The area of the portion of the circle `x^(2)+y^(2)=1` which lies inside the parabola `y^(2)=1-x,` is -

Options

A. `(pi)/(2)-(2)/(3)`
B. `(pi)/(2)+(2)/(3)`
C. `(pi)/(2)+(4)/(3)`
D. `(pi)/(2)-(4)/(3)`

Explanation

To find the area of the portion of the circle \(x^2 + y^2 = 1\) that lies inside the parabola \(y^2 = 1 - x\), we will follow these steps:

### Step 1: Find the intersection points of the circle and the parabola.
We substitute \(y^2\) from the parabola into the equation of the circle.

\[ x^2 + (1 - x) = 1 \]

This simplifies to:

\[ x^2 + 1 - x = 1 \]
\[ x^2 - x = 0 \]
Factoring gives:

\[ x(x - 1) = 0 \]

Thus, the solutions are \(x = 0\) and \(x = 1\).

### Step 2: Find the corresponding \(y\) values for the intersection points.
Using \(y^2 = 1 - x\):

1. For \(x = 0\):
\[ y^2 = 1 - 0 = 1 \implies y = \pm 1 \]
So, the points are \((0, 1)\) and \((0, -1)\).

2. For \(x = 1\):
\[ y^2 = 1 - 1 = 0 \implies y = 0 \]
So, the point is \((1, 0)\).

### Step 3: Identify the area to be calculated.
The area we need to find is the region of the circle that lies inside the parabola. The intersection points are \((0, 1)\), \((0, -1)\), and \((1, 0)\).

### Step 4: Calculate the area of the semicircle.
The area of the entire circle is given by:

\[ \text{Area of circle} = \pi r^2 = \pi (1^2) = \pi \]

Since we are interested in the upper half (semicircle), the area of the semicircle is:

\[ \text{Area of semicircle} = \frac{\pi}{2} \]

### Step 5: Calculate the area under the parabola from \(x = 0\) to \(x = 1\).
The parabola is given by \(y = \sqrt{1 - x}\). The area under the curve from \(x = 0\) to \(x = 1\) is calculated using integration:

\[ \text{Area} = \int_0^1 \sqrt{1 - x} \, dx \]

Using the substitution \(u = 1 - x\), we have \(du = -dx\). Changing the limits accordingly:

- When \(x = 0\), \(u = 1\)
- When \(x = 1\), \(u = 0\)

Thus, the integral becomes:

\[ \int_1^0 \sqrt{u} (-du) = \int_0^1 u^{1/2} \, du \]

Calculating the integral:

\[ = \left[ \frac{u^{3/2}}{3/2} \right]_0^1 = \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{2}{3} (1^{3/2} - 0^{3/2}) = \frac{2}{3} \]

### Step 6: Double the area under the parabola to account for both sides.
Since the parabola is symmetric about the x-axis, we multiply the area by 2:

\[ \text{Total area under the parabola} = 2 \times \frac{2}{3} = \frac{4}{3} \]

### Step 7: Calculate the area of the region inside the circle but outside the parabola.
The area of the portion of the circle that lies inside the parabola is:

\[ \text{Area inside circle} = \text{Area of semicircle} - \text{Area under parabola} \]

Substituting the values:

\[ \text{Area} = \frac{\pi}{2} - \frac{4}{3} \]

### Final Step: Simplify the expression.
To combine the two areas, we find a common denominator:

\[ \text{Area} = \frac{3\pi}{6} - \frac{8}{6} = \frac{3\pi - 8}{6} \]

Thus, the area of the portion of the circle that lies inside the parabola is:

\[ \boxed{\frac{3\pi - 8}{6}} \]