Establish a system of reference with origin in the center of the square such that the diameter #AB# lies on the #x# axis.
(a) For every #x in (-a,a)# the plane perpendicular to the #x# axis intercepts the solid in a square whose side has the length of the chord orthogonal to #AB# in the point #(0,x)#, which is:
#l = 2sqrt(a^2-x^2)#
So the area of the square is:
#S = l^2 = 4(a^2-x^2)#
The element of the volume of the solid generated by such squares between #x# and #x+dx# is then:
#dV = l^2dx = 4(a^2-x^2)dx#
Integrating over the interval:
#V = int_(-a)^a 4(a^2-x^2)dx#
using the linearity of the integral:
#V = 4 a^2 int_(-a)^a dx -4 int_(-a)^a x^2dx#
#V = [4 a^2x - 4x^3/3 ]_(-a)^a#
#V = 4a^3 -4/3a^3 +4a^3 -4/3 a^3 = 16/3a^3#
(b) In this case the length of the base is the same, so the area of the triangle is:
#S = sqrt3/4 l^2 = sqrt(3)(a^2-x^2)#
and the volume is:
#V = int_(-a)^a sqrt3 (a^2-x^2)dx = (4sqrt3)/3 a^3#
