The area (in sq. units) of the part of the circle x 2+ y 2=36, which is outside the parabola y 2=9 x , is

Question

The area (in sq. units) of the part of the circle x 2+ y 2=36, which is outside the parabola y 2=9 x , is (A) 24 π+3 √3 (B) 12 π-3 √3 (C) 24 π-3 √3 (D) 12 π+3 √3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7b262db21db10369f25cf1a78042d48e-.png" /> Required area =π ×(6)2-2 ∫ limits03 √9 x d x-∫ limits36 √36-x2 d x =36 π-12 √3-2(( x /2) √36- x 2+18 sin -1 ( x /6))6 =36 π-12 √3-2(9 π-3 π-(9 √3/2)) =24 π-3 √3

Q. The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36$ , which is outside the parabola $y^{2} = 9 x,$ is

$24 π + 33$

$12 π - 33$

$24 π - 33$

$12 π + 33$

Required area
$= π \times (6)^{2} - 2 0 \int 3 9 x d x - 3 \int 6 36 - x^{2} d x$
$= 36 π - 123 - 2 (\frac{x}{2} 36 - x^{2} + 18 sin^{- 1} \frac{x}{6})^{6}$
$= 36 π - 123 - 2 (9 π - 3 π - \frac{9 3}{2})$
$= 24 π - 33$

Q. The area (in sq. units) of the part of the circle x2+y2=36, which is outside the parabola y2=9x, is

24π+33​

12π−33​

24π−33​

12π+33​