— 288. 
si verifica facilmente che si ha 
di q S4F SR 
Gad  (2dy 
Ora, 
ax=2% YE2% 
F (x +2, y2a)= f Sf pd yi) de dy = 
0 
\Yr2 ra L2Q0 qae24 YTZA 
FIST 
(1) 
essendo 
a y 
=.{ (Fa) vd 
scie 
ni nT+T2X 
Psa )= fo bi; 4%, 7) dx dyi= L Nes y S) (2, y)dr'dy, 
0 0 
nr Yy+-2% yT2A Fe 
(0, y=E22)= J fe 4(0, 7) da'dy= 5 Sf fa J 4(0,y/) da dy. 
‘0 0 0 “0 Yy 
Perciò: 
F(r+2x, y+22)+F (0+2a,y—2a)+—F(o—-22, y+2a)+F(a—-22.y=2a)— 
2F (2+-22,y) — 2Fa—22, y) — 2F (2,9+2@) — 2F (x,y—22)+ 4F (2,y) = 
A TT2% YZ dn n DH2% n y_-2% Nana - y2% 
SCSESCSESTSESTST para 
Y y 
"PR i dBdy— E fee8 y+7) dBdy— 
iù 0 ‘0 ‘0 
f° ft-bv-pagan= (* [tera = 
250 *0 dA DA 
va 24 24 
! / [ È (r+B,y+y)—4(e—By+y)—4(0+By—y)+ deh, yy) [647 
"0 ‘0 l F, 
D'altra parte, , 
g(cpyty)= il Hi fi f @, y)da' dy= 
‘0 
"0 
x Y x UPS nese) Y v+8 yx) 
f f da f f 35 f f "a { f f(e',y)da' dy', 
0 0 ‘0 y E; I) n %) 
