Question

The area included between the curve $x^2+y^2=a^2$ and $\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a}(a>0)$ is:

• A. $(\pi +\frac{2}{3})a^2$
• B. $(\pi -\frac{2}{3})a^2$
• C. $\frac{2}{3}{a}^2$
• D. $\frac{2\pi }{3}{a}^2$

Answer 1

Answer: (B)

$(\pi -\frac{2}{3})a^2$

The graphs $\left|x\right|+\left|y\right|=a$ and ${\left|x\right|}^2+{\left|y\right|}^2=a^2$ are as shown in the figure

From the figure, it is clear that when powers of $|x|$ and $|y|$ both are reduced to half the straight lines get stretched inside.

Thus, the required area

$=4$ $[$shaded area in the first quadrant$]$

$=4[\frac{\pi a^2}{4}-{\int }_0^a{(\sqrt{a}-\sqrt{x})}^{2}dx]=(\pi -\frac{2}{3})a^2$

$[$In the first quadrant $x,y>0,$ therefore, $\sqrt{\left|x\right|}+\sqrt{\left|y\right|}=\sqrt{a}\Rightarrow \sqrt{x}+\sqrt{y}=\sqrt{a}\Rightarrow y={(\sqrt{a}-\sqrt{x})}^2]$