If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.
Have a look here: Conic section - Wikipedia, the free encyclopedia
If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.
Have a look here: Conic section - Wikipedia, the free encyclopedia
It is a lot easier to do it the other way but here it goes.
The equation of a cone is .
Suppose the radius of the base at is . Then we have or
Now we let
Next we have the equation of a general plane . We want to construct a plane
This would mean that or .
Now lets intersect this plane with our cone. This gives us the curve
or
Now at which will be our limit of integration. Keep in mind that the vertex of the cone is at . We know that therefore
Next we find the area and I'm going to stop here as it is getting quite messy. I guess that helps to a point but I don't see any point in continuing.
Of course you are right and I apologize for the confusion.
Obviously a hyperbolic cross-section is hard to handle as fobos3 has demonstrated. So I suggest to use parabolic cross-sections.
According to Archimedes a segment of a parabola has an area
Use the proportion
Use Pthagorian theorem to calculate the length of
Caution: This step probably contains a mistake which causes a wrong final result! I wasn't able to spot my error.
Now the colume of the cone should be:
My computer tells me that which is definitely too large.
(... but it looks like a very near miss)