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:
- (1) $6\pi-8$
- (2) $3\pi-8$
- (3) $6\pi-16$
- (4) $3\pi+8$
Solution:
Equation of circle: $(x-2\sqrt{3})^2 + y^2 = (2\sqrt{3})^2$.
Centre $C(2\sqrt{3}, 0)$ and radius $r = 2\sqrt{3}$. This circle passes through the origin $(0,0)$.
Equation of parabola: $y^2 = 2\sqrt{3}x$.
Intersection points:
Substitute $y^2$ in circle equation:
$$ (x-2\sqrt{3})^2 + 2\sqrt{3}x = 12 $$
$$ x^2 – 4\sqrt{3}x + 12 + 2\sqrt{3}x = 12 $$
$$ x^2 – 2\sqrt{3}x = 0 \Rightarrow x(x-2\sqrt{3}) = 0 $$
So, $x=0$ and $x=2\sqrt{3}$.
The region “inside circle and outside parabola” is the area of the semicircle (right half) minus the area of the parabola from $x=0$ to $x=2\sqrt{3}$.
$$ \text{Area} = \text{Area of Semicircle} – \int_{0}^{2\sqrt{3}} (\text{upper} – \text{lower parabola}) dx $$
$$ A = \frac{\pi r^2}{2} – 2\int_{0}^{2\sqrt{3}} \sqrt{2\sqrt{3}x} \, dx $$
$$ A = \frac{\pi(12)}{2} – 2\sqrt{2\sqrt{3}} \int_{0}^{2\sqrt{3}} x^{1/2} \, dx $$
$$ A = 6\pi – 2\sqrt{2\sqrt{3}} \left[ \frac{2}{3}x^{3/2} \right]_{0}^{2\sqrt{3}} $$
$$ A = 6\pi – \frac{4\sqrt{2\sqrt{3}}}{3} (2\sqrt{3})^{3/2} $$
$$ A = 6\pi – \frac{4}{3} \sqrt{2\sqrt{3}} \cdot (2\sqrt{3}) \cdot \sqrt{2\sqrt{3}} $$
$$ A = 6\pi – \frac{4}{3} (2\sqrt{3}) (2\sqrt{3}) = 6\pi – \frac{4}{3}(12) = 6\pi – 16 $$
Ans. (3)