Answer
$\iint_D\cos\sqrt{x^2+y^2}dA=(4\sin 2+2\cos 2-2)\pi$
Work Step by Step
In polar coordinates, the region $D$ is written as $\{(r,\theta)|0\leq r\leq 2,0\leq\theta\leq 2\pi\}$.
Evaluate the given integral:
$\iint_D\cos\sqrt{x^2+y^2}dA=\int_0^{2\pi}\int_0^2\cos(\sqrt{r^2})rdrd\theta$
$=\int_0^{2\pi}\int_0^2r\cos rdrd\theta$
$=\int_0^{2\pi}[r\sin r+\cos r]_0^2 d\theta$
$=\int_0^{2\pi}(2\sin 2+\cos 2-1)d\theta$
$=(2\pi-0)(2\sin 2+\cos 2-1)$
$=4\sin 2+2\cos 2-2)\pi$
