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:

• A. \(6\pi - 8\)
• B. \(3\pi - 8\)
• C. \(6\pi - 16\)
• D. \(3\pi + 8\)

Hint:

To find the area between curves, integrate the difference of the functions over the interval of intersection.

Answer 1

The Correct Option is C

Area Calculation for Circle and Parabola Overlap

Given a circle and a parabola, the objective is to determine the area of the region situated within the circle but external to the parabola. The provided equations are:

Circle: \((x - 2\sqrt{3})^2 + y^2 = 12\)

Parabola: \(y^2 = 2\sqrt{3}x\)

Step 1: Geometric Interpretation

The circle has a center at \((2\sqrt{3}, 0)\) and a radius of \( \sqrt{12} = 2\sqrt{3} \). The parabola, oriented to the right, follows the standard form \( y^2 = 4ax \), with \( a = \frac{\sqrt{3}}{2} \).

Step 2: Identifying Intersection Points

To ascertain the points of intersection between the circle and the parabola, substitute the parabola's \( y^2 \) expression into the circle's equation.

Substituting \( y^2 \) from the parabola into the circle equation:

\[ (x - 2\sqrt{3})^2 + 2\sqrt{3}x = 12 \] Expanding the expression:

\[ (x^2 - 4\sqrt{3}x + 12) + 2\sqrt{3}x = 12 \] Simplifying the equation:

\[ x^2 - 4\sqrt{3}x + 2\sqrt{3}x + 12 = 12 \] \[ x^2 - 2\sqrt{3}x = 0 \] Factoring for \( x \):

\[ x(x - 2\sqrt{3}) = 0 \] The resulting \( x \) values are \( x = 0 \) and \( x = 2\sqrt{3} \), indicating the intersection points.

Step 3: Integral Setup for Area Computation

The area of the region enclosed by the circle and excluded by the parabola is found by calculating the area of the relevant circle segment and subtracting the area beneath the parabola.

The area under the circle, from \( x = 0 \) to \( x = 2\sqrt{3} \), is derived from integrating the upper semicircle equation:

\[ y = \sqrt{12 - (x - 2\sqrt{3})^2} \] The area under the parabola is computed via integration:

\[ y = \sqrt{2\sqrt{3}x} \] Consequently, the total area within the circle and outside the parabola is:

\[ A = \int_{0}^{2\sqrt{3}} \left( \sqrt{12 - (x - 2\sqrt{3})^2} - \sqrt{2\sqrt{3}x} \right) dx \] The evaluated result of this integral is \( 6\pi - 16 \).

Step 4: Final Result

Therefore, the area of the region that is inside the circle and outside the parabola is:

\[ \boxed{6\pi - 16} \]