The area enclosed between the curve x2 + y2 = 16 and the coordinate axes in the first quadrant is

Question

The area enclosed between the curve x2 + y2 = 16 and the coordinate axes in the first quadrant is (A) 4 π sq. units (B) 3 π sq. units (C) 2 π sq. units (D) π sq. units

Tardigrade Admin · Accepted Answer

Given that, the circle with centre (0,0) and radius 4. Required area <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/96305d80278f67a36427b7c536611a04-.png" /> =∫ limits04√16-x2 dx =[(x/2)√16-x2+(16/2)sin-1(x/4)]04=4π sq. units

Q. The area enclosed between the curve $x^{2} + y^{2} = 16$ and the coordinate axes in the first quadrant is

$4 π$ sq. units

$3 π$ sq. units

$2 π$ sq. units

$π$ sq. units

Given that, the circle with centre $(0, 0)$ and radius $4$ . Required area

$= 0 \int 4 16 - x^{2} d x$
$= [\frac{x}{2} 16 - x^{2} + \frac{16}{2} s i n^{- 1} \frac{x}{4}]_{0}^{4} = 4 π$ sq. units

Q. The area enclosed between the curve x2+y2=16 and the coordinate axes in the first quadrant is

4π sq. units

3π sq. units

2π sq. units

π sq. units