The area lying in the first quadrant and boun | vedclass

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

Question

(A) The given circle is $x^{2}+y^{2}=4$,which is a circle centered at the origin $(0,0)$ with radius $r=2$.In the first quadrant,$y = \sqrt{4-x^{2}}$.

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) The given circle is $x^{2}+y^{2}=4$,which is a circle centered at the origin $(0,0)$ with radius $r=2$.In the first quadrant,$y = \sqrt{4-x^{2}}$.The area bounded by the circle,the $y$-axis $(x=0)$,and the line $x=2$ is the area of the quarter circle in the first quadrant.Area $= \int_{0}^{2} y \, dx = \int_{0}^{2} \sqrt{4-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}ight)$:Area $= \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}ight) ight]_{0}^{2}$.Substituting the limits:$= \left( \frac{2}{2} \sqrt{4-4} + 2 \sin^{-1} \left(\frac{2}{2}ight) ight) - \left( 0 + 2 \sin^{-1}(0) ight)$.$= (0 + 2 \sin^{-1}(1)) - (0) = 2 	imes \frac{\pi}{2} = \pi$ square units.Thus,the correct option is $A$.

The area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x=0$ and $x=2$ is

Similar Questions

The area bounded by the curve $y = x(1 - \ln x)$,the line $x = e^{-1}$,and the positive $X$-axis between $x = e^{-1}$ and $x = e$ is:

If the area above the $x$-axis,bounded by the curves $y = 2^{kx}$,$x = 0$,and $x = 2$ is $\frac{3}{\ln 2}$,then the value of $k$ is

The area of the figure bounded by $y = e^x$,$y = e^{-x}$,and the straight line $x = 1$ is

If the area of the region $\{(x, y): -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a > 0\}$ is $\frac{e^2 + 8e + 1}{e}$,then the value of $a$ is:

The area bounded by the curve $|y| + \frac{1}{2} = e^{-|x|}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module