The area of the region, inside the circle and outside the parabola is : [2025]

Question

The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

Smart Achievers · Accepted Answer

(3)

We have, ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ , a circle with centre $(2 \sqrt{3}, 0)$ and radius $\sqrt{12}$ and a parabola $y^{2} = 2 \sqrt{3} x$ .

$∴$ Required Area = $\frac{π (12)}{2} - 2 \int_{0}^{2 \sqrt{3}} {(2 \sqrt{3})}^{\frac{1}{2}} x^{1 / 2} d x$

$= 6 π - 2 {(2 \sqrt{3})}^{1 / 2} {[\frac{x^{3 / 2}}{3} \cdot 2]}_{0}^{2 \sqrt{3}}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{1 / 2} {(2 \sqrt{3})}^{3 / 2}$

$= 6 π - \frac{4}{3} {(2 \sqrt{3})}^{2} = 6 π - 16$ .

Solution

Q.
The area of the region, inside the circle ${(x - 2 \sqrt{3})}^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 2 \sqrt{3} x$ is : [2025]

1 $3 π + 8$

2 $3 π - 8$

3 $6 π - 16$

4 $6 π - 8$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The area of the region, inside the circle (x–23)2+y2=12 and outside the parabola y2=23x is : [2025]

1 3π+8

2 3π–8

3 6π–16

4 6π–8