If the infinite curve y = e^−7x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface.

Since the region is being revolved about the x-axis, the surface area equals
∫ 2πy √(1 + (dy/dx)^2) dx
= ∫(x = 0 to ∞) 2π * e^(-7x) * √(1 + (-7e^(-7x))^2) dx
= ∫(x = 0 to ∞) 2π √(1 + (7e^(-7x))^2) * e^(-7x) dx
= ∫(x = 7 to 0) 2π √(1 + w^2) * (-1/49) dw, letting w = 7e^(-7x)
= (2π/49) * ∫(x = 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, via tan t = w/1 and 'sohcahtoa'.

Link for integrating sec^3(t):
http://en.wikipedia.org/wiki/Integral_of…
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Finally, the surface area equals
(2π/49) * (1/2) [w√(1 + w^2) + ln |√(1 + w^2) + w|] {for w = 0 to 7}
= (π/49) [w√(1 + w^2) + ln |√(1 + w^2) + w|] {for w = 0 to 7}
= (π/49) [7√50 + ln (7 + √50)]
= (π/49) [14√2 + ln (7 + 5√2)].

I hope this helps!

youd get infinity area because the when rotated the function goes to infinity