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

Answer 1

The correct option is B

$\frac{2}{3}$ sq units

Explanation for correct option

Finding the area of the region bounded by the two equations

Step-1 Redefining the curve:

$$\begin{gathered} 1-y^2=\left|x\right|.....\left(1\right) \\ \left|x\right|+\left|y\right|=1.....\left(2\right) \end{gathered}$$

When $\left|x\right|+\left|y\right|=1$,

$\Rightarrow \left\{\begin{array}{ll}x+y=1,& x>0,y>0\\ x-y=1,& x>0,y<0\\ -x+y=1& x<0,y>0\\ -x-y=1& x<0,y<0\end{array}\right.$

When $1-y^2=\left|x\right|$,

$\Rightarrow \left\{\begin{array}{ll}1-y^2=x& x\ge 0\\ 1-y^2=-x& x<0\end{array}\right.$

Step-2 enclosed area calculations:

The area by the second equation in 1st quadrant is given by,

$$\begin{gathered} A_1={\int }_0^{1}xdy \\ ={\int }_0^{1}1-y^{2}dy \\ ={\left[y-\frac{y^3}{3}\right]}_0^1 \\ =\left(1-\frac{1}{3}\right)-(0-0) \\ =\frac{2}{3}\text{sq.units.} \end{gathered}$$

Now, we have to subtract the area enclosed by right-angled triangle to get the area enclosed by the curve in the first quadrant.

$$\begin{gathered} A_2=A_1-\frac{1}{2}\times l\times b \\ =\frac{2}{3}-\frac{1}{2}\times 1\times 1 \\ =\frac{2}{3}-\frac{1}{2} \\ =\frac{1}{6} \end{gathered}$$

Now, an area in each quadrant will be the same. So, the total area in all the four quadrants is given by,

$$\begin{gathered} A_{all}=4\left(A_2\right) \\ =4\times \frac{1}{6} \\ =\frac{4}{6} \\ =\frac{2}{3}\text{sq units} \end{gathered}$$
Hence, the correct answer is option (B).