Question

Using integration, find the area of the given region: $\left\{\right(x,y)||x-1|\le y\le \sqrt{5-x^2}\}$.

Answer 1

Given that,

$y_1=\left|x-1\right|$

and $y_2=\sqrt{5-x^2}$ are show in fly

let A be the area of hounded region

$A={\int }_{-1}^1\left(y_2-y_1\right)dx+{\int }_1^2\left(y_2-y_1\right)dx$

$A=\underset{-1}{\overset{1}{\int }}\left(\sqrt{5-x^2}+x-1\right)dx+\underset{1}{\overset{2}{\int }}\left(\sqrt{5-x^2}-x+1\right)dx$

$A=\underset{-1}{\overset{1}{\int }}\sqrt{5-x^2}dx+\underset{1}{\overset{2}{\int }}\sqrt{5-x^2}dx+\underset{-1}{\overset{1}{\int }}\left(x-1\right)dx+\underset{1}{\overset{2}{\int }}\left(-x+1\right)dx$

$A=\underset{-1}{\overset{2}{\int }}\sqrt{5-x^2}dx+{{\left[\frac{x^2}{2}-x\right]}_{-1}}^1+{\left[\frac{-x^2}{2}+x\right]}_1^2$

$A={\left[\frac{1}{2}x\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}\right]}_{-1}^2-\frac{5}{2}$

$A=1+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{5}}+1+\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{5}}-\frac{5}{2}$

$A=\frac{-1}{2}+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\times \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{3}}\sqrt{1-\frac{4}{5}}\right)$

$A=\frac{-1}{2}+\frac{5}{2}{\sin }^{-1}\left(1\right)$

Hence, we get ,

$\begin{array}{c}\frac{5\pi }{4}-\frac{1}{2}\\ \frac{5\pi -2}{4}\end{array}$