Question

Consider the following regions in the $XY$plane:
$R_1=\left\{\right(x,y):0\le x\le 1\text{and}0\le y\le 1\}$
$R_2=\left\{\right(x,y):x^2+y^2\le \frac{4}{3}\}$
The area of the region $R_1\cap R_2$ can be expressed as $\frac{a\sqrt{3}+b\pi }{9},$ where $a$ and $b$ are integers. Then the value of $(a+b)$ is

Answer 1

Required area $=\frac{1}{\sqrt{3}}+A_1,$
where $A_1=\underset{1/\sqrt{3}}{\overset{1}{\int }}\sqrt{\frac{4}{3}-x^2}dx$
Put $x=\frac{2}{\sqrt{3}}\sin \theta$
$\Rightarrow dx=\frac{2}{\sqrt{3}}\cos \theta d\theta$
$A_1=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{2}{\sqrt{3}}\cos \theta \cdot \frac{2}{\sqrt{3}}\cos \theta d\theta$
$=\frac{2}{3}\underset{\pi /6}{\overset{\pi /3}{\int }}(\cos 2\theta +1)d\theta$
$=\frac{2}{3}{[\frac{\sin 2\theta }{2}+\theta ]}_{\pi /6}^{\pi /3}$
$=\frac{2}{3}[\frac{1}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\pi }{3}-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}-\frac{\pi }{6}]$
$=\frac{\pi }{9}$
Hence, required area $=\frac{1}{\sqrt{3}}+\frac{\pi }{9}=\frac{3\sqrt{3}+\pi }{9}$
$a=3,b=1$
$\therefore a+b=4$