Key Concepts and Formulas

• Area of a Circle: The area of a circle with radius $r$ is given by $A = \pi r^2$.
• Area Under a Curve: The area under a curve $y = f(x)$ between $x = a$ and $x = b$ is given by $\int_a^b f(x) \, dx$.
• Intersection of Curves: To find the points of intersection of two curves, solve their equations simultaneously.

Step-by-Step Solution

Step 1: Find the area of the circle.

The equation of the circle is $x^2 + y^2 = 2$, so its radius is $r = \sqrt{2}$. Therefore, the area of the circle is $A_\text{circle} = \pi r^2 = \pi (\sqrt{2})^2 = 2\pi.$ This represents the region $C$.

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

The equations are $y^2 = x$ and $y = x$. Substituting $y = x$ into $y^2 = x$, we get $x^2 = x$, which gives $x^2 - x = 0$, or $x(x - 1) = 0$. Thus, $x = 0$ or $x = 1$. When $x = 0$, $y = 0$, and when $x = 1$, $y = 1$. The points of intersection are $(0, 0)$ and $(1, 1)$.

Step 3: Find the area of the region bounded by the parabola and the line.

The area of the region bounded by $y^2 = x$ and $y = x$ is given by $A_\text{parabola-line} = \int_0^1 (\sqrt{x} - x) \, dx = \left[ \frac{2}{3} x^{3/2} - \frac{1}{2} x^2 \right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6}.$ This represents the area of the region bounded by the parabola and the line.

Step 4: Find the points of intersection of the circle and the parabola.

The equations are $x^2 + y^2 = 2$ and $y^2 = x$. Substituting $y^2 = x$ into the circle's equation, we get $x^2 + x = 2$, or $x^2 + x - 2 = 0$. This factors as $(x + 2)(x - 1) = 0$, so $x = -2$ or $x = 1$. Since $y^2 = x$, $x$ must be non-negative, so we take $x = 1$. Then $y^2 = 1$, so $y = \pm 1$. The intersection points are $(1, 1)$ and $(1, -1)$.

Step 5: Find the points of intersection of the circle and the line.

The equations are $x^2 + y^2 = 2$ and $y = x$. Substituting $y = x$ into the circle's equation, we get $x^2 + x^2 = 2$, or $2x^2 = 2$, so $x^2 = 1$, which gives $x = \pm 1$. When $x = 1$, $y = 1$, and when $x = -1$, $y = -1$. The points of intersection are $(1, 1)$ and $(-1, -1)$.

Step 6: Find the area of the region common to the circle and the region bounded by the parabola and the line.

The region bounded by the parabola and the line lies inside the circle. Thus, the area of $C \cap R$ is just the area of the region bounded by the parabola and the line, which is $\frac{1}{6}$.

Step 7: Find the area of the region enclosed by the circle that is not common to the region bounded by the parabola and the line.

The required area is the area of the circle minus the area of the region bounded by the parabola and the line: $A_\text{required} = A_\text{circle} - A_\text{parabola-line} = 2\pi - \frac{1}{6} = \frac{12\pi - 1}{6} = \frac{1}{6}(12\pi - 1).$

Common Mistakes & Tips

• Be careful with signs when solving for intersection points.
• Remember to consider both positive and negative roots when taking square roots.
• Always sketch the curves to visualize the region whose area you are calculating.

Summary

We first calculated the area of the circle. Then, we found the area of the region bounded by the parabola and the line. Since the region bounded by the parabola and the line is completely contained within the circle in the region of interest, the required area is the area of the circle minus the area bounded by the parabola and the line. This gives us $\frac{1}{6}(12\pi - 1)$.

Final Answer

The final answer is $\boxed{\frac{1}{6}(12\pi - 1)}$, which corresponds to option (D). Wait! The provided answer is option (A) which is $\frac{1}{6}(24\pi - 1)$. Let's re-evaluate the problem. The area of the circle is $2\pi$. The area between the line and the parabola is $\frac{1}{6}$. The question asks for the area of the circle not common to the area bounded by the line and the parabola. So, we need to subtract the intersection of the circle with the region bounded by the line and parabola. The region bounded by the line and the parabola is inside the circle. So, the intersection is just the area bounded by the line and parabola which is $\frac{1}{6}$. Thus, the area is $2\pi - \frac{1}{6} = \frac{12\pi - 1}{6}$.

The correct answer provided is (A) $\frac{1}{6}(24\pi - 1)$. We made an error. The region bounded by y=x and $y^2=x$ is inside the circle. However, the question asks for the area of the circle not common to the region bounded by the parabola and the line. The area of the circle is $2\pi$. The area between the parabola and the line is $\int_0^1 (\sqrt{x} - x) dx = \frac{1}{6}$. Thus, the required area is $2\pi - \frac{1}{6} = \frac{12\pi - 1}{6}$.

The question is subtly different. It is not asking for the area of the circle MINUS the area between the line and the parabola. We need to find the correct intersection.

Let's reconsider the area of the circle that is not common to the region bounded by the parabola and the line. Area of circle = $2\pi$. Area between the line and parabola is $\frac{1}{6}$. The area of the circle not common to the region is $2\pi - \frac{1}{6} = \frac{12\pi - 1}{6}$. There must be an error in the question or the provided answer.

Let's assume the correct answer is indeed (A) $\frac{1}{6}(24\pi - 1)$. Then, let A be the area of the circle, which is $2\pi$. Let B be the area between y=x and $y^2 = x$, which is $\frac{1}{6}$. If $A - B = \frac{24\pi - 1}{6}$, then $2\pi - \frac{1}{6} = \frac{24\pi - 1}{6}$, which gives $\frac{12\pi - 1}{6} = \frac{24\pi - 1}{6}$, which is false. The only way we can get $24\pi$ is if the area of the circle is $4\pi$, which is not the case here.

Let's assume the question meant something else.

Consider the region bounded by $x^2 + y^2 = 2$, $y^2 = x$ and $y = x$. Then the required area is $2\pi - \frac{1}{6}$. This equals $\frac{12\pi - 1}{6}$, which is not the answer.

The area of the circle is $2\pi$. The area bounded by the line and parabola is $\frac{1}{6}$. The area of the circle not common to the region bounded by the parabola and line is $2\pi - \frac{1}{6} = \frac{12\pi - 1}{6}$.

The correct answer provided is $\frac{24\pi - 1}{6}$. This is incorrect. The area of the circle is $2\pi$.

I suspect that there is a typo in the options, and the question is indeed asking for the area of the circle minus the region bounded by the parabola and the line.

Final Answer Since the provided correct answer cannot be derived, I suspect there's an error in the options. Based on the problem statement and the given curves, the correct area is $\frac{1}{6}(12\pi - 1)$. Although this doesn't match any of the options, if we assume that the correct answer is $\frac{1}{6}(12\pi - 1)$ then the question would be asking for the area of the circle minus the area between the line and the parabola. The final answer is \boxed{\frac{1}{6}(12\pi - 1)}.