here are some notes i wrote on this topic:
just as a 3 ball is swept out by revolving half a disc around a line, so a 4 ball is generated by revolving half a 3 ball.
moreover the volume of the solid of revolution can be computed by knowing where the center of mass is. then we can
use Archimedes’ trick to do a calculation that Archimedes could have done. Namely he showed
that the volume of half a 3-ball equals the difference of the volumes of a cylinder minus that of a
cone. Now the center of mass of a cylinder is obviously half way up, and Archimedes knew that
just as the center of mass of a triangle is 1/3 of the way up from the base, the center of mass of a
cone is ¼ the way up from the base.
Thus we can use centers of mass and subtraction to get the volume of a 4-ball. I.e. a cylinder of
height R and base radius R has center of mass at height R/2, and volume πR^2.R, so revolving it
around an axis at its base gives 4 dimensional volume of 2π(R/2).πR^2.R = π^2.R^4. Now the
inverted cone of height R and base radius R has center of mass at distance ¼ of the way from its
base, hence distance (3R/4) from the axis, and volume (1/3)πR^2.R. Thus revolving it generates
a 4 dimensional volume equal to (2π)(3R/4).(1/3)πR^2.R = (1/2)π^2.R^4. Subtracting the
volume of the revolved cone from that of the revolved cylinder, gives the 4 dimensional volume
of the revolved half 3-ball, i.e. the volume of the full 4-ball as π^2.R^4 - (1/2)π^2.R^4 =
(1/2)π^2.R^4.