Q.Find the area in the first quadrant enclosed by the $x$-axis, the line $x = \sqrt{3}\,y$ and the circle $x^2 + y^2 = 4$, using definite integral.

The line $x=\sqrt3\,y$ (slope $30^\circ$) and circle $x^2+y^2=4$ meet at $(\sqrt3,1)$; the enclosed first-quadrant region is a $30^\circ$ sector, and integrating gives area $\dfrac{\pi}{3}$ square units.

Concept. Compute an area bounded by lines and a circle by splitting the $x$-range at the intersection and integrating the top boundary (line, then circular arc) minus the $x$-axis — the standard NCERT Class 12 mathematics definite-integral method. Geometrically the answer is a sector of angle $\theta$: area $=\tfrac12R^2\theta$.

Geometry. $x=\sqrt3\,y\Rightarrow y=\dfrac{x}{\sqrt3}$, so $\tan\theta=\dfrac{y}{x}=\dfrac{1}{\sqrt3}\Rightarrow\theta=30^\circ=\dfrac{\pi}{6}$. The circle has radius $R=2$. Line meets circle: $3y^2+y^2=4\Rightarrow y=1,\ x=\sqrt3$, i.e. $(\sqrt3,1)$.

Set up the integral. For $0\le x\le\sqrt3$ the top boundary is the line $y=\dfrac{x}{\sqrt3}$; for $\sqrt3\le x\le2$ it is the arc $y=\sqrt{4-x^2}$:

$A=\int_0^{\sqrt3}\frac{x}{\sqrt3}\,dx+\int_{\sqrt3}^{2}\sqrt{4-x^2}\,dx.$

Evaluate the first integral.

$\int_0^{\sqrt3}\frac{x}{\sqrt3}\,dx=\frac{1}{\sqrt3}\cdot\frac{x^2}{2}\Big|_0^{\sqrt3}=\frac{1}{\sqrt3}\cdot\frac{3}{2}=\frac{\sqrt3}{2}.$

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