Area bounded by $y=4-x^2,y=0$ and $y=3$ is

The given curves are $y=4-x^2,y=0$ and the line $y=3$

The rough sketch of the required area is as below

The area of the required region is the area of the shaded region

Area of the shaded region is two times the area of the region bounded by the curve $y=4-x^2$ and the lines $y=0,y=3$ and $y-$ axis

it implies that $A=2{\int }_0^3\sqrt{4-y}dy$

Hence, 

$\begin{array}{rcl}A& =& 2{\int }_0^3\sqrt{4-y}dy\\ & =& 2{\left(-\frac{{\left(4-y\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right)}_0^3\\ & =& \frac{4}{3}\left(-1+8\right)\\ & =& \frac{28}{3}\end{array}$

Therefore, the area of the required region is $\boxed{\frac{28}{3}}$ square units