The area bounded by the curves |x| + |y| geq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The region $|x| + |y| \leq 1$ represents a square with vertices at $(1, 0), (0, 1), (-1, 0), (0, -1)$.The region $x^2 + y^2 \leq 1$ represents a c

Vedclass · Accepted Answer

$(\pi - 2) 	ext{ sq. units}$

Vedclass · Answer

(C) The region $|x| + |y| \leq 1$ represents a square with vertices at $(1, 0), (0, 1), (-1, 0), (0, -1)$.The region $x^2 + y^2 \leq 1$ represents a circle with radius $r = 1$ centered at the origin.The area of the circle is $A_c = \pi r^2 = \pi(1)^2 = \pi$.The area of the square $|x| + |y| \leq 1$ is $A_s = \frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 2 	imes 2 = 2$.Since the square is inscribed within the circle,the area bounded by the curves $x^2 + y^2 \leq 1$ and $|x| + |y| \geq 1$ is the area of the circle minus the area of the square.Required Area $= A_c - A_s = \pi - 2 	ext{ sq. units}$.

The area bounded by the curves $|x| + |y| \geq 1$ and $x^2 + y^2 \leq 1$ is

Similar Questions

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is

The area enclosed by the curves $y=\ln(x+e^{2})$,$x=\ln(2/y)$ (which is $y=2e^{-x}$) and $x=\ln 2$,above the line $y=1$ is:

The area (in sq. units) enclosed between the curves $y=x^2$ and $y=|x|$ is

If the area of the region $\{(x, y) : |x-5| \leq y \leq 4 \sqrt{x}\}$ is $A$,then $3A$ is equal to . . . . . . .

The area of the region enclosed by the parabola $(y-2)^2=x-1$,the line $x-2y+4=0$,and the positive coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module