Question

Find the area bounded by the curves
$x^2+y^2=25$, $4y=|4-x^2|$ and x = 0 above the X-axis.

Answer 1

Given curves, $x^2+y^2=25$, $4y=|4-x^2|$ could be
sketched as below, whose points of intersection are
$x^2+\frac{(4-x^2{)}^2}{16}=25$
$\Rightarrow (x^2+24)$ $(x^2-16)$=0 &\Rightarrow & $x=\pm 4$
$\therefore$ Required area=$2[{\int }_0^4\sqrt{25-x^2}dx-{\int }_0^2(\frac{4-x^2}{4})dx-{\int }_2^4(\frac{x^2-4}{4}\left)dx\right]$
=$2\left[\right[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{sin}^{-1}\frac{x}{5}{]}_0^4$
$-\frac{1}{4}[4x-\frac{x^3}{3}{]}_0^2-\frac{1}{4}[\frac{x^3}{3}-4x{]}_2^4]$
$\left[\right[6+\frac{25}{2}si{n}^{-1}\left(\frac{4}{5}\right)]$-$\frac{1}{4}[8-\frac{8}{3}]$
$-\frac{1}{4}\left[\right(\frac{64}{3}-16)-(\frac{8}{3}-8\left)\right]]$
$=2[6+\frac{25}{2}si{n}^{-1}(\frac{4}{5})-\frac{4}{3}-\frac{4}{3}-\frac{4}{3}]$
= $[4+25si{n}^{-1}(\frac{4}{5}\left)\right]$ sq units