Question

Sketch the region bounded by the curves $y=\sqrt{5-x^2}$ and $y=|x-1|$ and find its area using integration.

Answer 1

The given curves are :

$x^2+y^2=5$

$[\because y=\sqrt{5-x^2}\Rightarrow y^2=5-x^2\Rightarrow x^2+y^2=5]$

and $y=\{\begin{array}{c}1-x,ifx<1\\ x-1,ifx>1\end{array}$

The required region is shown as shaded in the following figure :



$y=x-1$ meets $x^2+y^2=5$ at $B(2,1)$.

$y=1-x$ meets $x^2+y^2=5$ at $C(-1,2)$

$y=x-1$ and $y=1-x$ meet at $A(1,0)$.

$\therefore$ $\text{Reqd. area = area (MCBLM)- area (CMAC)- area (ALBA)}$

= ${\int }_{-1}^2\sqrt{5-x^2}dx-{\int }_{-1}^1(1-x)dx-{\int }_1^2(x-1)dx$

= ${[\frac{x\sqrt{5-x^2}}{2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}]}_{-1}^2-{[x-\frac{x^2}{2}]}_{-1}^1-{[\frac{x^2}{2}-x]}_1^2$

= $[(1+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{5}})-(-\frac{1}{2}\times 2+\frac{5}{2}{\sin }^{-1}(-\frac{1}{\sqrt{5}}))]$

$-[(1-\frac{1}{2})-(-1-\frac{1}{2})]-\left[\right(2-2)-(\frac{1}{2}-1)]$

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

= $-\frac{1}{2}+\frac{5}{2}[{\sin }^{-1}\frac{2}{\sqrt{5}}-{\sin }^{-1}(-\frac{1}{\sqrt{5}})]$

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

$\because {\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}[x.\sqrt{1-y^2}]+{\sin }^{-1}[y.\sqrt{1-x^2}]$
$x\ge 0,y\ge 0x^2+y^2\le 0$

= $-\frac{1}{2}+\frac{5}{2}\left[{\sin }^{-1}(\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}})\right]$

= $-\frac{1}{2}+\frac{5}{2}\left[{\sin }^{-1}(\frac{4}{5}+\frac{1}{5})\right]$

= $-\frac{1}{2}+\frac{5}{2}\left[{\sin }^{-1}1\right]=-\frac{1}{2}+\frac{5}{2}\times \frac{\pi }{2}$

= $(-\frac{1}{2}+\frac{5\pi }{4})$ sq. units