Answer
$=\frac{\pi \sin 1}{4}$
Work Step by Step
$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \cos \left(y^{2}+x^{2}\right) d x d y R=\left\{(r, \theta): 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 1\right\}$
convert to polar coordinates $R=\left\{(r, \theta): 0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant 1\right\}$
so
$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \cos \left(r^{2}\right) r d r d \theta$
integrate with respect to $=\frac{1}{2}=\int_{0}^{\frac{\pi}{2}}\left[\sin r^{2}\right]_{0}^{1} d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(-\sin 0+\sin 1) d \theta$
$=\frac{\sin 1}{2} \int_{0}^{\frac{\pi}{2}} d \theta$
integrate and evaluate
$=\frac{\sin 1}{2}[\theta]_{0}^{\frac{\pi}{2}}$
$=\frac{\sin 1}{2}\left[\frac{\pi}{2}-0\right]$
$=\frac{\pi \sin 1}{4}$
