Let R be the region (x, y): x^2+y^2 ≤ 16,\ y ≥ |x| . Which of the fol…

The correct answer is Option (4) → (A), (B) and (C) only

On the line y = x the circle gives

    \(2x^{2} = 16,\ \ x = \pm2\sqrt{2}\)

so above the x-axis the meeting points are \((2\sqrt{2}, 2\sqrt{2})\) on y = x and \((-2\sqrt{2}, 2\sqrt{2})\) on y = −x.

R lies above y = |x| and inside the circle. Each of the lines y = x and y = −x makes \(45^{\circ}\) with the y-axis, so R is a sector of angle \(90^{\circ}\), one quarter of the disc of radius 4:

    \( \displaystyle \frac{1}{4}\pi(4)^{2} = 4\pi \)

By integration, between \(x = -2\sqrt{2}\) and \(x = 2\sqrt{2}\) the top of R is \(y = \sqrt{16-x^{2}}\) and the bottom is y = |x|. By symmetry about the y-axis:

    \( \displaystyle \int_{-2\sqrt{2}}^{2\sqrt{2}}\left(\sqrt{16-x^{2}}-|x|\right)dx = 2\int_{0}^{2\sqrt{2}}\left(\sqrt{16-x^{2}}-x\right)dx \)

    \( \displaystyle = 2\left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\frac{x}{4}-\frac{x^{2}}{2}\right]_{0}^{2\sqrt{2}} \)

    \( \displaystyle = 2\left(4+8\times\frac{\pi}{4}-4\right) = 4\pi \)

The upper half-disc has area \(\frac{1}{2}\pi(4)^{2} = 8\pi\).

So the meeting points are \((\pm2\sqrt{2}, 2\sqrt{2})\), the area of R is \(4\pi\) (also given by the integral in (C)), and the upper half-disc has area \(8\pi\).

Checking the statements

(A) The lines y = x and y = −x meet the circle \(x^{2}+y^{2}=16\) above the x-axis at the points \((2\sqrt{2}, 2\sqrt{2})\) and \((-2\sqrt{2}, 2\sqrt{2})\) — Correct. \(8+8 = 16\) for both points.

(B) The area of R is \(4\pi\) sq. units — Correct. R is a quarter of the disc of area \(16\pi\).

(C) The area of R is \(\displaystyle\int_{-2\sqrt{2}}^{2\sqrt{2}}\left(\sqrt{16-x^{2}}-|x|\right)dx\) — Correct. Top minus bottom of R over its x-range; the integral equals \(4\pi\).

(D) The area of R is one-third of the area of the region \(\{(x, y): x^{2}+y^{2} \le 16,\ y \ge 0\}\) — Incorrect. \(4\pi\) is one-half of \(8\pi\).