Quote by willib
wouldent that be the integral from infinity to infinity of 1/Y dx..
since you are revolving around the x axis...

No, the formula for the surface area is S=2π∫r(x)[1+(f'(x))
^{2}]dx.
Teclo,
The integral is pretty easy if you do it by parts. Start from:
∫(1+x
^{4})
^{1/2}/x
^{3}dx
and let u=(1+x
^{4})
^{1/2} and dv=dx/x
^{3}.