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: Identify the intersection points
First, we need to find the points where the line \( y = x \) intersects the circle \( x^2 + y^2 = 32 \).

Substituting \( y = x \) into the circle's equation:
\[ x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4 \text{ (since we are in the first quadrant)} \]
Thus, the intersection point is \( (4, 4) \).

### Step 2: Determine the area under the line and the circle
We will calculate the area in two parts:
1. The area under the line \( y = x \) from \( x = 0 \) to \( x = 4 \).
2. The area under the circle from \( x = 4 \) to \( x = 4\sqrt{2} \).

### Step 3: Calculate the area under the line \( y = x \)
The area under the line from \( x = 0 \) to \( x = 4 \) can be calculated using the formula for the area of a triangle:
\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \]

### 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 \( x \):
\[ y = \sqrt{32 - x^2} \]
Now, we integrate this from \( x = 4 \) to \( x = 4\sqrt{2} \):
\[ \text{Area}_{\text{circle}} = \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx \]

### Step 5: Solve the integral
To solve the integral, we can use the formula for the area under a semicircle:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C \]
Here, \( a^2 = 32 \) so \( a = 4\sqrt{2} \).

Calculating the integral:
\[ \int \sqrt{32 - x^2} \, dx = \frac{x}{2} \sqrt{32 - x^2} + 16 \sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \]
Evaluating from \( 4 \) to \( 4\sqrt{2} \):
\[ \left[ \frac{x}{2} \sqrt{32 - x^2} + 16 \sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}} \]

At \( x = 4\sqrt{2} \):
\[ \frac{4\sqrt{2}}{2} \cdot 0 + 16 \cdot \frac{\pi}{4} = 4\pi \]

At \( x = 4 \):
\[ \frac{4}{2} \cdot \sqrt{16} + 16 \cdot \frac{\pi}{4} = 4 + 4\pi \]

Thus, the area under the circle is:
\[ 4\pi - (4 + 4\pi) = -4 \]

### Step 6: Total area calculation
Now, the total area is:
\[ \text{Total Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{circle}} = 8 + (4\pi - (4 + 4\pi)) = 8 - 4 = 4 \]

### Final Answer
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:
\[ \boxed{4} \]