Question

The area of the region bounded by $1-y^2=|x|$ and $|x|+|y|=1$ is

• A. $\frac{1}{3}$ sq unit
• B. $\frac{2}{3}$ sq unit
• C. $\frac{4}{3}$ sq unit
• D. $1$ sq unit

Answer 1

The correct option is B $\frac{2}{3}$ sq unit

Since, $|x|+|y|=1$
$\Rightarrow \{\begin{array}{ccc}x+y=1,& x>0,& y>0\\ x-y=1,& x>0,& y<0\\ -x+y=1,& x<0,& y>0\\ -x-y=1,& x<0,& y<0\end{array}$
and $1-y^2=|x|$
$\Rightarrow \{\begin{array}{cc}1-y^2=x,& x\ge 0\\ 1-y^2=-x,& x<0\end{array}$
Therefore, required area is
$=|2{\int }_0^1\sqrt{(1-x)}dx|+|2{\int }_{-1}^0\sqrt{(x+1)}dx|-4(\frac{1}{2}\cdot 1\cdot 1)$
$=\frac{2}{3}$ sq. unit