The area of region R = \(x,y): x^2 + y^2 ≤ 16, y ≤ √3x, x ≥ 0, y ≥ 0\ is: (in Sq. units) Option 1: (2)/3 Option 2: 8 Option 3: (8)/3 Option 4: 3

Answer: (8)/3. Solution: The line y=3 x makes an angle of 60^ with the x-axis, so the region is the circular sector of radius 4 from =0 to =/3. Area = 1/2r^2 = 1/2(16)(/3) = 8/3.

Calculus

Application of Integrals

Medium

core

The area of region $R = \{(x,y): x^2 + y^2 \le 16, y \le \sqrt{3}x, x \ge 0, y \ge 0\}$ is: (in Sq. units)

$\dfrac{2\pi}{3}$

$8\pi$

$\dfrac{8\pi}{3}$

$3\pi$

✅ Correct Option: 3

The line $y=\sqrt3 x$ makes an angle of $60^\circ$ with the x-axis, so the region is the circular sector of radius $4$ from $\theta=0$ to $\theta=\pi/3$ .

Area $= \frac{1}{2}r^2\theta = \frac{1}{2}(16)(\pi/3) = 8\pi/3$ .

2023: 23 May Shift 3

The area of the region $\{(x, y) : y \geq x^2 \text{ and } y \leq |x|\}$ is

2025: 22 May Shift 2

The area of the smaller region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and the straight line $3x + 4y = 12$ is:

2026: 22 May Shift 2

The area (in sq. units) of the region bounded by the line $2y=3x+6$ , x-axis and the ordinates x=-3 and x=1 is:

2026: 19 May Shift 2

The area (in sq. units) of region bounded by the curves $y^2 = 4x$ and $y = x$ is:

2022: 25 May Shift 1

The area enclosed by the ellipse $\frac{x^2}{9^2} + \frac{y^2}{6^2} = 1$ is:

2026: 22 May Shift 1

Area bounded by the curves

$y=\begin{cases}x+2, & x\ge-1\\-x, & x<-1\end{cases}$ , $x=-2, x=3$ and $y=0$ is:

The area of region R={(x,y):x2+y2≤16,y≤3x,x≥0,y≥0}R = \{(x,y): x^2 + y^2 \le 16, y \le \sqrt{3}x, x \ge 0, y \ge 0\}R={(x,y):x2+y2≤16,y≤3​x,x≥0,y≥0} is: (in Sq. units)

Solution

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The area of region $R = \{(x,y): x^2 + y^2 \le 16, y \le \sqrt{3}x, x \ge 0, y \ge 0\}$ is: (in Sq. units)