Hmmm...either there's a mistake or this is trivial:
$\displaystyle \displaystyle{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2=\left(4R^2(\cos^2\phi+\sin^2\p hi\right))=4R^2}$ , and then
$\displaystyle \displaystyle{\sqrt{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2}=2R}$ , so we have
$\displaystyle \displaystyle{2\pi\int\limits^{\pi/2}_0 2R\cos \phi\sin\phi\sqrt{\left(2R\cos \phi\right)^2+\left(-2R\sin\phi\right)^2}\,d\phi=4R^2\cdot 2\pi\int\limits^{\pi/2}_0\cos\phi\sin\phi\,d\phi}$
$\displaystyle \displastyle{=8R^2\pi\cdot \frac{1}{2}\left[\sin^2\phi\left]^{\pi/2}_0=4R^2\pi}$
Tonio