The area lying in the first quadrant and boun | vedclass

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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

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