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: x² + h² = r². This can be rewritten as x = sqrt(r²-h²). So the area of the cross-section is π(sqrt(r²-h²)²), which is equal to π(r²-h²). 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 (πr²-πy²). 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 (πr²-πy²) can be replaced with (πr²-πh²), which (when factored) comes out to π(r²-h²)-- 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.
Vcyl = πr²h = πr³
Vcone = (1/3)πr²h = (1/3)πr³
Vtotal = Vcyl-Vcone = (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.