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.[br]2. Find the radius of the off-centered disk slice on the sphere (hint: call the offset h and use the Pythagorean theorem).[br]3. Find the area of the off-centered disk on the sphere.[br]4. Prove that the cone angle shown in the diagram is 45 degrees. (Hint: look for an isosceles right triangle.)[br]5. Find the inner and outer radii of the annulus. (Again use h as the offset.)[br]6. Find the area of the annulus.[br]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.[br]8. Find the volume of the cylinder minus cones structure.[br]9. Simplify your answer for (8) to get a formula for the volume of a sphere.