The volume of the solid obtained by rotating the region bound by the curve #y=R(x)# and #x#-axis in the interval #(c,d)# is
#V=piint_c^d[R(x)]^2dx#
Here, we have #R(x)=x^2+1#, #c=2# and #d=3# and volume is
#piint_color(red)2^color(red)3[x^2+1]^2dx#
(once we have put #R(x)#, put #c# amd #d# as well)
= #piint_color(red)2^color(red)3[x^4+2x^2+1]dx#
= #color(red)(pi[x^5/5+(2x^3)/3+x+C]_2^3#
(once integral has been done no need to write it)
= #color(red)(pi[(3^5/5+(2xx3^3)/3+3+C)-(2^5/5+(2xx2^3)/3+2+C)]#
= #color(red)(pi[(243/5+54/3+3+C)-(32/5+16/3+2+C)]#
= #color(red)(pi[(243-32)/5+(54-16)/3+(3-2)]#
= #color(red)(pi[211/5+38/3+1]#
= #color(red)(pi[(211xx3+38xx5+15)/15]#
= #color(red)(pi[(633+190+15)/15]#
= #color(red)(pi[(838)/15]=55.86pi#
