Answer
$=-128$
Work Step by Step
$\int_{-4}^{0} \int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}} 3 x d y d x$
convert the limits to polar coordinates
$R=\left\{(r, \theta): \frac{\pi}{2} \leqslant \theta \leqslant \frac{3 \pi}{2}, 0 \leqslant r \leqslant 4\right\}$
$r \cos \theta=x \rightarrow 3 x=3 r \cos \theta$
so
$=\int_{\pi / 2}^{3 \pi / 2} \int_{0}^{4} 3 r^{2} \cos \theta d r d \theta$
integrate with respect to $r$
$=\int_{\pi / 2}^{3 \pi / 2}\left[r^{3} \cos \theta\right]_{0}^{4} d \theta$
$=\int_{\pi / 2}^{3 \pi / 2}\left(4^{3} \cos \theta-0^{3} \cos \theta\right) d \theta$
$=64 \int_{\pi / 2}^{3 \pi / 2} \cos \theta d \theta$
evaluate and integrate
$=64[\sin \theta]_{\pi / 2}^{3 \pi / 2}$
$64\left[-\sin \frac{\pi}{2}+\sin \frac{3 \pi}{2}\right]$
$=(-1-1)64$
$=-128$
