Question

Find the area bounded by the circle x² + y² = 16 and the line ${\displaystyle \sqrt{3}y=x}$ in the first quadrant, using integration.

Answer 1

The area bounded by the circle x² + y² = 16 , x = ${\displaystyle \sqrt{3}y=x}$ , and the x-axis is the area OAB.

Solving x² + y² = 16 , x = ${\displaystyle \sqrt{3}y=x}$ we have

${\displaystyle {\left(\sqrt{3}y\right)}^2+y^2=16}$

⇒3y² + y² = 16

⇒4y² = 16

⇒y² = 4 

⇒ y = 2 (In the first quadrant, y is positive)

When y = 2, x = ${\displaystyle 2\sqrt{3}}$

So, the point of intersection of the given line and circle in the first quadrant is ${\displaystyle (2\sqrt{3},2)}$

The graph of the given line and cirlce is shown below:

Required area =  Area of the shaded region = Area OABO = Area OCAO + Area ACB

Area OCAO = ${\displaystyle \frac{1}{2}\times 2\sqrt{3}\times 2=2\sqrt{3}}$ sq units

Area ABC = ${\displaystyle {\int }_{2\sqrt{3}}^{4}ydx}$

= ${\displaystyle {\int }_{2\sqrt{3}}^4\sqrt{16-x^2}dx}$

${\displaystyle ={[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\frac{{\sin }^{-1}x}{4}]}_{2\sqrt{3}}^4}$

${\displaystyle =[(0+8{\sin }^{-1}1)-(\frac{2\sqrt{3}}{3}\times 2+8\times \frac{{\sin }^{-1}\sqrt{3}}{2})]}$

${\displaystyle =8\times \frac{\pi }{2}-2\sqrt{3}-8\times \frac{\pi }{3}}$

= ${\displaystyle (\frac{4\pi }{3}-2\sqrt{3})}$ sq unit

∴ Required area = ${\displaystyle (\frac{4\pi }{3}-2\sqrt{3})+2\sqrt{3}=\frac{4\pi }{3}}$ sq units