﻿462 
  Dr. 
  F. 
  L. 
  Hitchcock 
  on 
  the 
  Operator 
  V 
  in 
  

  

  volume, 
  respectively. 
  On 
  the 
  surface 
  of 
  the 
  sphere 
  I, 
  ?n, 
  

  

  and 
  n 
  become 
  -, 
  -, 
  and 
  -. 
  Now 
  let 
  X= 
  — 
  — 
  , 
  Y= 
  -§- 
  , 
  and 
  

   -s-p 
  a 
  a 
  a 
  ox 
  oy 
  

  

  Z= 
  ^~. 
  Substituting 
  in 
  (2) 
  we 
  have 
  

  

  The 
  left 
  side 
  simplifies 
  by 
  Euler's 
  theorem 
  giving 
  

  

  ™JJ 
  F 
  ^ 
  S 
  = 
  ffjAF^V, 
  .... 
  (4) 
  

  

  where 
  F 
  m 
  is 
  any 
  homogeneous 
  scalar 
  function 
  of 
  positive 
  

   degree 
  m 
  and 
  'the 
  integrations 
  are 
  with 
  reference 
  to 
  the 
  

   sphere 
  of 
  radius 
  a 
  with 
  centre 
  at 
  the 
  origin. 
  

  

  Next 
  start 
  afresh 
  with 
  the 
  left 
  member 
  of 
  (A). 
  By 
  the 
  

   form 
  of 
  the 
  definition 
  of 
  a 
  homogeneous 
  function 
  which 
  was 
  

   laid 
  down 
  in 
  the 
  first 
  paper, 
  F 
  m 
  (p) 
  = 
  r 
  m 
  F(u), 
  where 
  F(u) 
  

   denotes 
  a 
  function 
  of 
  \Jp, 
  that 
  is, 
  a 
  function 
  of 
  a 
  point 
  on 
  

   a 
  unit 
  sphere. 
  Therefore 
  by 
  performing 
  the 
  integration 
  with 
  

   respect 
  to 
  r 
  (writing 
  e?S 
  = 
  ? 
  i2 
  dS 
  and 
  dV 
  = 
  r 
  2 
  <iSorfr), 
  we 
  shall 
  

   obtain 
  

  

  ffi/W.F, 
  B 
  (p)^=fV+y(r)*-.fJF( 
  u 
  VS 
  , 
  . 
  (5) 
  

  

  where 
  the 
  double 
  integral 
  is 
  carried 
  over 
  the 
  surface 
  of 
  a 
  

   sphere 
  of 
  unit 
  radius. 
  But 
  on 
  the 
  surface 
  of 
  the 
  sphere 
  of 
  

   radius 
  a 
  we 
  have 
  r 
  constant 
  and 
  equal 
  to 
  a, 
  hence 
  

  

  [fF(«)rfS„=^jjF 
  m 
  (p)rfS 
  (C) 
  

  

  the 
  integral 
  on 
  the 
  right 
  being 
  taken 
  over 
  the 
  sphere 
  of 
  

   radius 
  a. 
  The 
  formula 
  (A) 
  now 
  follows 
  at 
  once 
  by 
  elimi- 
  

   nation 
  of 
  the 
  double 
  integrals 
  between 
  (4), 
  (5), 
  and 
  (6). 
  

  

  By 
  (A) 
  we 
  may 
  solve 
  a 
  variety 
  of 
  simple 
  problems. 
  As 
  an 
  

   example, 
  if 
  the 
  density 
  at 
  a 
  point 
  in 
  the 
  sphere 
  is 
  equal 
  to 
  

   the 
  distance 
  from 
  the 
  centre, 
  the 
  moment 
  of 
  inertia 
  about 
  

   OZ 
  is 
  found 
  by 
  writing 
  /(r) 
  = 
  r, 
  and 
  F(p) 
  = 
  # 
  2 
  -f 
  y 
  2 
  , 
  when 
  (A) 
  

   becomes 
  

  

  3. 
  Extension 
  to 
  vector 
  functions. 
  — 
  In 
  proving 
  (A) 
  by 
  the 
  

   aid 
  of 
  Grauss's 
  theorem, 
  (Fp) 
  was 
  a 
  scalar 
  function 
  of 
  p. 
  

   There 
  is 
  nothing 
  in 
  the 
  result 
  itself 
  to 
  require 
  this 
  restriction. 
  

  

  