Question

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

Answer: (B)

$\frac{2}{3}$.sq.unit

$1-y^2=|x|$

$|x|+|y|=1$

$\begin{array}{cc}\text{Area}& ={\int }_0^{1}xdy\\ & ={\int }_0^1(1-y^2)dy\end{array}$

$={[y-\frac{y^3}{3}]}_0^1$

$=1-1/3$

$=\frac{2}{3}$ sq.unit (Area of one region)

$\begin{array}{cc}\text{Area}& =\frac{2}{3}-\frac{1}{2}\\ & =4[\frac{2}{3}-\frac{1}{2}]\\ & =\frac{4}{6}=\frac{2}{3}\text{squnits}\end{array}$

Correct option is (B).