The Calippo™ popsicle has a specific shape, that I would describe as a circle of radius $r$ and a line segment $l$, typically of length $2r$, that's at a distance $h$ from the circle, parallel to the plane the circle is on, with its midpoint on a perpendicular line that goes through the centre of the circle.
The shape itself consists of lines connecting the circle to line segment.

  1. Do I need to specify how the points on the circle map to a point on the line segment?
    Obviously, the points on the circle directly under the end points of the line segment should map "straight up" to those end points. The points on the circle half way between those, at $\tfrac{\pi}{2}$, should map to the mid point of the line segment.
    But intuitively there should be a mapping that gives the "most outer" shape such that even if every point on the circle is connected to every point on the line segment, those lines never leave "the popsicle".
    Thanks to a comment by Mark S. I now know this is called the convex hull of the circle and the line segment.

  2. Given this description, how do I calculate the surface and the volume of this shape?

  3. Does this shape have an official name?
    It's not the round chisel as shown in this answer, since it lacks the edge in the shape of half an ellipse.

A calippo

  • $\begingroup$ Your "most outer" idea is a good one, and it's called the "convex hull". It seems this shape is the convex hull of a particular circle and line segment. I'm posting this only as a comment since it doesn't answer your "volume/surface area" question. $\endgroup$
    – Mark S.
    Oct 24, 2016 at 2:10
  • $\begingroup$ Your shape is probably a right circular conoid. en.wikipedia.org/wiki/Conoid It is not an oloid, which has no straight segment. $\endgroup$
    – user65203
    Oct 24, 2016 at 7:17
  • $\begingroup$ @YvesDaoust Yes! That's it! If you flesh it out a bit in an answer, I'll accept that. (I'll award a bounty to Blue for the work they did). $\endgroup$
    – SQB
    Oct 24, 2016 at 7:50
  • 1
    $\begingroup$ Note that the conoid is a proper subset of the convex hull of the circle and the line segment. Maybe one should subject a Calippo to a convexity test. $\endgroup$ Oct 24, 2016 at 11:49

5 Answers 5


It's hard to tell from the picture, but from the description "The shape itself consists of lines connecting the circle to line segment", I might represent the shape thusly:

Suppose that the circle has radius $r$, and that the shape has height $h$. If the surface lines joining the segment to the circle lie in a plane perpendicular to that segment, then the figure looks like this:

enter image description here

Then, we can parameterize $P_\theta$ on the circle, and companion point $Q_\theta$ on the segment ...

$$P_\theta = (r \cos\theta, r \sin\theta, h) \qquad Q_\theta = ( r\cos\theta, 0, 0)$$

... so that the line between them has equation

$$P_\theta + t \; (Q_\theta - P_\theta) : \begin{cases} x = r \cos\theta \\ y = r (1-t) \sin\theta \\ z = h (1- t)\end{cases}$$

Eliminating the parameters $\theta$ and $t$ yields this formula for the surface:

$$x^2 z^2 + y^2 h^2 = r^2 z^2 \tag{1}$$

Note that the equation (as well as the figure) indicates that the level curves of the surface are ellipses, with constant major radius ($r$) and linearly-varying minor radius ($z r/h$).

Now, the volume and surface area of the shape can be determined from this function:

$$y = f(x,z) = \frac{z}{h} \sqrt{r^2-x^2} \quad \tag{2}$$ over the rectangular region determined by $-r \leq x \leq r$ and $0\leq z \leq h$.

Edit. OP has indicated, in comments with @YvesDaoust, that the desired shape is the conoid, which indeed fits the surface described by $(1)$. (The parameterization given in the Wikipedia article matches $(1)$, for $r = 1$ and $h = z_0$, and under the coordinate transformation $z \to z_0-z$.)

The Wikipedia entry states that the volume of the conoid is $\frac{\pi}{2}r^2 h$. This is easily confirmed from $(2)$ ...

$$V = \frac{2}{h} \int_{0}^{h} z \int_{-r}^{r} \sqrt{r^2-x^2}\,dx dz = \frac{2}{h} \int_{0}^{h} \frac{\pi}{2} r^2 z dz = \frac{2}{h}\cdot \frac{\pi}{4} r^2 h^2 = \frac{\pi}{2}r^2 h \tag{3}$$

(where we have recognized the $x$ integral as giving the area of the half-circle of radius $r$).

For surface area, we note that $$f_x = \frac{-xz}{h \sqrt{r^2-x^2}} \qquad\qquad f_z = \frac{1}{h}\sqrt{r^2-x^2}$$ so that $$\begin{align} S &= 2\;\int_{0}^{h} \int_{-r}^{r} \sqrt{(f_x)^2 + (f_z)^2 + 1\;}\; dx dz \\ &= \frac{2}{h}\;\int_{0}^{h} \int_{-r}^{r} \;\sqrt{\frac{ h^2 (r^2-x^2) + (r^2-x^2)^2 + x^2 z^2}{r^2 - x^2}\;} \; dx dz \end{align}$$

This is a bit trickier to evaluate symbolically. I'll have to return to it.


#1. Yes, you must specify how the circle and segment map to each other.

I've never seen a Calippo, so I can only guess that it uses the "obvious" map. That is, if we align the segment on the x-axis, the map sends every point on the circle (x,y) to (x,0). If so, calculating the volume is very easy (just note the difference in cross-sections perpendicular to the x-axis with the same cross-sections of a cylinder), though calculating the surface area isn't so simple.

However, if you used a less obvious map you can arrive at other results. For example, by mapping all but a small portion of the circle to a tiny portion of the segment, you can make the figure arbitrarily close in volume to a pyramid.

So until you define the map, questions #2 and #3 can't be answered. I'd find it fascinating to know under what circumstances could you explicitly calculate the surface area/volume given an explicit map, but that's way past my ability to answer.

  • $\begingroup$ Well, the points on the circle directly under the end points of the line segment should map straight up, while the points halfway between them should map to the mid point of the line segment, under any reasonable mapping. But what I meant is that, if I map each point on the circle to each point on the line segment, does the outer shape change? Or is there one mapping that gives for all points on the circle the "most outer" mapping, with all other mappings being "obscured" by it? I'll try to formulate it better and add it to the question. $\endgroup$
    – SQB
    Oct 23, 2016 at 20:28
  • $\begingroup$ I think the implicit map is that every pair of a point from the circle and a point from the segment are connected by a new line segment, and all these line segments combine to the shape. Of course this is equivalent to your obvious map. $\endgroup$ Oct 23, 2016 at 20:46
  • $\begingroup$ @pokep , you are fully right: changing the mapping changes the shapes of the intermediate cross- sections, and thus volume and surface. Given the map, the volume and surface integral can be easily formulated, but explicitly computed ... is another matter, specially for the surface which will possibly involve elliptic integrals. $\endgroup$
    – G Cab
    Oct 25, 2016 at 21:44
  • $\begingroup$ @SQB: sure that the surface is made up by four symmetrical parts, and the mapping should respect that. But different mappings will provide different intermediate cross-section shapes: you can easily realize that as soon as you write the equation of the line connecting each couple of points. $\endgroup$
    – G Cab
    Oct 25, 2016 at 21:53

While we are at it we may as well look at the convex hull $C$ of the disc $D_r$ and the segment $\sigma$ arranged as $$D_r:=\{(x,y,z)|x^2+y^2\leq r^2, z=0\}\>,\qquad\sigma:=\{(s,0,h)|-r\leq s\leq r\}\ .$$ The set $C$ is the union of all solid cones with tips on $\sigma$ and base $D_r$. The plane $z=th$, $0\leq t\leq1$, intersects any such cone in a disc of radius $r(1-t)$ with center on the line parallel to $(1,0,0)$ through the point $(0,0,th)$. The union of these discs is a stadium-shaped figure bounded by two semidiscs of radius $r(1-t)$ and two parallel segments of length $2rt$ at distance $2r(1-t)$ from each other. The area of this shape then computes to $$A(t)=\pi r^2(1-t)^2+4r^2t(1-t)\qquad(0\leq t\leq1)\ ,$$ so that we obtain $${\rm vol}(C)=h\int_0^1 A(t)\>dt={r^2 h\over3}(2+\pi)\approx1.714\>r^2h\ ,$$ whereas the conoid has volume $\approx1.571\>r^2h$, about $8.35\%$ less.

The surface $\partial C$ consists of two plane isosceles triangles, parts of the two outermost cone mantles, and $D_r$. It should be possible to compute ${\rm area}(\partial S)$ explicitly as well.

  • $\begingroup$ your model is plausible, as further discussed in this post $\endgroup$
    – G Cab
    Apr 27, 2017 at 19:12

I think the shape of "Calippo popsicle" is like a round chisel. There was a question like this before and this might help you: Is there a name for a 3D shape that looks like a circle when viewed from one axis, a square from another, and a triangle from the third?

  • $\begingroup$ I don't think it's a round chisel, since the calippo doesn't have that edge the shape of half an ellipse. It is closely related and would probably be a valid answer to that other question. $\endgroup$
    – SQB
    Oct 23, 2016 at 19:52

Two cutting planes shown symmetric to cone axis meeting below its vertex before removing Region above the planes. What remains is like the Calippo shape of different geometric proportion.


If $ 2r,2w, h $ ,concurrency of straight cone generators are indicated as being given, then the truncated shape is uniquely given as part of cone of semi vertical angle= $ \tan^{-1}\dfrac{r-w}{h}.$



Oh, may be you are referring to a shape like a long toothpaste tube right or left conoid? page 61, DJ Struik, ruled surface of parametrization

$$ (x,y,z)= ( r \cos\theta, r \sin \theta, f(\theta) \,) $$

$$ f(\theta) = a \theta + b $$

with metric

$$ ds^2 = dr^2 + (r^2+f^{\prime2})\, d\theta^2 $$

The special case $ \pi r = 2 w $ is easy, obtained by isometrically flattening a circle to a diameter during manufacture.

  • $\begingroup$ Not really. That's similar to the "round chisel" found by @Seyed and that's not it. $\endgroup$
    – SQB
    Oct 24, 2016 at 7:07
  • $\begingroup$ Above is a cone different from @Seyed's cylinder, Both have elliptic intersections, For large height/radius ratio it is difficult to tell the difference. $\endgroup$
    – Narasimham
    Oct 24, 2016 at 7:19

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