base is the line segment z = h, y = 0,
0 <= x <= a, and the semi-minor axis of the elliptical base is the
line segment z = h, x = 0, 0 <= y <= b. The volume of the cone is then
the triple integral:

V = Int{Int[Int(1)dz]dy}dx

where the limits of integration are:

-a <= x <= a
-b*sqrt(a^2-x^2)/a <= y <= b*sqrt(a^2-x^2)/a
h*sqrt(x^2/a^2+y^2/b^2) <= z <= h

The innermost integral can be done immediately. The remaining double
integral can be simplified a little by using the symmetry of the
ellipse, to:

V = 4*h*Int{Int[1-sqrt(x^2/a^2+y^2/b^2)]dy}dx

where the limits of integration are:

0 <= x <= a,
0 <= y <= b*sqrt(a^2-x^2)/a

This is how I would set it up. Mathematica(TM) concurs that, when

V = Pi*a*b*h/3

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus
College Higher-Dimensional Geometry
High School Calculus
High School Higher-Dimensional Geometry

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