OP ARTS AND SCIENCES. 



31; 



the terminator, and the constant factors expressing the intensity of the 

 light and the dimensions of the cylinder will be neglected. The short 

 arc BC may be represented by ofy, if we consider the vertex of an 

 angle equal to y as placed in the axis of the cylinder. The quantities 

 of light received from the slopes AB and AC will be respectively ex- 

 pressed by s cos (0 — /?) cos (y — ft) and s cos (9 -j- /3) cos (y -|- y8). 

 The value of s is g sec /? dy, and that of ^ is y — (tt — v). 



These expressions do not represent the consequences which result 

 from the opacity of the surfaces considered, so that the integration 

 must be confined to the limits within which the whole surface of every 

 slope is illuminated and visible. The slopes facing towards the ter- 

 minator for which — /8 < — ^ tt will not be illuminated, and the 

 slopes respectively opposite them will be partly in the shade. The 

 slopes facing towards the illuminated limb for which y -j- /8 > J tt 

 will not be visible, and those opjjosite to them will be partly hidden. 

 The required integral is therefore 



— 1 sec ft \ Tc'^or (y - ft) cos [vJr(y- P)] d{y-ft) 



IJ iff — v 



+ Teas (y + ft) COS b + (y + ft)] d (y + ft)] 



the terms of which have the same foi'm as that of an expression 

 already found (p. 312). If v = 2 ft, as well as when w < 2 /?, no 

 finite region exhibits slopes wholly visible and illuminated, so that this 

 formula is then inapplicable. For those cases in which y > 2 /3, 

 the sum of its two equal terms becomes, on reduction, 



I sec ft [cos 2 ft sin {v — 2 ft) —(v — 2ft) cos v]. 



When ft = 0, the limits of the integration coincide with those of the 

 phase, and the result becomes | (sin v — v cos v) , as before. When 

 ft = J TT, V cannot exceed 2 ft. 



To complete the solution, we have still to consider the partly visible 

 and the partly illuminated slopes. The first case is exhibited in Fig. 2, 



