Verified answer

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

∫ 2πy √(1 + (dy/dx)^2) dx

= ∫(x = 0 to 2) 2π * sin(πx/2) * √(1 + ((π/2) cos(πx/2))^2) dx

= ∫(x = 0 to 2) 2π * sin(πx/2) * √(1 + (π/2)^2 cos^2(πx/2)) dx

Letting u = cos(πx/2), du = (-π/2) sin(πx/2) dx yields

∫(u = 1 to -1) 2√(1 + (πu/2)^2) * -2 du

= ∫(u = -1 to 1) 4√(1 + (πu/2)^2) du

= ∫(u = 0 to 1) 8√(1 + (πu/2)^2) du, since the integrand is even

Now, let πu/2 = tan t, (π/2) du = sec^2(t) dt.

So, ∫ 8√(1 + (πu/2)^2) du

= ∫ 8 * sec t * (2/π) sec^2(t) dt

= ∫ 8(2/π) sec^3(t) dt

= 4(2/π) [sec t tan t + ln |sec t + tan t|] + C

= (8/π) [√(1 + (πu/2)^2) * (πu/2) + ln |√(1 + (πu/2)^2) + (πu/2)|] + C,

by tan t = (πu/2) / 1 and 'sohcahtoa'

Link for integrating sec^3(t):

http://en.wikipedia.org/wiki/Integral_of_secant_cu...

Finally, the surface area equals

(8/π) [√(1 + (πu/2)^2) * (πu/2) + ln |√(1 + (πu/2)^2) + (πu/2)|] {for u = 0 to 1}

= (8/π) [√(1 + π^2/4) * (π/2) + ln (√(1 + π^2/4) + π/2))]

= 4√(1 + π^2/4) + (8/π) ln (√(1 + π^2/4) + π/2))]

= 2√(4 + π^2) + (8/π) ln (√(4 + π^2)/2 + π/2)).

I hope this helps!