1. Tardigrade
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  4. Consider the following regions in the plane: R1= (x, y): 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 and R2= (x, y): x2+y2 ≤ 4 / 3 The area of the region R 1 ∩ R 2 can be expressed as ( a √3+ b π/9), where a and b are integers, then -

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Q. Consider the following regions in the plane: $R_1=\{(x, y): 0 \leq x \leq 1$ and $0 \leq y \leq 1\}$ and $R_2=\left\{(x, y): x^2+y^2 \leq 4 / 3\right\}$ 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 -

Application of Integrals

Solution:

image
$A=\frac{1}{\sqrt{3}}+\int\limits_{1 / \sqrt{3}}^1 \sqrt{\frac{4}{3}-x^2} d x$
$=\frac{1}{\sqrt{3}}+\left[\frac{x}{2} \sqrt{\frac{4}{3}-x^2}+\frac{2}{3} \sin ^{-1}\left(\frac{x \sqrt{3}}{2}\right)\right]_{1 / \sqrt{3}}^1$
$=\frac{1}{\sqrt{3}}+\left[\left(\frac{1}{2 \sqrt{3}}-\frac{1}{2 \sqrt{3}}\right)+\frac{2}{3}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\right]=\frac{3 \sqrt{3}+\pi}{9}$