Find the exact area of the surface obtained by rotating the curve about the x-axis.

y=sqrt(1+e^x) for x gt-eq 0 and lt-eq 1

Question

Find the exact area of the surface obtained by rotating the curve about the x-axis. 

y=sqrt(1+e^x) for x gt-eq 0 and lt-eq 1

johnt · Answer

$$y = \sqrt{1 + e^x},    0 \le x \le 1$$

johnt · Answer

I know the formula is$$S = \int\limits_{a}^{b} 2\pi f(x) \sqrt{1 + f'(x)^2}  dx$$

johnt · Answer

and $$y' = \frac{ e^x }{ 2\sqrt{1 + e^x} }$$

johnt · Answer

But this gives an enormous integral that I can't get started solving.$$S = 2\pi \int\limits_{0}^{1} \sqrt{1+ e^x} \sqrt{1 + (\frac{ e^x }{ 2\sqrt{1 + e^x} })^2}$$

anonymous · Answer

you can do it just consentrate

johnt · Answer

ok I was able to simplify it to this: $$2\pi \int\limits_{0}^{1} \frac{ e^x + 2 }{ 2 } dx$$