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

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Solution

The correct option is C

$A r e a = \int_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} d x + \int_{\sqrt{3}}^{2} \sqrt{4 - x^{2}} d x = \frac{π}{3} s q . u n i t s$

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Similar questions

x = \sqrt{3} y

x^{2} + y^{2} = 4

\sqrt{3} y

x = \sqrt{3} y

x^{2} + y^{2} = 4

x = \sqrt{3} y

Find the area of the region in the first quadrant enclosed by X-axis and $x = \sqrt{3}$ y and the circle $x^{2} + y^{2} = 4$

y = x

x^{2} + y^{2} = 32.

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