Are your dimensions those shown as r1, r2 and h below?
.\ /\
. 
. \ 
. 
. \ 
. 
. \ 
. 
   APEX
/\  r1 \ 
  \ 
  \ 
h  \ 
  \ 
  \ 
\/  r2 \ \/
  

If so, you may find it easier to find the hypothetical apex height of the original cone first:
APEX = h * r2 / (r2  r1)
Then the height of the removed cone with radius r1, is APEX  h.
Volume and curved surface area best found by subtraction as WHOLE CONE minus REMOVED CONE.
Volume (for a nontruncated cone) = PI.R
^{2}.H/3
Curved surface area (for a nontruncated cone) = PI.R.L, where L is slant length.
Slant length L found by Pythagoras. For the total surface area you presumably wish to add the circular areas at top and bottom.
After a lot of algebra, and some fortuitous(?) cancellation, I get
V = PI (r
_{1}^{2}+r
_{2}^{2}+r
_{1}r
_{2})h/3
A = PI [ (r
_{1}+r
_{2})S + r
_{1}^{2} + r
_{2}^{2} ] (including top and bottom circles)
where
S = sqrt[ h
^{2} + (r
_{2}r
_{1})
^{2} ]
is the slant length of the truncated cone.
I would always write
R * R
, not
pow(R,2)
.
Your BRACKETING is wrong. If you space out your variables much more, you will find it easier to get brackets in the correct place  this is particularly true of your sqrt() calculation for slant height and the collection of terms for volume.
You might also like to make PI a little more accurate, whilst "Interger" is actually spelt "double".