Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D \sin\sqrt{x^2+y^2}\,dx\,dy, $$ where $D$ is the annulus $\pi<\sqrt{x^2+y^2}<2\pi$.

Step 1 · Convert to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, the annulus becomes $\pi<r<2\pi$, $0\leqslant\theta<2\pi$, and $\sqrt{x^2+y^2}=r$:

$$ I = \int_0^{2\pi}\int_{\pi}^{2\pi} r\sin r\,dr\,d\theta. $$

Step 2 · Integrate with respect to $r$

By parts with $u=r$, $dv=\sin r\,dr$ ($du=dr$, $v=-\cos r$):

$$ \int r\sin r\,dr = -r\cos r + \int\cos r\,dr = -r\cos r+\sin r. $$

Evaluating from $\pi$ to $2\pi$:

$$ \Big[-r\cos r+\sin r\Big]_\pi^{2\pi} = \big(-2\pi\cos2\pi+\sin2\pi\big) - \big(-\pi\cos\pi+\sin\pi\big) = -2\pi - \pi = -3\pi. $$

Step 3 · Integrate with respect to $\theta$

$$ I = \int_0^{2\pi} (-3\pi)\,d\theta = 2\pi\cdot(-3\pi) = -6\pi^2. $$

Additional Notes

• The result is negative because $\sin r$ is negative throughout most of the interval $r\in(\pi,2\pi)$ (it is negative on $(\pi,2\pi)$ entirely, in fact), and this negativity dominates the weighted radial integral.
• Verified numerically: direct double integration over the annulus gives $I\approx-59.2176$, matching $-6\pi^2\approx-59.2176$.