The area of the region cut off by the line `x-y=0`, X-axis from the circle `x^(2)+y^(2)=16`, in first quadrant is

Options

A. `4pi`
B. `2pi`
C. `16pi`
D. `8pi`

Answer

To find the area of the region cut off by the line \(x - y = 0\) (which is the line \(y = x\)) and the X-axis from the circle \(x^2 + y^2 = 16\) in the first quadrant, we will follow these steps:

### Step 1: Identify the curves and their intersection point
The given equations are:
1. Circle: \(x^2 + y^2 = 16\)
2. Line: \(x - y = 0 \Rightarrow y = x\)

To find the intersection point, we substitute \(y = x\) into the circle's equation:
\[ x^2 + x^2 = 16 \Rightarrow 2x^2 = 16 \Rightarrow x^2 = 8 \Rightarrow x = 2\sqrt{2} \]
Since we are in the first quadrant, \(y\) will also be \(2\sqrt{2}\). Therefore, the intersection point \(A\) is \((2\sqrt{2}, 2\sqrt{2})\).

### Step 2: Set up the area calculation
We need to calculate the area of the region \(OAP\) where \(O\) is the origin \((0,0)\), \(A\) is the intersection point \((2\sqrt{2}, 2\sqrt{2})\), and \(P\) is the point where the circle intersects the X-axis, which is \((4, 0)\).

The area can be divided into two parts:
1. Area under the line from \(O\) to \(A\).
2. Area under the circle from \(A\) to \(P\).

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

### Step 4: Calculate the area under the circle
The area under the circle from \(A\) to \(P\) is calculated using integration. The equation of the circle can be expressed as:
\[ y = \sqrt{16 - x^2} \]
We will integrate from \(x = 2\sqrt{2}\) to \(x = 4\):
\[ \text{Area}_{\text{circle}} = \int_{2\sqrt{2}}^{4} \sqrt{16 - x^2} \, dx \]

### Step 5: Solve the integral
Using the formula for the area under a semicircle:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
where \(a = 4\):
\[ \text{Area}_{\text{circle}} = \left[ \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1}\left(\frac{x}{4}\right) \right]_{2\sqrt{2}}^{4} \]

Evaluating at the limits:
1. At \(x = 4\):
\[ = \frac{4}{2} \cdot 0 + 8 \cdot \frac{\pi}{2} = 4\pi \]
2. At \(x = 2\sqrt{2}\):
\[ = \frac{2\sqrt{2}}{2} \cdot 2\sqrt{2} + 8 \sin^{-1}\left(\frac{2\sqrt{2}}{4}\right) = 2 \cdot 2 + 8 \cdot \frac{\pi}{4} = 4 + 2\pi \]

Thus, the area under the circle from \(A\) to \(P\) is:
\[ 4\pi - (4 + 2\pi) = 2\pi - 4 \]

### Step 6: Total area
The total area \(OAP\) is:
\[ \text{Total Area} = \text{Area}_{\text{triangle}} + \text{Area}_{\text{circle}} = 4 + (2\pi - 4) = 2\pi \]

### Final Answer
The area of the region cut off by the line \(x - y = 0\) and the X-axis from the circle \(x^2 + y^2 = 16\) in the first quadrant is:
\[ \boxed{2\pi} \]