The area bounded by the circle x^2 + y^2 = 4, | vedclass

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(C) The given circle is $x^2 + y^2 = 2^2$,which has a radius $R = 2$. The line is $y = \frac{x}{\sqrt{3}}$,which makes an angle $	heta$ with the $x$-

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) The given circle is $x^2 + y^2 = 2^2$,which has a radius $R = 2$. The line is $y = \frac{x}{\sqrt{3}}$,which makes an angle $	heta$ with the $x$-axis such that $	an 	heta = \frac{1}{\sqrt{3}}$,so $	heta = \frac{\pi}{6}$.The area bounded by the circle,the line,and the $x$-axis in the first quadrant is the area of the sector of the circle with radius $R = 2$ and central angle $	heta = \frac{\pi}{6}$.The area of a sector is given by the formula $A = \frac{1}{2} R^2 	heta$.Substituting the values,we get $A = \frac{1}{2} 	imes (2)^2 	imes \frac{\pi}{6} = \frac{1}{2} 	imes 4 	imes \frac{\pi}{6} = \frac{\pi}{3}$.Thus,the correct option is $C$.

The area bounded by the circle $x^2 + y^2 = 4$,the line $x = \sqrt{3}y$,and the $x$-axis lying in the first quadrant is:

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