Question

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

Answer 1

$4y=|4-x^2|$ , $x^2+y^2=25$ , $x=0$

Intersection points

$\Rightarrow$ $x^2+y^2=25$ and $4y=|x^2-4|$

Solving

$y=3$ , $x=4$

As it is with $x=0$ we need to find area $DABCD$

$A=(0,1)$ , $B=(2,0)$ , $C=(4,3)$ , $D=(0,5)$ , $E=(-4,3)$ , $F=(-2,0)$

$DABCD={\int }_0^4\sqrt{25-x^2}-\frac{1}{4}|4-x^2|dx$

$={[\frac{x\sqrt{25-x^2}}{2}+\frac{25{\sin }^{-1}c/5}{2}]}_0^4-\frac{1}{4}{\int }_0^2[4-x^2]dx-\frac{1}{4}{\int }_2^4-[4-x^2]dx$

$\Rightarrow$ $\frac{4}{2}\times 3+\frac{25{\sin }^{-1}\left(4/5\right)}{2}-\frac{1}{4}{[4x-\frac{x^3}{3}]}_0^2+\frac{1}{4}{[4x-\frac{x^3}{3}]}_2^4$

$\Rightarrow$ $6+\frac{25{\sin }^{-1}\left(4/5\right)}{2}-16$

$\Rightarrow$ $\frac{25{\sin }^{-1}\left(4/3\right)}{2}-10$