58 IRROTATIONAL MOTION. &quot; [CHAP. III. 



in question leads to a useful expression for the kinetic energy in 

 any case of irrotational motion, we give the following proof of it. 



Let u, v, w, (f&amp;gt; be any functions which are finite, continuous, 

 and single-valued at all points of a connected region S completely 

 bounded by one or more closed surfaces ; let dS be an element of 

 any one of these surfaces, I, m, n the direction-cosines of the normal 

 to it drawn inwards. We shall prove that 



mu) dS 



&amp;lt;Bn d.6v d. 



where the double-integral is taken over the whole boundary of S, 

 and the triple-integral throughout its interior. 



If we conceive a series of surfaces drawn so as to divide S into 

 any number of separate parts, the integral 



I !( 



dS .................. (18), 



taken over the boundary of S, is equal to the sum of the similar 

 integrals taken each over the whole boundary of one of these parts. 

 For, for every element do- of a dividing surface, we have, in the 

 integrals corresponding to the parts lying on the two sides of this 

 surface, elements $ (lu + mv + nw) da, and &amp;lt;/&amp;gt; (I u + m v + nw) dcr, 

 respectively. But the normals to which I, in, n, I , m, n refer being 

 drawn inwards in each case, we have I = I, m = m, n = n ; 

 so that in forming the sum of the integrals spoken of the elements 

 due to the dividing surfaces disappear, and we have left only those 

 due to the original boundary of S. 



Now let us suppose the dividing surfaces to consist of three 

 infinite series of planes parallel to yz, zx y xy } respectively. Let 

 x, y, z be the co-ordinates of the centre of one of the rectangular 

 spaces thus formed, dx, dy, dz the lengths of its edges, and let us 

 calculate the value of (18) taken over the boundary of this space. 

 As in Art. 8 the part of the integral due to the y^-face nearest the 



origin is f fat, J 3 dxj dijdz, and that due to the opposite face 



