From the simplified quadratic equation, we find two possible values for x: x=0 and x=2√3. Substitute these values back into the parabola equation y² = 2√3 x to find the corresponding y -coordinates. This gives us the intersection points (0,0), (2√3, 2√3), and (2√3, -2√3).

From the simplified quadratic equation, we find two possible values for x: x=0 and x=2√3. Substitute these values back into the parabola equation y² = 2√3 x to find the corresponding y -coordinates. This gives us the intersection points (0,0), (2√3, 2√3), and (2√3, -2√3).

Let u = x - 2√3, so du = dx. When x=0, u = -2√3. When x=2√3, u = 0. The integral becomes ∫_-2√3^(0) √(12 - u²) du. This represents the area of a quarter circle of radius √12 = 2√3. Thus, the integral evaluates to (1)/(4) π (2√3)² = (1)/(4) π (12) = 3π.

Let u = x - 2√3, so du = dx. When x=0, u = -2√3. When x=2√3, u = 0. The integral becomes ∫_-2√3^(0) √(12 - u²) du. This represents the area of a quarter circle of radius √12 = 2√3. Thus, the integral evaluates to (1)/(4) π (2√3)² = (1)/(4) π (12) = 3π.

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:

Answer: $6\pi - 16$

Step-by-step solution

Step 1: Identify the equations and find intersection points

We are given the equations of a circle and a parabola. The circle is centered at $(2\sqrt{3}, 0)$ with radius $\sqrt{12} = 2\sqrt{3}$ . The parabola opens to the right with its vertex at the origin. To find the area of the region, we first need to find the points where the circle and parabola intersect.

Step 2: Solve for intersection points

Substitute the equation of the parabola, $y^2 = 2\sqrt{3} x$ , into the equation of the circle. This eliminates $y$ and allows us to solve for the $x$ -coordinates of the intersection points. We expand the circle equation and simplify to get a quadratic equation in $x$ .

Step 3: Determine intersection coordinates

From the simplified quadratic equation, we find two possible values for $x$ : $x=0$ and $x=2\sqrt{3}$ . Substitute these values back into the parabola equation $y^2 = 2\sqrt{3} x$ to find the corresponding $y$ -coordinates. This gives us the intersection points $(0{,}0)$ , $(2\sqrt{3}, 2\sqrt{3})$ , and $(2\sqrt{3}, -2\sqrt{3})$ .

Step 4: Set up the integral for the area

The required area is the region inside the circle and outside the parabola. Due to symmetry about the $x$ -axis, we can calculate the area in the upper half and multiply by 2. The upper boundary is given by the circle $y = \sqrt{12 - (x - 2\sqrt{3})^2}$ and the lower boundary by the parabola $y = \sqrt{2\sqrt{3} x}$ . The integration limits are from $x=0$ to $x=2\sqrt{3}$ .

Step 5: Evaluate the integral for the circular segment

Let $u = x - 2\sqrt{3}$ , so $du = dx$ . When $x=0$ , $u = -2\sqrt{3}$ . When $x=2\sqrt{3}$ , $u = 0$ . The integral becomes $\int_{-2\sqrt{3}}^{0} \sqrt{12 - u^2} du$ . This represents the area of a quarter circle of radius $\sqrt{12} = 2\sqrt{3}$ . Thus, the integral evaluates to $\frac{1}{4} \pi (2\sqrt{3})^2 = \frac{1}{4} \pi (12) = 3\pi$ .

Step 6: Evaluate the integral for the parabolic segment

Now we evaluate the integral for the parabolic part. We integrate $x^{1/2}$ and apply the limits of integration. This gives us $\sqrt{2\sqrt{3}} \cdot \frac{2}{3} (2\sqrt{3})^{3/2} = \frac{2}{3} (2\sqrt{3})^2 = \frac{2}{3} (12) = 8$ .

Step 7: Calculate the total area

The total area is twice the difference between the area under the circular arc and the area under the parabolic arc from $x=0$ to $x=2\sqrt{3}$ . Substituting the calculated values, we get $2(3\pi - 8) = 6\pi - 16$ .

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The area of the region, inside the circle (x−23)2+y2=12(x - 2\sqrt{3})^2 + y^2 = 12(x−23​)2+y2=12 and outside the parabola y2=23xy^2 = 2\sqrt{3} xy2=23​x is: