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 ? 

Say $x(y) = \sqrt{3-y^2}$.  This is a formula for a semi-circle of radius $\sqrt{3}$, valid from $y=-\sqrt{3}$ to $y = \sqrt{3}$.
The area of this semi-circle is $\pi r^2 / 2 = 3\pi/2$, which is $\int_{-\sqrt{3}}^{\sqrt{3}}x(y)\,dy$.  
Your integral has an extra factor of four in it, so you need to multiply the result by 4 to get $6\pi$.
