226 Prof. Challis on the Solution of a Problem in the 



nates are continuous with the curve, and that the curve is a semi- 

 circle, the radius of which is c. Hence, A 2 and c being given, 

 the length of the ordinates and the content of the solid may be 

 calculated. 



The method of solution adopted in the August Number of 

 1861, gave for the connecting line in the last example a portion 

 of the curve described by the focus of an hyperbola rolling on a 

 straight line. The foregoing argument has shown that the fal- 

 lacy of this determination consists in making use of an integral 

 obtained by means of a factor, and not subsequently taking into 

 account what was indicated by the factor. 



Although it results from this new method that the curve just 

 mentioned has no application in the solution of the problem that 

 has been under discussion, I think it worth while to point out 

 that the length and area of that curve are erroneously calculated 

 in the August Number, and to obtain the correct values, as the 

 process will serve to illustrate a part of the reasoning in the pre- 

 sent communication. The semiaxes of the hyperbola being a 

 and b, the differential equation of the curve is 



«•-«£.)' 



which remains the same whether y be positive or negative. But 

 for the present purpose it will suffice to consider only the posi- 

 tive values ofy. By supposing that j» 2 = 0, and substituting 

 a^ — a 2 for b 2 , we obtain a(e — 1) for the positive minimum 

 value of y, and a(e+\) for the positive maximum value. Also 

 the curve is evidently an oval symmetrically disposed about the 

 maximum ordinate, and the extreme abscissae are those corre- 

 sponding to the ordinates y — b. For determining the length of 

 the curve, we have 



ds= + 



2ay dy 



4/4ay-(fl*-y 8 )* 



the + sign applying from the minimum to the maximum ordi- 

 nate, and the — sign from the maximum to the minimum or- 

 dinate. Taking the positive sign and integrating, 



e+'M 





Hence, if the integral be taken from y?=a(e— 1) to y = a(e + l) 

 and be doubled, the periphery of the oval is found to be 2ira. 

 By integrating from y — b to y = a{e-\-l) } and from y—a{e— 1) 



