Demonstration of the volume of a sphere based on a modification of Archimedes proof. Archimedes realized that the volume of a sphere equals the volume of a cylinder (of the same height and radius) minus the volume of a cone (of the same height and radius). The version of the proof illustrated here uses a pair of cones whose apexes meet in the middle instead of a single cone. (It is easy to show that a single cone and pair of cones with the same radius and half the height have the same volume. Try it.) Any horizontal slice across the sphere and the [cylinder minus cones structure] intersects equal areas: a reduced size circular disk, in one case, and an annulus (washer-shape) in the other. You could in principle make stacks of thin cards that make a sphere in one case and the cylinder minus cones shape on the other. Since the volumes of cylinders (area of base times height) and cones (1/3 of area of base times height) are already established (at the time in the course when this is introduced) doing the algebra (2/3 area of central cross section times diameter) results in the formula for the volume of a sphere (4/3 pi R cubed).

1. Show that the volume of a cone equal the volume of two cones of the same radius but half the height.
2. Find the radius of the off-centered disk slice on the sphere (hint: call the offset h and use the Pythagorean theorem).
3. Find the area of the off-centered disk on the sphere.
4. Prove that the cone angle shown in the diagram is 45 degrees. (Hint: look for an isosceles right triangle.)
5. Find the inner and outer radii of the annulus. (Again use h as the offset.)
6. Find the area of the annulus.
7. Do whatever algebra is necessary to show that the areas of the disk cross section of the sphere and the annulus cross section of the other shape are equal.
8. Find the volume of the cylinder minus cones structure.
9. Simplify your answer for (8) to get a formula for the volume of a sphere.