Area of the region in the first quadrant exclosed by the X-axis, the line y=x and the circle `x^(2)+y^(2)=32` is

Options

A. `16pi" sq units"`
B. `4pi" sq units"`
C. `32pi" sq units"`
D. `24 pi " sq units"`

Explanation

To find the area of the region in the first quadrant enclosed by the x-axis, the line \( y = x \), and the circle \( x^2 + y^2 = 32 \), we can follow these steps:

### Step 1: Understand the equations and points of intersection
1. The equation of the circle is \( x^2 + y^2 = 32 \). This can be rewritten in the standard form where the radius \( r = \sqrt{32} = 4\sqrt{2} \).
2. The line \( y = x \) intersects the circle. To find the intersection point, substitute \( y = x \) into the circle's equation:
\[ x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4 \]
Thus, the intersection point is \( (4, 4) \).

### Step 2: Set up the area calculation
The area we need to calculate consists of two parts:
1. The area of the triangle formed by the points \( (0, 0) \), \( (4, 0) \), and \( (4, 4) \).
2. The area under the circle from \( x = 4 \) to \( x = 4\sqrt{2} \).

### Step 3: Calculate the area of the triangle
The area \( A_t \) of the triangle can be calculated using the formula:
\[ A_t = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \]

### Step 4: Calculate the area under the circle
To find the area under the circle from \( x = 4 \) to \( x = 4\sqrt{2} \), we first express \( y \) in terms of \( x \):
\[ y = \sqrt{32 - x^2} \]
The area \( A_c \) under the curve from \( x = 4 \) to \( x = 4\sqrt{2} \) can be computed using the integral:
\[ A_c = \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx \]

### Step 5: Evaluate the integral
Using the integral formula for \( \int \sqrt{a^2 - x^2} \, dx \):
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
where \( a = 4\sqrt{2} \).

Now, we compute:
\[ A_c = \left[ \frac{x}{2} \sqrt{32 - x^2} + \frac{32}{2} \sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}} \]

### Step 6: Substitute the limits
1. For \( x = 4\sqrt{2} \):
\[ A_c = \left[ \frac{4\sqrt{2}}{2} \cdot 0 + 16 \cdot \frac{\pi}{2} \right] = 8\pi \]
2. For \( x = 4 \):
\[ A_c = \left[ \frac{4}{2} \cdot \sqrt{16} + 16 \cdot \frac{\pi}{4} \right] = 4 \cdot 4 + 4\pi = 16 + 4\pi \]

### Step 7: Combine the areas
The total area \( A \) is given by:
\[ A = A_t + A_c = 8 + (8\pi - (16 + 4\pi)) = 8 + 8\pi - 16 - 4\pi = 4\pi - 8 \]

### Final Area Calculation
Thus, the area of the region in the first quadrant enclosed by the x-axis, the line \( y = x \), and the circle \( x^2 + y^2 = 32 \) is:
\[ \text{Area} = 8 + 4\pi \]