Question

Using integration find the area of the following region $(x,y):|x+2|\le y\le \sqrt{20-x^2}$.

Answer 1

$y\ge |x+2|$

$P,Q,R$ are points of intersection $y=\sqrt{20-x^2}$ with $y=x+2$ & $x=0$

from graph

$P=(2,4)$

$Q=(0,2)$

$R=(0,2\sqrt{5})$

area of shaded region $={\int }_0^2(\sqrt{20-x^2}-(x+2)0dx$

$={\int }_0^2\sqrt{20-x^2}dx-{\int }_0^2(x+2)dx$

$={[\frac{x}{2}\sqrt{20-x^2}+10{\sin }^{-1}\frac{x}{2\sqrt{5}}]}_0^2-{[\frac{x^2}{2}+2x]}_0^2$

$=4+10{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)-6$

Area $=10{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)-2$