Question

If the area bounded by the curves(lies above the $x-$axis) $x^2+y^2=25$ and $4y=|4-x^2|$ is $a{\sin }^{-1}\frac{b}{5}+b\text{sq.units}$. Then the value of $(\sqrt{a}+b)=$

Answer 1

The bounded region is shown in the figure :
So, Required area $=2\underset{0}{\overset{4}{\int }}(\sqrt{25-x^2}-|1-\frac{x^2}{4}|)dx=2\left[\underset{0}{\overset{4}{\int }}\sqrt{25-x^2}dx-\underset{0}{\overset{2}{\int }}(1-\frac{x^2}{4})dx+\underset{2}{\overset{4}{\int }}(1-\frac{x^2}{4})dx\right]=2[6+{\sin }^{-1}\frac{4}{5}-\frac{4}{3}-\frac{8}{3}]=25{\sin }^{-1}\frac{4}{5}+4=a{\sin }^{-1}\frac{b}{5}+b$
$\therefore \sqrt{a}+b=9$