Key Concepts and Formulas

• Area Between Curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by $\int_{a}^{b} |f(x) - g(x)| dx$.
• Equation of a Circle: The general equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
• Equation of a Parabola: The equation of a rightward-opening parabola with vertex at $(h, k)$ is $(y-k)^2 = 4a(x-h)$.

Step-by-Step Solution

Step 1: Rewrite the equations in standard form.

• The equation of the parabola is given as $y^2 = 8x + 4$. We can rewrite this as $y^2 = 8(x + \frac{1}{2})$. This is a rightward-opening parabola with vertex at $(-\frac{1}{2}, 0)$.
• The equation of the circle is given as $x^2 + y^2 + 4\sqrt{3}x - 4 = 0$. Completing the square for the $x$ terms, we have $(x^2 + 4\sqrt{3}x) + y^2 = 4$. Adding $(2\sqrt{3})^2 = 12$ to both sides, we get $(x^2 + 4\sqrt{3}x + 12) + y^2 = 4 + 12$, which simplifies to $(x + 2\sqrt{3})^2 + y^2 = 16$. This is a circle with center $(-2\sqrt{3}, 0)$ and radius $4$.

Step 2: Find the points of intersection of the parabola and the circle.

• We need to solve the system of equations: $y^2 = 8x + 4 \qquad (1)$ $(x + 2\sqrt{3})^2 + y^2 = 16 \qquad (2)$
• Substitute equation (1) into equation (2): $(x + 2\sqrt{3})^2 + (8x + 4) = 16$ $x^2 + 4\sqrt{3}x + 12 + 8x + 4 = 16$ $x^2 + (4\sqrt{3} + 8)x = 0$ $x(x + 4\sqrt{3} + 8) = 0$ So, $x = 0$ or $x = -8 - 4\sqrt{3}$.
• If $x = 0$, then from equation (1), $y^2 = 8(0) + 4 = 4$, so $y = \pm 2$. The intersection points are $(0, 2)$ and $(0, -2)$.
• If $x = -8 - 4\sqrt{3}$, then $x < -8 - 4(1.7) \approx -8 - 6.8 = -14.8$. Since the parabola $y^2 = 8x+4$ only exists for $x \geq -1/2$, the intersection point $x = -8 - 4\sqrt{3}$ is extraneous.
• Thus, the only intersection points are $(0, 2)$ and $(0, -2)$.

Step 3: Set up the integral to find the area.

• We need to find the area enclosed by $y^2 = 8x + 4$ and $(x + 2\sqrt{3})^2 + y^2 = 16$. Expressing $x$ in terms of $y$ for both equations, we have:
  • Parabola: $x = \frac{y^2 - 4}{8}$
  • Circle: $(x + 2\sqrt{3})^2 = 16 - y^2$, so $x + 2\sqrt{3} = \pm \sqrt{16 - y^2}$. Therefore, $x = -2\sqrt{3} \pm \sqrt{16 - y^2}$. Since we need the smaller region, we choose $x = -2\sqrt{3} + \sqrt{16 - y^2}$ as the right boundary for the enclosed area.
• The area is given by $A = \int_{-2}^{2} \left[ (-2\sqrt{3} + \sqrt{16 - y^2}) - \left( \frac{y^2 - 4}{8} \right) \right] dy$ $A = \int_{-2}^{2} \left[ -2\sqrt{3} + \sqrt{16 - y^2} - \frac{y^2}{8} + \frac{1}{2} \right] dy$ $A = 2 \int_{0}^{2} \left[ -2\sqrt{3} + \sqrt{16 - y^2} - \frac{y^2}{8} + \frac{1}{2} \right] dy$ $A = 2 \left[ -2\sqrt{3}y + \frac{y}{2}\sqrt{16 - y^2} + 8\sin^{-1}\left(\frac{y}{4}\right) - \frac{y^3}{24} + \frac{1}{2}y \right]_{0}^{2}$ $A = 2 \left[ -4\sqrt{3} + \sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right) - \frac{8}{24} + 1 \right]$ $A = 2 \left[ -4\sqrt{3} + 2\sqrt{3} + 8\left(\frac{\pi}{6}\right) - \frac{1}{3} + 1 \right]$ $A = 2 \left[ -2\sqrt{3} + \frac{4\pi}{3} + \frac{2}{3} \right]$ $A = -\frac{12\sqrt{3}}{3} + \frac{8\pi}{3} + \frac{4}{3} = \frac{4 - 12\sqrt{3} + 8\pi}{3}$ $A = \frac{1}{3}(4 - 12\sqrt{3} + 8\pi)$

Step 4: Simplify the result.

The area is $\frac{1}{3}(4 - 12\sqrt{3} + 8\pi)$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs, especially when completing the square and substituting equations.
• Choosing the Correct Region: Visualizing the curves is crucial to determine which region's area is required. Ensure you're integrating the correct 'top' and 'bottom' functions (or 'right' and 'left' functions).
• Extraneous Solutions: Always check for extraneous solutions when solving simultaneous equations involving square roots.

Summary

We found the area of the smaller region enclosed by the given parabola and circle by first rewriting their equations in standard form. We then found the intersection points of the curves and set up a definite integral to represent the area. After evaluating the integral, we obtained the area as $\frac{1}{3}(4 - 12\sqrt{3} + 8\pi)$.

Final Answer

The final answer is \boxed{\frac{1}{3}(4-12 \sqrt{3}+8 \pi)}, which corresponds to option (C).