In modern parlance, how did he arrive at these calculations? Why is the volume of the inscribed figure, Cone or Pyramid, 1/3 the Volume of its "parent" figure, Cylinder or Prism?

If the Prism is a Rectangular Prism in which a Rectangular Pyramid is inscribed, Eudoxus partitioned, for example, the inscribed Rectangular Pyramid with "n" small Rectangles having "x" width and height "h". The total area of the partitioned-inscribed Rectangular Pyramid is the sum of all the areas of its composit Rectangles. Since the measure of Area is expressed in "squared units", i.e., 1 cm X 1 cm = 1 cm ², the Volume of the inscribed rectangular Pyramid is the Sum of all the "squared composit smaller Areas:

• Volume = (Sum of all areas of the smaller Rectangles)(h). The Sum of Squares is equal to:
  • (n(n +1)(2n +1))/6; for volumetric calculations: (n(n +1)(2n +1))/(6n³)

• Volume = (n(n +1)(2n +1))/(6n³)(Area)(h) = (2n³ + 3n² + n)/(6n³)(Area)(h) = (1/3 + 1/2n + 1))/(6n²)(Area)(h): as "n" becomes very large, 1/2n + 1/6n² = 0, so that:

• Volume = 1/3(Area of the Base)(h)

GENERALLY:

1. For a Cylinder: Volume = (r²)(h)
   • For a Cone: Volume = 1/3(r²)(h)

2. For Rectangular Prism: Volume = (l)(w)(h)
   • For a inscribed Rectangular Pyramid: Volume = 1/3((l)(w)(h))

3. For a Triangular Prism: Volume = 1/2(altitude)(base)(h)
   • For and inscribed Triangular Pyramid: Volume = 1/3( 1/2(altitude)(base)(h)

4. For a Trapezoidal Prism: Volume = 1/2 (altitude)(base + base'))(h))
   • For and inscribed Trapezoidal Pyramid: Volume = 1/3((1/2 (altitude)(base + base'))(h))

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There is sn interesting relationship between the Volume a Cylinder, and an inscribed Double Cone and Sphere:

If the height of each figure is equal to "2r" so the the radius of the Sphere is equal to "r", the Volume of the Cylinder and Inscribed Double Cone is:

• Volume of Cylinder = 2r³;
• Volume of the Inscribed Double Cone = 2(1/3r³) = 2/3(r³)

The remaining 1/3 of the Volume within the Cylinder can be calculated as follows:

• Volume = (1/3((1/2(r x 2r)2r))2 = 4/3(r³), i.e., 2r³ - 2/3r³ = 6/3r³ - 2/3r³ = 4/3r³. The formula for the Volume of a Sphere is 4/3r³.

Thus, the Volume of a Cylinder is:

• Volume of Cylinder (2r³) = Volume of Double Cone (2/3(r³) + Volume of the Sphere (4/3(r³) = 6/3r³ = 2r³