**18.** 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) \( 3\pi + 8 \)

1. Identify the two curves

• Circle: ((x-2\sqrt{3})^2 + y^2 = 12).
  Centre at $(2\sqrt{3},0)$, radius $R = \sqrt{12}=2\sqrt{3}$.
• Parabola: $y^2 = 2\sqrt{3}\,x$.
  Opens to the right with vertex at the origin.

2. Find the intersection points

Set $y^2 = 2\sqrt{3}x$ into the circle:

which simplifies to

Thus intersection points are

• $(0,0)$ (bottom-most contact)
• $(2\sqrt{3},\;\pm\,2\sqrt{3})$ (two symmetric top points).

3. Decide which side is called “outside the parabola”

For a right-opening parabola, left of the curve (smaller $x$) is usually described as “outside”. Try a test point, say $(0,1)$; it lies left of the parabola and (once checked) inside the circle, so this is the required region.

4. Set up a horizontal-strip integral

Take a strip at height $y$ (there is symmetry about the $x$-axis, so we integrate $0\to R$ and double):

• Left boundary (circle):
  $x_{\text{circle}} = 2\sqrt{3} - \sqrt{12 - y^2}$
• Right boundary (parabola):
  $x_{\text{para}} = \frac{y^2}{2\sqrt{3}}$
• Width kept:
  $w(y)=x_{\text{para}}-x_{\text{circle}} = -2\sqrt{3}+\sqrt{12-y^2}+\frac{y^2}{2\sqrt{3}}.$

The required area is therefore

5. Evaluate each part separately

1. Constant part: $\int_{0}^{2\sqrt{3}}(-2\sqrt{3})dy = -12$
2. Quarter-circle part: $\int_{0}^{2\sqrt{3}}\sqrt{12-y^2}\,dy = \tfrac14(\pi R^2)=3\pi$
3. Parabola part: $\int_{0}^{2\sqrt{3}}\frac{y^2}{2\sqrt{3}}dy = 4$ Putting them together and doubling for symmetry:

Thus, option (3) is correct.

What’s happening here?

Imagine you have a big round pizza (the circle) and over it you draw a curvy banana-like line (the right-opening parabola).

• We only want the part of the pizza that is still inside the round crust but outside the banana line (so the banana part is to the right, we keep the left sliver that the banana did not cover).
• To know how much pizza that is, we cut the pizza into very thin horizontal strips, measure how wide each strip is that we keep, and add (integrate) all those tiny widths.

The maths tells us that little sliver’s area turns out to be option (3) :
$6\pi - 16$

Step-by-Step Solution

1. Circle data: centre $(2\sqrt{3},0)$, radius $2\sqrt{3}$.
2. Parabola data: $y^2 = 2\sqrt{3}x$ opens to the right.
3. Intersections: Solve simultaneously to get $(0,0),\;(2\sqrt{3},\;2\sqrt{3}),\;(2\sqrt{3},\;-2\sqrt{3}).$
4. Region chosen: “Outside the parabola” means left of it (test point $(0,1)$ works).
5. Horizontal strip width:
   $w(y)=\frac{y^2}{2\sqrt{3}}-\Bigl(2\sqrt{3}-\sqrt{12-y^2}\Bigr).$
6. Area integral (double for symmetry): $$A = 2\int_{0}^{2\sqrt{3}}\left[-2\sqrt{3}+\sqrt{12-y^2}+\frac{y^2}{2\sqrt{3}}\right]dy.$$
7. Evaluate: a) $\displaystyle\int_{0}^{2\sqrt{3}}(-2\sqrt{3})dy=-12$
   b) $\displaystyle\int_{0}^{2\sqrt{3}}\sqrt{12-y^2}\,dy = 3\pi$ (quarter circle)
   c) $\displaystyle\int_{0}^{2\sqrt{3}}\frac{y^2}{2\sqrt{3}}dy = 4$
   Combine: $-12+3\pi+4=-8+3\pi$. Double:
   $A = 2(-8+3\pi) = 6\pi - 16.$

Final Answer: $(3)\;6\pi - 16$