The area (in sq. units) of the region inside the circle \(x^{2}+y^{2}=16\) but outside the region \(|x|+|y| \le 4\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(16\pi-32\)
The region \(|x|+|y| \le 4\) is a square with corners (±4, 0) and (0, ±4). These corners lie on the circle of radius 4, so the square lies inside the circle, and the required area is the circle area minus the square area.
In the first quadrant the square is the triangle under the line y = 4 − x, and the square is symmetric about both axes:
\( \displaystyle \int_{0}^{4}(4-x)\,dx = \left[4x-\frac{x^{2}}{2}\right]_{0}^{4} = 8 \)
Square: \(4\times8 = 32\)
Circle: \(\pi\times4^{2} = 16\pi\)
\(A = 16\pi-32\)