Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., `x=x(t), y=(t),` then the area of the figure is evaluated by one of the three formulas :
`S=-int_(alpha)^(beta)y(t)x'(t)dt,`
`S=int_(alpha)^(beta)x(t)y'(t)dt,`
`S=(1)/(2)int_(alpha)^(beta)(xy'-yx')dt,`
Where `alpha and beta` are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
The area of the region bounded by an are of the cycloid `x=a(t-sin t), y=a(1- cos t)` and the x-axis is

Options

A. `6pia^(2)` sq. units
B. `3pia^(2)` sq. units
C. `4pia^(2)` sq. units
D. None of these

Explanation

To find the area of the region bounded by the cycloid defined by the parametric equations \( x = a(t - \sin t) \) and \( y = a(1 - \cos t) \) and the x-axis, we will use the formula for area under a parametric curve.

### Step-by-Step Solution:

1. **Identify the Parametric Equations**:
We have the equations:
\[ x(t) = a(t - \sin t) \]
\[ y(t) = a(1 - \cos t) \]

2. **Determine the Limits of Integration**:
For one complete arch of the cycloid, the parameter \( t \) varies from \( 0 \) to \( 2\pi \).

3. **Choose the Area Formula**:
We will use the formula:
\[ S = -\int_{\alpha}^{\beta} y(t) x'(t) \, dt \]
where \( \alpha = 0 \) and \( \beta = 2\pi \).

4. **Calculate \( x'(t) \)**:
First, we need to find the derivative \( x'(t) \):
\[ x'(t) = \frac{d}{dt}[a(t - \sin t)] = a(1 - \cos t) \]

5. **Set Up the Integral**:
Substitute \( y(t) \) and \( x'(t) \) into the area formula:
\[ S = -\int_{0}^{2\pi} a(1 - \cos t) \cdot a(1 - \cos t) \, dt \]
This simplifies to:
\[ S = -a^2 \int_{0}^{2\pi} (1 - \cos t)^2 \, dt \]

6. **Expand the Integrand**:
Expand \( (1 - \cos t)^2 \):
\[ (1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t \]
Using the identity \( \cos^2 t = \frac{1 + \cos 2t}{2} \), we have:
\[ (1 - \cos t)^2 = 1 - 2\cos t + \frac{1 + \cos 2t}{2} = \frac{3}{2} - 2\cos t + \frac{1}{2}\cos 2t \]

7. **Integrate Each Term**:
Now we can integrate:
\[ S = -a^2 \left[ \int_{0}^{2\pi} \left( \frac{3}{2} - 2\cos t + \frac{1}{2}\cos 2t \right) dt \right] \]
The integrals evaluate as follows:
- \( \int_{0}^{2\pi} dt = 2\pi \)
- \( \int_{0}^{2\pi} \cos t \, dt = 0 \)
- \( \int_{0}^{2\pi} \cos 2t \, dt = 0 \)

Thus,
\[ S = -a^2 \left[ \frac{3}{2} \cdot 2\pi \right] = -3\pi a^2 \]

8. **Take the Modulus**:
Since area cannot be negative, we take the modulus:
\[ S = 3\pi a^2 \]

### Final Answer:
The area of the region bounded by the cycloid and the x-axis is:
\[ \boxed{3\pi a^2} \text{ square units} \]