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