Surface area of revolution is given by, #A= 2piint_a^bf[x]sqrt[1+[f'[x]]^2# dx.
From the question, #2x=y^2#, differentiating implicitly, #2=2ydy/dx#, i.e, #dy/dx= 1/y# , but #y=sqrt2x,# so #dy/dx=1/sqrt[2x]# and so #[dy/dx]^2 = 1/[2x]#.
Therefore #A= 2piint_a^b sqrt2xsqrt[1+1/[2x]#dx= #2piint_a^bsqrt2xsqrt[[2x+1]/[2x]# dx.
#A= 2piint_a^b[sqrt2x]/[sqrt2x]sqrt[2x+1]dx# = #2piint_a^bsqrt[2x+1]#dx.........#[1]#.
Let #u= 2x+1#, so #[du]/[dx]=2#, ie, #dx=[du]/[2]#.
substituting for #u# and #du# in ......#[1]# we have ,
#A= 2piintsqrtu[du]/[2]# = # piintsqrtudu#, and after integrating by the power rule,
In terms of #u# area #A=[2pi]/[3]sqrt[u^3#, and substituting back for #u = 2x+1#
#A = [2pi]/[3]sqrt[[2x+1]^3]#, evaluated for #x= 2, x=0#, will give the answer above.
