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 -

Question

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 - (A) a =3 (B) a =1 (C) b=1 (D) b=3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/451f9de625341b8a2b4243a078722af8-.png" /> A=(1/√3)+∫ limits1 / √31 √(4/3)-x2 d x =(1/√3)+[(x/2) √(4/3)-x2+(2/3) sin -1((x √3/2))]1 / √31 =(1/√3)+[((1/2 √3)-(1/2 √3))+(2/3)((π/3)-(π/6))]=(3 √3+π/9)

Q. Consider the following regions in the plane: $R_{1} = {(x, y) : 0 \leq x \leq 1$ and $0 \leq y \leq 1}$ and $R_{2} = {(x, y) : x^{2} + y^{2} \leq 4/3}$ The area of the region $R_{1} \cap R_{2}$ can be expressed as $\frac{a 3 + bπ}{9}$ , where $a$ and $b$ are integers, then -

$a = 3$

$a = 1$

$b = 1$

$b = 3$

$A = \frac{1}{3} + 1/ 3 \int 1 \frac{4}{3} - x^{2} d x$
$= \frac{1}{3} + [\frac{x}{2} \frac{4}{3} - x^{2} + \frac{2}{3} sin^{- 1} (\frac{x 3}{2})]_{1/ 3}^{1}$
$= \frac{1}{3} + [(\frac{1}{2 3} - \frac{1}{2 3}) + \frac{2}{3} (\frac{π}{3} - \frac{π}{6})] = \frac{3 3 + π}{9}$

Q. 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 R1​∩R2​ can be expressed as 9a3​+bπ​, where a and b are integers, then -

a=3

a=1

b=1

b=3