Last week’s Quiz Question was to determine the volume common to three intersecting cylinders of 3-inch diameter. Mike has suggested two solutions, but I am sure they are not correct. I have not, myself, worked this one.

So this week’s Quiz Question will address possibly ways to attack the problem. Start with the following diagram. It shows two intersecting cylinders.

From above the intersection looks like the above. If you join cylinders thusly, you will see the straight lines crossing in the middle as shown. The common volume is four of the shapes shown in the next diagram, a set of three orthogonal projections of the piece out.

These are what you get when you take a cylinder…

and make two cuts as shown.

The diameter of the cylinder is 3 inches. What’s the volume of the cut out piece shown. Post your answer in the comments section below.

## Update and solution

Consider the shape depicted in the set of orthographic projections above.

Compute the volume of this shape based on a cut that passes through the center of the cylinder. View this cut from above, and it is apparent the shape of the cut is triangle **ABC**. The length of triangle base is **b**, and **b** is **R sin ∝**.

Now compute the volume of the shape by summing up all the small pyramids generated as shown. **α** is the angle of the cut measured from the top edge. **R** is the radius of the cylinder. **Δα** is the vanishingly small angular width of such a pyramid.

Revert to integral calculus to compute the total volume. Here are the equations. This is based on the volume of a pyramid of height **R** and base of area **R**^{2} sin ∝ Δα, using the volume of a pyramid being **AR/3**. The height of the pyramid is the radius **R**.

And I get for the volume of the shape the following:

This is ¼ the volume of the intersection of two cylinders. Compare this to the volume of a sphere of radius R. The volume of the intersection should be more than the volume of the sphere.