152 U. S. COAST AND GEODETIC SURVEY. 



of the expression is the transcendental function known as 

 the integral sine; it is represented by the series 



/. 



sin X J _ Q? 7? 0^ x^ 

 o X ^^~^ 3.3! "^5.5! 7-7! "^9. 9! 



The value of this series for a; = 7r is approximately 1.852. 



To aid in the evaluation of the integrated part, we shall 

 restore the value of a; = tt sin <p 



r . , _ sm (tt sm (p) , ^ cos (tt sm^n^ 



— 4t cosec ^ + 2 =— ^ ^^ + 27r = P 



L sm^ (p sm (p Jo 



2 sin>(7r sin <p) +27r sin <p cosjir sin <p) — 4x sin <p 

 sin^ <p 



y . r 2 sin (t sin <p) +27r sin <p cos (tt sin <p) — 47r sin <p ~] 







<P=0 



14-.^ •+ r^y cosy cos(y sin y)+2ir cos v?cos (ff sin y) —27r2siny cosy stn(ir sin v?)—4ir co sy 1 

 = limit; L 2sinyCoi^ J 



<p=0 

 ~<^=oL sin^ J 



I* •, P— 2 7r' COS y sin (y SJn y)— 7r» COS y Sin (y SJn y)— ff* Stn y COS y COS (y SJn y) *] 



= limit [^ ^^ J 



<p=0 



= 0. 



Therefore 



^ = [ _ 47r - 27r + (2^2 + 4) 1 .852] a^ 

 = [-67r+(27r2 + 4) 1.852]a2 

 = [-67r + 23.74xl.852]a2 

 = (- 18.85 + 43.97)a2 

 = 25.12 a2. 



Area of the sphere = 4x^2 = 12.57 a^. 



Area of map 25.12 ^ , 



Area of sphere^ 12:57^2 very nearly. 



The area is therefore increased approximately in the ratio 

 of 2 : 1. 



