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

Step 1: Understand the equations.
The equation of the circle is \((x - 2\sqrt{3})^2 + y^2 = 12\).

The center of the circle is \((2\sqrt{3}, 0)\) and the radius is \(\sqrt{12} = 2\sqrt{3}\).

The equation of the parabola is \(y^2 = 2\sqrt{3}x\).

Step 2: Find the intersection points.

Substitute \(y^2 = 2\sqrt{3}x\) into the equation of the circle:

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

\(x^2 - 4\sqrt{3}x + 12 + 2\sqrt{3}x = 12\)

\(x^2 - 2\sqrt{3}x = 0\)

\(x(x - 2\sqrt{3}) = 0\)

So, \(x = 0\) or \(x = 2\sqrt{3}\).

When \(x = 0\), \(y^2 = 0\), so \(y = 0\).

When \(x = 2\sqrt{3}\), \(y^2 = 2\sqrt{3}(2\sqrt{3}) = 12\), so \(y = \pm 2\sqrt{3}\).

The intersection points are \((0, 0)\), \((2\sqrt{3}, 2\sqrt{3})\), and \((2\sqrt{3}, -2\sqrt{3})\).

Step 3: Calculate the area.

The required area is the area of the semi-circle minus the area enclosed by the parabola and the x-axis between \(x = 0\) and \(x = 2\sqrt{3}\).

Area of the semi-circle = \(\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (12) = 6\pi\).

Area under the parabola between \(x = 0\) and \(x = 2\sqrt{3}\) is:

\(2\int_0^{2\sqrt{3}} \sqrt{2\sqrt{3}x} \, dx = 2\sqrt{2\sqrt{3}} \int_0^{2\sqrt{3}} x^{1/2} \, dx\)

\(= 2\sqrt{2\sqrt{3}} \left[ \frac{2}{3} x^{3/2} \right]_0^{2\sqrt{3}} = 2\sqrt{2\sqrt{3}} \cdot \frac{2}{3} (2\sqrt{3})^{3/2}\)

\(= \frac{4}{3} \sqrt{2\sqrt{3}} \cdot 2\sqrt{3} \sqrt{2\sqrt{3}} = \frac{4}{3} \cdot 2\sqrt{3} \cdot 2\sqrt{3} = \frac{4}{3} \cdot 12 = 16\)

Required area = \(6\pi - 16\).