The volume of a sphere can also be very elegantly found by means of a proposition known as Cavalieri's Theorem. This asserts that if two solids lie between parallel planes, and are such that the two sections made by any plane parallel to the given planes are equal in area, the solids are themselves equal in volume. Thus, if these solids have the same altitude, _a_, and if _S_ and _S'_ are equal sections made by a plane parallel to _MN_, then the solids have the same volume. The proof is simple, since prisms of the same altitude, say _a_/_n_, and on the bases _S_ and _S'_ are equivalent, and the sums of _n_ such prisms are the given solids; and as _n_ increases, the sums of the prisms approach the solids as their limits; hence the volumes are equal. [Illustration] This proposition, which will now be applied to finding the volume of the sphere, was discovered by Bonaventura Cavalieri (1591 or 1598-1647). He was a Jesuit professor in the University of Bologna, and his best known work is his "Geometria Indivisilibus," which he wrote in 1626, at least in part, and published in 1635 (second edition, 1647). By means of the proposition it is also possible to prove several other theorems, as that the volumes of triangular pyramids of equivalent bases and equal altitudes are equal. [Illustration] To find the volume of a sphere, take the quadrant _OPQ_, in the square _OPRQ_. Then if this figure is revolved about _OP_, _OPQ_ will generate a hemisphere, _OPR_ will generate a cone of volume (1/3)[pi]_r_^3, and _OPRQ_ will generate a cylinder of volume [pi]_r_^3. Hence the figure generated by _ORQ_ will have a volume [pi]_r_^3 - (1/3)[pi]_r_^3, or (2/3)[pi]_r_^3, which we will call _x_. Now _OA_ = _AB_, and _OC_ = _AD_; also (_OC_)^2 - (_OA_)^2 = (_AC_)^2, so that (_AD_)^2 - (_AB_)^2 = (_AC_)^2, and [pi](_AD_)^2 - [pi](_AB_)^2 = [pi](_AC_)^2. But [pi](_AD_)^2 - [pi](_AB_)^2 is the area of the ring generated by _BD_, a section of _x_, and [pi](_AC_)^2 is the corresponding section of the hemisphere. Hence, by Cavalieri's Theorem, (2/3)[pi]_r_^3 = the volume of the hemisphere. [therefore] (4/3)[pi]_r_^3 = the volume of the sphere. In connection with the sphere some easy work in quadratics may be introduced even if the class has had only a year in algebra. For example, suppose a cube is inscribed in a hemisphere of radius _r_ and we wish to find its edge, and thereby its surface and its volume. If _x_ = the edge of the cube, the diagonal of the base must be _x_[sqrt]2, and the projection of _r_ (drawn from the center of the base to one of the vertices) on the base is half of this diagonal, or (_x_[sqrt]2)/2. Hence, by the Pythagorean Theorem, _r_^2 = _x_^2 + ((_x_[sqrt]2)/2)^2 = (3/2)_x_^2 [therefore] _x_ = _r_[sqrt](2/3), and the total surface is 6_x_^2 = 4_r_^2,