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)` is:

Options

A. (a) `ab`
B. (b) `pi ab`
C. (c) `(pi)/(2) ab`
D. (d) `2 pi ab`

Answer

To find the area of the ellipse defined by the parametric equations \( x = a \cos t \) and \( y = b \sin t \) for \( t \) ranging from \( 0 \) to \( 2\pi \), we can use the formula:

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

### Step 1: Determine the derivatives \( x' \) and \( y' \)

Given:
- \( x = a \cos t \)
- \( y = b \sin t \)

We calculate the derivatives:
\[ x' = \frac{dx}{dt} = -a \sin t \]
\[ y' = \frac{dy}{dt} = b \cos t \]

### Step 2: Set the limits of integration

For the ellipse, the limits are:
- \( \alpha = 0 \)
- \( \beta = 2\pi \)

### Step 3: Substitute into the area formula

Now substitute \( x \), \( y \), \( x' \), and \( y' \) into the area formula:
\[ S = \frac{1}{2} \int_{0}^{2\pi} \left( a \cos t \cdot b \cos t - b \sin t \cdot (-a \sin t) \right) dt \]

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

### 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

The integral \( \int_{0}^{2\pi} 1 \, dt \) evaluates to:
\[ \int_{0}^{2\pi} 1 \, dt = 2\pi \]

### Step 7: Final calculation

Substituting back, we have:
\[ S = \frac{ab}{2} \cdot 2\pi = ab\pi \]

Thus, the area of the ellipse is:
\[ \boxed{ab\pi} \]