Determine the area of the surface of revolution about the x-axis of the intersection between the curves r=sinx and r=cosx.

I drew a picture and found they intersected at 0 and .5 and I know the equation is:

$\displaystyle S = \int_a^b {2\pi y\sqrt {1 + {{(\frac{{dy}}

{{dx}})}^2}} dx}

$

but I don't know if b is .5 or pi/4

and I tried this:

$\displaystyle S = \int {2\pi \cos x\sqrt {1 + {{( - \sin x)}^2}} dx - \int {2\pi (\sin x)\sqrt {1 + {{(\cos x)}^2}} dx} }

$

but I couldn't integrate it correctly