The area bounded by $y=4-x^2$ and $y=0$ is 

The given curve is $y=4-x^2$ and the lines are $y=0,y=3$

$y=0\Rightarrow x=\pm 2$ 

The shaded area is double the area of the region bounded by the curve $x=\sqrt{4-y}$, the lines $y=0,y=3$ and $y-$ axis.

Hence, the required area is $A=2{\int }_0^2\sqrt{4-y}dy$

it gives 

$\begin{array}{rcl}A& =& 2{\int }_0^2\sqrt{4-y}dy\\ & =& 2{\left[-\frac{{\left(4-y\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^2\\ & =& -\frac{4}{3}\left(0-4^{\frac{3}{2}}\right)\\ & =& \frac{4}{3}\left(8\right)\\ & =& \frac{32}{3}\end{array}$

Therefore, the required area is $\boxed{\frac{32}{3} \mathrm{square} \mathrm{units}}$