Polar Integration Question

Hi all,

The question -

Calculate $\displaystyle V = \int \int_R \sin{\sqrt{x^2 + y^2}} dA $

R is the region inside the circle $\displaystyle x^2 + y^2 = 4\pi^2$

My attempt (converting to polar) - $\displaystyle V = \int \int_R z dA $

$\displaystyle z = \sin{\sqrt{x^2 + y^2}} $

Assuming we can find a quarter of the circle, than multiply by 4 for the total area...

$\displaystyle V = 4 \int^{\frac{\pi}{2}}_0 \int^{2\pi}_0 \sin{r} r dr d\theta $

$\displaystyle = 4 \int^{\frac{\pi}{2}}_0 [\sin{r} - r\cos{r}]^{2\pi}_{0} d\theta $

$\displaystyle = 4 \int^{\frac{\pi}{2}}_0 [\sin{2\pi} - 2\pi\cos{2\pi}] d\theta $

Here I'm stuck and quite frankly it's looking a lot more complicated than I imagine it should, can someone please point out the error in my working?

On the off chance I am correct so far, can I treat those $\displaystyle 2\pi$ sections as radians? I don't believe I can as that wasn't the context they were used originally, so what would I do?

Thanks for your time folks.