Answer
$=\pi (1-\cos 9) $
Work Step by Step
$\int_{R} \int \sin \left(y^{2}+x^{2}\right) d A$
$R: 9=x^{2}+y^{2}$
So
\}$R=\{(r, \theta): 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 3\}$
$\sin r^{2}=\sin \left(y^{2}+x^{2}\right)$
$\int_{R} \int \sin \left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{3}\left(\sin r^{2}\right) r d r d \theta$
integrate with respect to $r$
$=-\frac{1}{2} \int_{0}^{2 \pi}\left[\cos r^{2}\right]_{0}^{3} d \theta$
$=-\frac{1}{2} \int_{0}^{2 \pi}[-\cos 0+\cos 9] d \theta$
$=-\frac{1}{2} \int_{0}^{2 \pi}(-1+\cos 9) d \theta$
integrate with respect to $\theta$
$=-\frac{1}{2}(\cos 9-1)[\theta]_{0}^{2 \pi}$
$=\pi (1-\cos 9) $
