Derivation:

The volume of a hemisphere with radius *r* is the same as the volume of a cylinder with radius and height *r*, with a right cone of radius and height *r* with the vertex at the center of the base removed (the vertex is down-- think ice cream cone). To prove this, we will find that the cross-sectional areas of the hemisphere and of the "cylinder less a cone" are equal at any height *h* (Cavalieri's Principle).

**The Hemisphere**

Any cross-section of a hemisphere parallel to the great circle will also be a circle; let's assign it a radius of *x*. If that circle is height *h* above the base, then it can also be described as part of a right triangle whose legs are *h* and *x*, and whose hypotenuse is *r*, the radius of the hemisphere. We can then describe *x* with the Pythagorean theorem: *x*² + *h*² = *r*². This can be rewritten as *x* = sqrt(*r*²-*h*²). So the area of the cross-section is π(sqrt(*r*²-*h*²)²), which is equal to π(*r*²-*h*²). Remember that.

**The "Cylinder Less Cone"**

The shape of a cross-section of the cylinder less cone is a little trickier, but it's easy to figure out: the area of the base of the cylinder, minus a cross-section of the cone at height *h*-- we'll give this cross-section a radius of *y*. So the cross-sectional area is (π*r*²-π*y*²). But look at this:

___________
/___________\
|\ | /|
| \___y___| / |
| **\** **|**h / |
| l**\ |** / |
|____**\|**/____|
/___________\

The triangle formed by *h*, *y*, and the line segment between where *y* meets the slant height of the cone and the center of the base (*l* in the diagram) is a 45-45-90 isosceles triangle. Therefore, *h* = *y*. So our previous equation (π*r*²-π*y*²) can be replaced with (π*r*²-π*h*²), which (when factored) comes out to π(*r*²-*h*²)-- the same as the area of a cross-section of the hemisphere at the same height! Therefore, the two solids are of the same volume.

**Almost There**

We now have to find the volume of the cylinder less cone.

V_{cyl} = πr²h = πr³

V_{cone} = (1/3)πr²h = (1/3)πr³

V_{total} = V_{cyl}-V_{cone} = (2/3)πr³

But remember, this is the volume of a hemisphere. Double it, and we get, at long last, the volume of a sphere:

4/3πr³

Q.E.D.