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 loop described as
`x=(t)/(3)(6-t),y=(t^(2))/(8)(6-t)` is

Options

A. `(27)/(5)` sq. units
B. `(24)/(5)` sq. units
C. `(27)/(6)` sq. units
D. `(21)/(5)` sq. units

Explanation

To find the area of the loop described by the parametric equations \( x = \frac{t}{3}(6 - t) \) and \( y = \frac{t^2}{8}(6 - t) \), we will use the formula for the area \( S \) given by:

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

### Step 1: Find \( x'(t) \) and \( y'(t) \)

First, we need to differentiate \( x(t) \) and \( y(t) \) with respect to \( t \).

1. **Differentiate \( x(t) \)**:
\[ x(t) = \frac{t}{3}(6 - t) = \frac{6t - t^2}{3} \]
\[ x'(t) = \frac{d}{dt} \left( \frac{6t - t^2}{3} \right) = \frac{1}{3}(6 - 2t) = \frac{6 - 2t}{3} \]

2. **Differentiate \( y(t) \)**:
\[ y(t) = \frac{t^2}{8}(6 - t) = \frac{6t^2 - t^3}{8} \]
\[ y'(t) = \frac{d}{dt} \left( \frac{6t^2 - t^3}{8} \right) = \frac{1}{8}(12t - 3t^2) = \frac{3t(4 - t)}{8} \]

### Step 2: Set the limits of integration \( \alpha \) and \( \beta \)

Next, we need to determine the values of \( t \) that correspond to the beginning and end of the traversal of the loop.

To find these values, we can analyze the parametric equations. The loop typically occurs when \( t \) ranges from 0 to 6 (as \( t \) approaches 6, both \( x \) and \( y \) approach 0).

Thus, we set:
\[ \alpha = 0, \quad \beta = 6 \]

### Step 3: Substitute into the area formula

Now we can substitute \( x(t) \), \( y(t) \), \( x'(t) \), and \( y'(t) \) into the area formula.

\[ S = \frac{1}{2} \int_{0}^{6} \left( x(t) y'(t) - y(t) x'(t) \right) dt \]

Substituting the expressions we found:
\[ S = \frac{1}{2} \int_{0}^{6} \left( \frac{6t - t^2}{3} \cdot \frac{3t(4 - t)}{8} - \frac{6t^2 - t^3}{8} \cdot \frac{6 - 2t}{3} \right) dt \]

### Step 4: Simplify the integrand

Now we simplify the integrand:
1. For the first term:
\[ \frac{6t - t^2}{3} \cdot \frac{3t(4 - t)}{8} = \frac{(6t - t^2)t(4 - t)}{8} \]
2. For the second term:
\[ \frac{6t^2 - t^3}{8} \cdot \frac{6 - 2t}{3} = \frac{(6t^2 - t^3)(6 - 2t)}{24} \]

Combine these two terms and simplify further.

### Step 5: Evaluate the integral

After simplifying, evaluate the integral from 0 to 6. This will give the area \( S \).

### Final Step: Calculate the area

After performing the integration and substituting the limits, you will arrive at the final area of the loop.