Verified answer

Since we are revolving the region about the x-axis, the surface area equals

∫ 2πy √(1 + (dy/dx)^2) dx.

Letting y = e^(-7x) for x ≥ 0 yields

∫(x = 0 to ∞) 2πe^(-7x) √(1 + (-7e^(-7x))^2) dx

= ∫(x = 0 to ∞) 2πe^(-7x) √(1 + (7e^(-7x))^2) dx

= ∫(w = 7 to 0) 2π √(1 + w^2) * dw/(-49), letting w = 7e^(-7x)

= (2π/49) ∫(w = 0 to 7) √(1 + w^2) dw.

Now, let w = tan t, dw = sec^2(t) dt.

So, ∫ √(1 + w^2) dw

= ∫ sec t * sec^2(t) dt

= ∫ sec^3(t) dt

= (1/2) [sec t tan t + ln |sec t + tan t| + C

= (1/2) [w √(1 + w^2) + ln |√(1 + w^2) + w|] + C, by tan t = w/1 and 'sohcahtoa'

Hence, the surface area equals

(2π/49) * (1/2) [w √(1 + w^2) + ln |√(1 + w^2) + w|] {for w = 0 to 7}

= (π/49) [7√50 + ln(√50 + 7)]

= (π/49) [35√2 + ln(7 + 5√2)].

I hope this helps!