The area in square units of the region bounde | vedclass

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

Question

(D) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.We need to find the area

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.We need to find the area bounded by $x = 0$ and $x = 4$ above the $X$-axis.This region represents one-quarter of the circle with radius $r = 4$.The area of a full circle is $A = \pi r^2 = \pi(4)^2 = 16\pi$.The area of the required region is $\frac{1}{4} 	imes 16\pi = 4\pi$ square units.

The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

Similar Questions

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

The area under the curve $y = x^2 - 4x$ within the $x$-axis and the line $x = 2$ is:

Find the area bounded by the curves $\{(x, y): y \geq x^{2} \text{ and } y=|x|\}$.

The area of the circle $x^{2}+y^{2}=16$ exterior to the parabola $y^{2}=6x$ is

The area of the region bounded by the $y$-axis,$y = \cos x$,and $y = \sin x$ for $0 \leq x \leq \frac{\pi}{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module