Computing area with parametrically represented boundaries
If the boundary of a figure is represented by parametric equations x = x (t) , y = y(t) , then the area of the figure is evaluated by one of the three formulae
`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 traversal of the contour .
The area of ellipse enclosed by x = a cos t , y = b sint `(0 le t le 2pi)`

Options

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

Answer

To compute the area of 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 use one of the formulas provided. Here, we will use the formula:

\[ S = \frac{1}{2} \int_{\alpha}^{\beta} (x y' - y x') dt \]

### Step 1: Identify the derivatives
First, we need to find \( x' \) and \( y' \):

- \( x = a \cos t \) implies \( x' = -a \sin t \)
- \( y = b \sin t \) implies \( y' = b \cos t \)

### Step 2: Set the limits of integration
The limits for \( t \) are given as \( \alpha = 0 \) and \( \beta = 2\pi \).

### Step 3: Substitute into the area formula
Now, substitute \( x \), \( y \), \( x' \), and \( y' \) into the area formula:

\[ S = \frac{1}{2} \int_{0}^{2\pi} \left( (a \cos t)(b \cos t) - (b \sin t)(-a \sin t) \right) dt \]

This simplifies to:

\[ S = \frac{1}{2} \int_{0}^{2\pi} \left( ab \cos^2 t + ab \sin^2 t \right) dt \]

### Step 4: Factor out constants
Factor out the constant \( ab \):

\[ S = \frac{ab}{2} \int_{0}^{2\pi} (\cos^2 t + \sin^2 t) dt \]

### Step 5: Use the Pythagorean identity
Using the identity \( \cos^2 t + \sin^2 t = 1 \):

\[ S = \frac{ab}{2} \int_{0}^{2\pi} 1 \, dt \]

### Step 6: Evaluate the integral
Now, evaluate the integral:

\[ S = \frac{ab}{2} \cdot [t]_{0}^{2\pi} = \frac{ab}{2} (2\pi - 0) = \frac{ab}{2} \cdot 2\pi = ab\pi \]

### Final Result
Thus, the area of the ellipse is:

\[ \boxed{ab\pi} \]