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I showed my calculus students two circular cones with the same base and height, but one of them "right" and the other slanted, and asked which had a greater volume. They all answered correctly that both have the same volume, but a lot of them had just memorized the formula for the volume of a cone, so the question mostly failed at causing them to think. What would be a good example of two solids that have the same volume by Cavalieri's principle, but where this fact is not otherwise obvious or deducible from frequently memorized formulae?

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    $\begingroup$ The wikipedia article you cite contains the napkin ring problem, which seems to qualify easily. $\endgroup$ Commented Sep 25, 2014 at 20:41

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One interesting example along these lines involves the following three solids:

  1. Half of a sphere (the upper half) with a radius of $r$.

  2. A circular cone (vertex at the bottom) with base of radius $r$ and height $r$.

  3. A circular cylinder with base of radius $r$ and height $r$.

The statement that's true is that the areas of the cross sections of shape #1 plus the areas of the cross sections of shape #2 are equal to the areas of the cross sections of shape #3. It follows that the volumes add, i.e. $$ \frac{2}{3}\pi r^3 \,+\, \frac{1}{3}\pi r^3 \;=\; \pi r^3. $$ This shows that you can derive the formula for the volume of a sphere from the formula for the volume of a cone, and vice-versa.

If you like, you make this into an example of the kind you requested by having one of the solids be the complement of either #1 or #2 in #3. That is, have the first solid be a half-sphere and the second be the complement of a cone in a cylinder, or vice-versa.

Edit: Another interesting example is the following pair of solids:

  1. The paraboloid obtained by rotating the region $x^2 \leq y \leq 1$ in the $xy$-plane around the $y$-axis, with circular cross sections.

  2. A triangular prism on a $1$-$1$-$\sqrt{2}$ triangle, with depth $\pi$, using cross sections parallel to one of the $1\times \pi$ rectangular faces.

This example lets you compute the volume of a paraboloid without using calculus (assuming you don't count Cavalieri's principle as calculus).

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I've always been surprised by this way of measuring the area of an ellipse. Instead of the classical use of the semi-major and semi-minor axes, you can actually do something similar to what you do with a parallelogram:

Draw two parallel lines that just touch the ellipse on the outside. We'll say these lines are horizontal. We'll call the distance between these lines h. Halfway between them, measure across your ellipse horizontally. Let's call half this width r. Then the area of the ellipse is $\pi r (\frac12 h)$. It's particularly impressive when $h=2r$ because then the area is just $\pi r^2$.

This works because your ellipse is a shear of another ellipse with the same height and horizontal radius. In essence because of the 2D version of Cavalieri's principle.

circle and ellipse

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  • $\begingroup$ Nice! How hard is it to show that any two ellipses with the same height and horizontal radius have all the same horizontal cross-sections? $\endgroup$ Commented Sep 24, 2014 at 17:15
  • $\begingroup$ Hmmm. Not sure at this stage. My understanding is actually in the other direction, which is much easier to show: a shear of any shape preserves horizontal cross-sections, and a shear of an ellipse is still an ellipse. $\endgroup$ Commented Sep 24, 2014 at 22:43
  • $\begingroup$ Ok; am I wrong, then, that you'd need to know that any two ellipses with the same height and horizontal radius have the same horizontal cross-sections in order to actually calculate the area of an ellipse this way? $\endgroup$ Commented Sep 25, 2014 at 23:24
  • $\begingroup$ What about this: given an ellipse with particular height and horizontal radius, show that there is a shear of this ellipse which is a "standard" ellipse. Since shearing preserves horizontal cross-sections, the area must be the same by Cavalieri's principle. $\endgroup$ Commented Sep 26, 2014 at 3:43
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Spiral paper blocks and regular paper blocks

This image search should provide some example pictures of spiral paper blocks.

I would not know of a technique of calculating the volume of the shape of a spiral paper block which does not use Cavalieri’s principle or something higher. However, with the principle it becomes directly clear that the volume is the same as that of a regular paper block (with the same paper size and height). As a bonus, the individual sheets correspond to the individual slices, not to forget that the principle itself is nicely illustrated as both blocks contain the same amount of paper.

Didactically, you might consider introducing the shape abstractly first, without presenting or mentioning the real-life thing – but whether that’s a good idea depends on several factors.

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  • $\begingroup$ If anybody knows (or can produce) an image of a spiral paper block that can be embedded here, feel free to do so. $\endgroup$
    – Wrzlprmft
    Commented Mar 24, 2014 at 14:04
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This example here I discovered myself:

Consider a cylinder of radius $r$ and height $h$, circumscribing a paraboloid $y=h\left(\dfrac{x}{r}\right)^2$ whose apex is at the center of the bottom base of the cylinder and whose base is the top base of the cylinder.
Also consider the paraboloid $y=h-h\left(\dfrac{x}{r}\right)^2$, with equal dimensions but with its apex and base flipped.

For every height $0\le y\le h$, the disk-shaped cross-sectional area $\pi\left(\sqrt{1-\frac{y}{h}}\,r\right)^2$ of the flipped paraboloid is equal to the ring-shaped cross-sectional area $\pi r^2-\pi\left(\sqrt{\frac{y}{h}}\,r\right)^2$ of the cylinder part ''outside'' the inscribed paraboloid.

Therefore, the volume of the flipped paraboloid is equal to the volume of the cylinder part ''outside'' the inscribed paraboloid. In other words, the volume of the paraboloid is $\dfrac{\pi}{2}r^2h$, half the volume of its circumscribing cylinder. The volume of the paraboloid is half the volume of its circumscribing cylinder.

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