The area of the region bounded by the curve `y = sqrt(16-x^(2))` and `X`-axis is

Options

A. `8pi" sq units"`
B. `20pi" sq units"`
C. `16pi" sq units"`
D. `256 pi " sq units"`

Explanation

To find the area of the region bounded by the curve \( y = \sqrt{16 - x^2} \) and the X-axis, we can follow these steps:

### Step 1: Identify the curve and its boundaries
The equation \( y = \sqrt{16 - x^2} \) represents the upper half of a circle with radius 4 centered at the origin. The full equation of the circle is \( x^2 + y^2 = 16 \).

### Step 2: Determine the limits of integration
The curve intersects the X-axis where \( y = 0 \). Setting \( \sqrt{16 - x^2} = 0 \) gives:
\[ 16 - x^2 = 0 \implies x^2 = 16 \implies x = \pm 4 \]
Thus, the area we want to find is bounded between \( x = -4 \) and \( x = 4 \).

### Step 3: Set up the integral for the area
The area \( A \) can be computed using the integral:
\[ A = \int_{-4}^{4} \sqrt{16 - x^2} \, dx \]

### Step 4: Use the property of even functions
The function \( \sqrt{16 - x^2} \) is even, meaning \( f(-x) = f(x) \). Therefore, we can simplify the integral:
\[ A = 2 \int_{0}^{4} \sqrt{16 - x^2} \, dx \]

### Step 5: Solve the integral
To solve \( \int \sqrt{16 - x^2} \, dx \), we can use the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a = 4 \). Thus, we have:
\[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}\left(\frac{x}{4}\right) + C \]

### Step 6: Evaluate the definite integral
Now we evaluate:
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]

Calculating at the upper limit \( x = 4 \):
\[ = \frac{4}{2} \sqrt{16 - 16} + 8 \sin^{-1}(1) = 0 + 8 \cdot \frac{\pi}{2} = 4\pi \]

Calculating at the lower limit \( x = 0 \):
\[ = \frac{0}{2} \sqrt{16 - 0} + 8 \sin^{-1}(0) = 0 + 0 = 0 \]

Thus, the definite integral evaluates to:
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = 4\pi \]

### Step 7: Calculate the total area
Now, substituting back into our area formula:
\[ A = 2 \cdot 4\pi = 8\pi \]

### Final Answer
The area of the region bounded by the curve \( y = \sqrt{16 - x^2} \) and the X-axis is:
\[ \boxed{8\pi} \]