Compute the area enclosed by the ellipse `x= a cos t, y= b sin t(0 le t le 2pi)`

Answer

To compute the area enclosed by the ellipse defined by the parametric equations \( x = a \cos t \) and \( y = b \sin t \) for \( 0 \leq t \leq 2\pi \), we can follow these steps:

### Step 1: Understanding the Parametric Equations
The given parametric equations describe an ellipse centered at the origin with semi-major axis \( a \) along the x-axis and semi-minor axis \( b \) along the y-axis.

### Step 2: Area Calculation Using Integration
The area \( A \) enclosed by the ellipse can be computed using the formula for area in parametric form:
\[ A = \int_{t_1}^{t_2} y \frac{dx}{dt} \, dt \]
where \( y = b \sin t \) and \( \frac{dx}{dt} = -a \sin t \).

### Step 3: Setting Up the Integral
Substituting the values into the area formula, we have:
\[ A = \int_{0}^{2\pi} (b \sin t)(-a \sin t) \, dt \]
This simplifies to:
\[ A = -ab \int_{0}^{2\pi} \sin^2 t \, dt \]

### Step 4: Evaluating the Integral
Using the identity \( \sin^2 t = \frac{1 - \cos(2t)}{2} \), we can rewrite the integral:
\[ A = -ab \int_{0}^{2\pi} \frac{1 - \cos(2t)}{2} \, dt \]
This can be split into two integrals:
\[ A = -\frac{ab}{2} \left( \int_{0}^{2\pi} 1 \, dt - \int_{0}^{2\pi} \cos(2t) \, dt \right) \]

### Step 5: Calculating the Integrals
The first integral evaluates to:
\[ \int_{0}^{2\pi} 1 \, dt = 2\pi \]
The second integral evaluates to:
\[ \int_{0}^{2\pi} \cos(2t) \, dt = 0 \]
Thus, we have:
\[ A = -\frac{ab}{2} (2\pi - 0) = -\frac{ab}{2} \cdot 2\pi = -ab\pi \]

### Step 6: Final Area Calculation
Since area cannot be negative, we take the absolute value:
\[ A = ab\pi \]

### Conclusion
The area enclosed by the ellipse is given by:
\[ \boxed{ab\pi} \]