48 IRROTATIONAL MOTION. [CHAP. Ill 



completely bounded by one or more closed surfaces $; let BS be 

 an element of any one of these surfaces, and I, m, n the direction- 

 cosines of the normals to it drawn inwards. We shall prove in the 

 first place that 



where the triple-integral is taken throughout the region, and the 

 double-integral over its boundary. 



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

 region into any number of separate parts, the integral 



ff(lU+mV+nW)dS .................. (2), 



taken over the original boundary, is equal to the sum of the 

 similar integrals each taken over the whole boundary of one of 

 these parts. For, for every element So- of a dividing surface, 

 we have, in the integrals corresponding to the parts lying on 

 the two sides of this surface, elements (IU + mV+nW)$<r, 

 and (l'U+m'V+riW)Str, respectively. But the normals to 

 which I, m, n, and 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 the region. 



Now let us suppose the dividing surfaces to consist of three 

 systems of planes, drawn at infinitesimal intervals, parallel to 

 yz, zx, xy, respectively. If x, y, z be the co-ordinates of the 

 centre of one of the rectangular spaces thus formed, and 

 &e, Sy, $z the lengths of its edges, the part of the integral 

 (2) due to the yz-facs nearest the origin is 



and that due to the opposite face is 



The sum of these is dU/dx. SasSyfa. Calculating in the same 

 way the parts of the integral due to the remaining pairs of faces, 

 we get for the final result 

 dU 



