Question

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

Answer 1

Given curves are

$x^2+y^2=25$

$4y=|4-x^2|$

$x=0$

and above x-axis

Solving (1) and (2), we get

$4y+4+y^2=25$

$\Rightarrow (y+2{)}^2=5^2$

$\Rightarrow y=3,-7$

$y=-7$ is rejected, $y=3$ gives the points above x-axis.

When $y=3,x=\pm 4$

Hence, the points of intersection are $P(4,3)$ and $Q(-4,3)$

The required area is

$=2\left[{\int }_0^4\sqrt{25-x^2}dx-\frac{1}{4}{\int }_0^2(4-x^2)dx-\frac{1}{4}{\int }_2^4(x^2-4)dx\right]$

$=2{[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5})}_0^4-\frac{1}{4}{(4x-\frac{x^3}{3})}_0^2-\frac{1}{4}{(\frac{x^3}{3}-4x)}_2^4$

$=2[6+\frac{25}{2}{\sin }^{-1}\frac{4}{5}]-\frac{1}{4}[8-\frac{8}{3}]-\frac{1}{4}[(\frac{64}{3}-16)-(\frac{8}{3}-8)]$

$=4+25{\sin }^{-1}\frac{4}{5}sq.units$