62 Staties. 



= | PM X G^ 2 , on account of the right-angled triangle CPM 5 

 consequently, 



CG = 



But the surface of the sector CMAM', is equal to the arc MAM 



CM 

 Geom. multiplied by , therefore 



290. 



rr - PM x ^** 



~ 



X CM Jtf jJJW' MAM 



2 



That is, //ie distance of the centre of a circle from the centre of 

 gravity of any one of its sectors, is a fourth proportional to the arc, 

 the radius, and two thirds of the chord of the arc. 



The formulas above found may be applied to any other curve, 

 as the parabola, &c. 



108. We now proceed to the consideration of curved surfa- 

 ces, confining ourselves to those of solids of revolution. Reason- 

 ing, then, as in the preceding articles, it will be perceived that 

 the centre of gravity of each elementary zone, is in the axis of 



Fig. 46. revolution CA, and that it must be regarded as at the centre P 

 of one of the bases of this zone, considered as having an infinite- 

 ly small breadth. But we have seen that the expression for this 



Cal. 97. zone j s 2 a y ^/d-x 2 + dy*, TI representing the ratio of the diam- 



eter to the circumference. We shall have, therefore, (denoting 

 always by ft, the distance AC of A, the origin of the abscissas, 



from JVW, the axis of the moments) 2 n (ft x) y \/d x 2 

 for the moment of this zone ; from which it follows that the dis- 

 tance of G, the centre of gravity of the surface, from the point C, 

 designating this surface by tf, will be 



/2 71 (fe a?) y\/dx* 



109. Let us suppose, in order to apply this formula, that it 

 is proposed to find the centre of gravity of the convex surface of 

 47> the right cone ANN' ; we denote AP by or, PM by </, the height 

 AC by /i, CN the radius of the base by a, and the side AN by e. 

 On account of the similar triangles, ^CJV", Mrm, we have 

 AC : AN :: Mr : Mm-, 



