Area bounded by the semi-sircle `y=sqrt(4-x^(2))`, and its diameter y = 0, is

Options

A. `2pi`
B. `pi`
C. `(pi)/(2)`
D. `2pi^(2)`

Answer

To find the area bounded by the semicircle \( y = \sqrt{4 - x^2} \) and its diameter \( y = 0 \), we can follow these steps:

### Step 1: Identify the equation of the semicircle
The given equation of the semicircle is \( y = \sqrt{4 - x^2} \). This represents the upper half of the circle defined by the equation \( x^2 + y^2 = 4 \) (which is a circle with radius 2 centered at the origin).

### Step 2: Determine the limits of integration
The semicircle intersects the x-axis (where \( y = 0 \)) at points where \( \sqrt{4 - x^2} = 0 \). Solving this gives:
\[ 4 - x^2 = 0 \implies x^2 = 4 \implies x = -2 \text{ and } x = 2 \]
Thus, the limits of integration are from \( x = -2 \) to \( x = 2 \).

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

### Step 4: Use the formula for the integral of the semicircle
The integral \( \int \sqrt{a^2 - x^2} \, dx \) can be evaluated using the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a = 2 \). Therefore, we have:
\[ \int \sqrt{4 - x^2} \, dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \left( \frac{x}{2} \right) + C \]

### Step 5: Evaluate the definite integral
Now we need to evaluate the definite integral from \( -2 \) to \( 2 \):
\[ A = \left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \left( \frac{x}{2} \right) \right]_{-2}^{2} \]

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

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

### Step 6: Combine results
Now, we subtract the lower limit from the upper limit:
\[ A = \pi - (-\pi) = \pi + \pi = 2\pi \]

### Final Answer
Thus, the area bounded by the semicircle and its diameter is:
\[ \boxed{2\pi} \]