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).
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.
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: