Monthly Archives: May 2017

The Volume of a Sphere

Here is a quick explanation for the formula of the volume of a sphere. This is based on the proof known to the Ancient Greeks.

For this, consider three objects:

  1. A cylinder;
  2. A cone with the point  on the bottom;
  3. A hemisphere with the flat side down.

Each object has a height of r and a radius of r.

Take a slice of each object at some height h. The exposed surface (cross-section) will be a circle.

For the cylinder, the radius of this exposed circle will be r, because the radii of all circular cross-sections is r. So the area of the circle for the cylinder at height h is πr2.

For the cone, the radius of this exposed circle will be h, so the area is πh2.

By definition, each point on the hemisphere is r units away from the center. Each point on the exposed circle is at a height of h. Using the Pythagorean Theorem, the radius of the exposed circle is the square root of (r2-h2), so the area is π(r2-h2).

Note that this is the difference between the cylinder and the cone: This is the key.

Since, for each cross-section of the three objects, the area for the cylinder is equal to that of the cone plus that of the hemisphere, it must be the case that the volume of a cylinder is equal to the volume of a cone and the volume of a hemisphere, when all objects have the same radius and height.

We know the formula for the volume of a cylinder: πr2h. If h = r, then πr3.

We know the volume of a cone is one-third of this, so the volume of a hemisphere is two-thirds of this, 2πr3/3. The volume of a sphere is twice that of a hemisphere, that is, 4πr3/3.

Week of May 15, 2017

Presentations: May 19 May 18 May 17 May 16 May 15


Practice: Naming arcs, arc measures, and lengths
Discussion: Finding pi using the perimeter of polygons


Notes and practice: Area of Circles and Sectors


Practice: Area of Circles and Sectors, review all unit material


Notes and Practice: Area of a sphere (review: cylinder and cone)


Practice: All unit material
Notes: Circle tangents