Please pardon me if I don't use the correct terminology. Part of why I cannot solve this problem is that I don't even know what to research!

Given a circle placed on top of the cone, the shape drawn on the cone is not a simple conic section, it's a weird shape I'm not familiar with and I don't know what it's called.

The goal is to calculate the volume above the cone with any arbitrary placement of the circle. In the diagram above I've simplified things by only changing the offset of the circle in one dimension, but I'd like to know how to calculate for two dimensional offsets as well.

The red circle is placed directly in the center of the cone. Assuming we know $\theta$ and $d$ we can calculate the area above the red cross section as the conic volume minus the volume of it's contained cylinder:

$$\frac{2}{3}\pi r^2h$$

However as soon as the center point of the circular cross section is not the top of the cone I have no idea how to calculate the volume above the cross section

  • $\begingroup$ Suggestion: get the equation expressing $z$ in terms of a point $x,y$ located at a random point in the circle which is the base of the cone. Then assume the circular region is the vertical product of a circle lying inside the circle of the cone's base, and parametrize that circle. Finally choose a third parameter to represent height above the "base circle" of the circular slab, and plug the resulting $z$ coordinate into the cone equation. That gives a start. One question: are you trying to find area, or volume, of the shape. If area it's on the cone surface, volume may be easier. $\endgroup$
    – coffeemath
    Nov 21, 2013 at 4:42
  • $\begingroup$ I'm trying to calculate the volume. I'm not sure I follow your explanation : ( Could you detail it out a bit more? Or point me in the right direction to understand what your saying lol $\endgroup$
    – Parad0x13
    Nov 21, 2013 at 6:01
  • 1
    $\begingroup$ One more thing: is it the volume over the plane and under the inverted cone which also lies inside the cylinder which is vertical and generated by a circle which is contained in the circle making the base of the inverted cone? Or is it some kind of volume determined which is above the cone and bounded by the cylinder? $\endgroup$
    – coffeemath
    Nov 21, 2013 at 14:36
  • $\begingroup$ I've put up an answer with some idea of how to get the volume under the cone and inside the vertical cylinder. Let me know if I got the region right which you seek volume of... $\endgroup$
    – coffeemath
    Nov 21, 2013 at 15:29

1 Answer 1


I'll assume you want the volume above the plane on which the cone sits and below the inverted cone. I'll also assume that the circle bounding the vertical cylinder is viewed as drawn in the $x,y$ plane with its center at $(a,0)$ and a radius of $r_2$, in such a way that the entire bounding circle of the cylinder lies inside the big circle bounding the bottom of the cone.

Now a parametrization of the region inside the circle bounding the vertical cylinder is given by the set of points $(a+t \cos \theta,t \sin \theta)$ where $0 \le t \le r_2$ and $0 \le \theta \le 2\pi.$ Setting up polar coordinates on this circle to do the integral for volume means we have to use the polar volume element $t dt d\theta$ in the integral.

What we need now is the height. If $k=d/2$ and $h=(k/2)\tan \theta$ [NOTE I'm using this $\theta$ only temporarily, and the later use for the smaller circle is unrelated] then the height $z$ of the cone at a point at distance $w$ from the origin is $z=h-wh/k$ (a linear function going down from $h$ when $w=0$ to $0$ when $w=k$ since $k$ is the radius of the big circle.

You should now be integrating the height up to the cone multiplied by the polar area element, over the region inside the smaller circle. Except for the issue of keeping notation straight, this integral is easy to set up and I'll leave that to you.

When I plugged the double integral into maple the closed form results were extremely messy, and one of the two iterations involved a nonelementary function. So maybe the best thing is to just set up the integral, and use numerical methods to approximate it in particular cases.

  • $\begingroup$ Thank you for your input, I'm so confused it's not even funny. For instance I don't know anything about integrals or what the word polar means, so I have a good deal of research ahead of me $\endgroup$
    – Parad0x13
    Nov 22, 2013 at 5:51
  • $\begingroup$ Area and volume integrals are usually done in calculus, second semester. If you're not through that, it's unlikely to find a simple expression for the volume, given how complicated the answer turns out to be when done using integrals. Of course maybe the integrals themselves could be simplified in this case and there could then be a simple answer. $\endgroup$
    – coffeemath
    Nov 22, 2013 at 11:15
  • $\begingroup$ I'll do a double take this afternoon. You've inspired me to get into the weeds of integrals today! BTW this is all to help solve a Project Euler problem projecteuler.net/problem=431 $\endgroup$
    – Parad0x13
    Nov 25, 2013 at 1:17

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