# Application of integration: Volume

1. Dec 4, 2005

### nyyfan0729

Find the volume of the right cone of height 9 whose base is an ellipse. Major axis 12 and minor axis 6.

2. Dec 4, 2005

### arildno

use the formula for the volume of a cone.

3. Dec 4, 2005

### HallsofIvy

I didn't know there was a formula for the volume of a cone with a non-circular base!

Hmm- I guess there will be once we do the integral!

nyfan0729: At height "y" above the base of the cone, a cross section will be an ellipse with major axis of length (2/3)(9- y) and minor axis of length (1/3)(9-y): that would have area
$$\frac{2\pi}{9}(9-y)^2$$

It should be easy to integrate that from y= 0 to 9.

4. Dec 4, 2005

### arildno

If it is done correctly, it is one third of the product of the base area and the height (a right cone is, of course, assumed):

Let (x,y) be an arbitrary point in the base area A lying in the plane z=0, and for simplicity, let the position of the apex be (0,0,h), where h is the height,

Thus, the cone may be represented as:
$$\vec{F}(x,y,t)=((x,y,0)-(0,0,h))t+(0,0,h)=(xt,yt,h(1-t)), (x,y)\in{A}, 0\leq{t}\leq{1}$$
The Jacobian is [itex]ht^{2}[/tex], from which the volume formula follows:
$$V=\int{dV}=\iint_{A}\int_{0}^{1}t^{2}hdtdA=\frac{h}{3}\iint_{A}dA=\frac{Ah}{3}$$

5. Dec 4, 2005

### Tide

Arildno,

I believe the result is even more general than that. You don't have to assume a "right cone." As long as the horizontal cross sections of the cone are self-similiar then you'll get the same result.

6. Dec 5, 2005

### arildno

I assumed a "right cone" since that was assumed in the exercise.
You are right, of course..

7. Dec 5, 2005

### Martin Rattigan

But it's obviously half the volume of a right circular cone of diameter 12 because it's just compressed to half size in the direction of the minor axis, so $$\frac{\pi 6^2}{2.3}$$

Last edited: Dec 5, 2005