Question

The area of the smaller region bounded by the circle $x^2+y^2=1$ and $\left|y\right|=x+1$ is

• A. $\frac{\pi }{4}-\frac{1}{2}$
• B. $\frac{\pi }{2}-1$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{2}+1$

Answer 1

The correct option is B $\frac{\pi }{2}-1$

The equation of circle is $x^2+y^2=1$

$\therefore$ $r=1$

$|y|=x+1$

The first line is $y=x+1$

when $y=0$ we get, $x=-1$

When $x=0$ we get, $y=1$

So we get points $A(-1,0),B(0,1)$

The second line is $-y=x+1$

when $y=0$ we get, $x=-1$

When $x=0$ we get, $y=-1$

So we get points $A(-1,0),B^{\prime }(0,-1)$

$\Rightarrow$ Required area $=2|{\int }_{-1}^0\sqrt{1-x^2}-(x+1\left)dx\right|$

$=2{|\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-\frac{x^2}{2}-x|}_{-1}^0$

$=2|0-(\frac{1}{2}{\sin }^{-1}-\frac{(-1{)}^2}{2}+1)|$

$=2|-(\frac{-\pi }{4}+\frac{1}{2})|$

$=\frac{\pi }{2}-1$