Question

The area(in sq. units) of the region $A=\left\{\left(x,y\right):\left|x\right|+\left|y\right|\le 1,2y^2=\left|x\right|\right\}$

• A. $\frac{1}{6}$
• B. $\frac{5}{6}$
• C. $\frac{1}{3}$
• D. $\frac{7}{6}$

Answer 1

The correct option is B

$\frac{5}{6}$

Explanation for the correct option:

Finding the area:

Given, the curves are $\left|\mathrm{x}\right|+\left|\mathrm{y}\right|\le 1,2{\mathrm{y}}^2=\left|\mathrm{x}\right|$

Plotting the curves,

Total area,

$$\begin{gathered} =4{\int }_0^{\frac{1}{2}}\left[\left(1-\mathrm{x}\right)-\left(\sqrt{\frac{\mathrm{x}}{2}}\right)\right]d\mathrm{x} \\ =4{\left[\mathrm{x}-\frac{{\mathrm{x}}^2}{2}-\frac{1}{\sqrt{2}}\frac{{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^{\frac{1}{2}} \\ =4\left[\left(\frac{1}{2}\right)-\left(\frac{1}{8}\right)-\left(\frac{\sqrt{2}}{3}\right){\left(\frac{1}{2}\right)}^{\frac{3}{2}}\right] \\ =\frac{4\times 5}{24} \\ =\frac{5}{6}\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, option (B) is the correct answer