Key Concepts and Formulas

• Area under a curve: The area bounded by the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by $\int_a^b |f(x)| dx$.
• Area between two curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by $\int_a^b |f(x) - g(x)| dx$. We need to determine which function is greater in the interval $[a, b]$.
• Equation of a circle: The equation of a circle with center at the origin and radius $r$ is $x^2 + y^2 = r^2$.

Step-by-Step Solution

Step 1: Sketch the region

We are given two inequalities: $x^2 + y^2 \le 1$ $y^2 \le 1 - x$

The first inequality represents the region inside and on the circle with center (0, 0) and radius 1. The second inequality can be rewritten as $x \le 1 - y^2$, which represents the region to the left of and on the parabola $x = 1 - y^2$.

Step 2: Find the points of intersection

To find the points of intersection of the circle $x^2 + y^2 = 1$ and the parabola $x = 1 - y^2$, substitute $x = 1 - y^2$ into the equation of the circle: $(1 - y^2)^2 + y^2 = 1$ $1 - 2y^2 + y^4 + y^2 = 1$ $y^4 - y^2 = 0$ $y^2(y^2 - 1) = 0$ $y = 0, y = 1, y = -1$

When $y = 0$, $x = 1 - 0^2 = 1$. When $y = 1$, $x = 1 - 1^2 = 0$. When $y = -1$, $x = 1 - (-1)^2 = 0$.

Therefore, the points of intersection are (1, 0), (0, 1), and (0, -1).

Step 3: Set up the integral for the area

The required area can be found by integrating with respect to $y$. The area is symmetric about the x-axis, so we can calculate the area in the first and second quadrants and multiply by 2. In the upper half-plane, the region is bounded by $x = \sqrt{1 - y^2}$ (from the circle) on the right and $x = 1 - y^2$ (from the parabola) on the left, from $y = 0$ to $y = 1$. Therefore, the area in the upper half plane is $A_{upper} = \int_0^1 (\sqrt{1 - y^2} - (1 - y^2)) dy$

The total area is then $A = 2 A_{upper} = 2 \int_0^1 (\sqrt{1 - y^2} - (1 - y^2)) dy$ $A = 2 \left[ \int_0^1 \sqrt{1 - y^2} dy - \int_0^1 (1 - y^2) dy \right]$

Step 4: Evaluate the integrals

The first integral, $\int_0^1 \sqrt{1 - y^2} dy$, represents the area of a quarter of the unit circle. So, $\int_0^1 \sqrt{1 - y^2} dy = \frac{1}{4} \pi (1)^2 = \frac{\pi}{4}$

The second integral is: $\int_0^1 (1 - y^2) dy = \left[ y - \frac{y^3}{3} \right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}$

Step 5: Calculate the total area

Substituting the values of the integrals back into the expression for the total area: $A = 2 \left[ \frac{\pi}{4} - \frac{2}{3} \right] = \frac{\pi}{2} - \frac{4}{3}$

Common Mistakes & Tips

• Incorrectly identifying the region: Make sure to sketch the curves accurately to correctly identify the region whose area needs to be calculated.
• Sign errors: Pay attention to the order of subtraction when integrating between two curves. The upper curve minus the lower curve (when integrating with respect to x) or the right curve minus the left curve (when integrating with respect to y).
• Using symmetry: If the region is symmetric, use symmetry to simplify the calculation.

Summary

We first sketched the region defined by the given inequalities, which were a circle and a parabola. We found the points of intersection of the two curves. Then, we set up an integral to calculate the area of the region, integrating with respect to $y$, and using symmetry to simplify the calculation. The final area is $\frac{\pi}{2} - \frac{4}{3}$.

Final Answer

The final answer is $\frac{\pi}{2} - \frac{4}{3}$, which corresponds to option (D).