solve $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$ $\int_{-\sqrt{3}}^{\sqrt{3}} 4 \sqrt{3-y^2}dy$
trig sub
$y = \sqrt{3}\sin(u)$
$dy = \sqrt{3}\cos(u)du$
\begin{align}\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}& 4 \sqrt{3-3\sin^2(u)}\sqrt{3}\cos(u)\,du = 12 \int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} \cos^2(u)\,du\\
&= 6\int_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})} (1+\cos(2u))du\\
&=\left[6u + 3\sin(2u)\right]_{\sqrt{3}\sin(-\sqrt{3})}^{\sqrt{3}\sin(\sqrt{3})}
\end{align}
If I plug in the limits to you I get some insanely low number while the answer should be $6 \pi$
what do I do ? 

Your endpoints are wrong.  If $y = \pm \sqrt{3}$, $\sin(u) = \pm 1$.  So take $u$ from $-\pi/2$ to $\pi/2$.
