perpendicular to the base, and cutting it in the lines OA, OB, will be faces of a triangular prism, whose base will be the triangle AOB, and whose altitude will be the altitude of the cylinder. The volume of this prism will be equal to the area of the triangle AOB by the height of the cylinder. Hence, dividing the circle into elementary triangles, the cylinder will be equal to the sum of all the prisms, and therefore its volume will be equal to the area of the base multiplied by the altitude. Cor. 1.—If r be the radius, and h the height of the cylinder, Cor. 2.—If ABCD be a rectangle; X a line in its plane parallel to the side AB; O the middle point of the rectangle; the volume of the solid described by the revolution of ABCD round X is equal to the area of ABCD multiplied by the circumference of the circle described by O. Dem.—Produce the lines AD, BC to meet X in the points E, F. Then when the rectangle revolves round X, the rectangles ABFE, DCFE will describe cylinders whose bases will be circles having AE, DE as radii, and whose common altitude will be AB. Hence the difference between the volumes of these cylinders will be equal to the differences between the areas of the bases multiplied by AB, that is = π(AE2 − DE2).AB. Therefore the volume described by ABCD = π.AB.(AE + DE)(AE − DE); but AE + DE = 2OG, and AE − DE = AD. Hence volume described by the rectangle ABCD = 2π.OG.AB.AD. = rectangle ABCD multiplied by the circumference of the circle described by its middle point O. Observation.—This Cor. is a simple case of Guldinus’s celebrated theorem. By its assistance we give in the two following corollaries original methods of finding the volumes of the cone and sphere, and it may be applied with equal facility to the solution of several other problems which are usually done by the Integral Calculus. Cor. 3.—The volume of a cone is one-third the volume of a cylinder having the same base and altitude.