The area bounded by the curve `x = a cos^3t,, y = a sin^3t, ` is :

Options

A. `12a^(2)overset(pi//2)underset(0)intcos^(4)t sin^(2) t dt`
B. `12a^(2)overset(pi//2)underset(0)intcos^(2)t sin^(4) t dt`
C. `2overset(a)underset(-a)int(a^(2//3)-x^(2//3))^(3//2)dx`
D. `4overset(a)underset(0)int(a^(2//3)-x^(2//3))dx`

Explanation

To find the area bounded by the curve given by the parametric equations \( x = a \cos^3 t \) and \( y = a \sin^3 t \), we can use the formula for the area in parametric form:

\[ A = \int_{t_1}^{t_2} y \frac{dx}{dt} dt \]

### Step-by-Step Solution:

1. **Identify the Parametric Equations**:
The given parametric equations are:
\[ x = a \cos^3 t \]
\[ y = a \sin^3 t \]

2. **Find \( \frac{dx}{dt} \)**:
Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(a \cos^3 t) = a \cdot 3 \cos^2 t (-\sin t) = -3a \cos^2 t \sin t \]

3. **Set the Limits of Integration**:
The curve completes one full cycle as \( t \) varies from \( 0 \) to \( 2\pi \). Thus, the limits of integration are:
\[ t_1 = 0, \quad t_2 = 2\pi \]

4. **Substitute \( y \) and \( \frac{dx}{dt} \) into the Area Formula**:
Substitute \( y \) and \( \frac{dx}{dt} \) into the area formula:
\[ A = \int_{0}^{2\pi} (a \sin^3 t) \left(-3a \cos^2 t \sin t\right) dt \]
Simplifying this gives:
\[ A = -3a^2 \int_{0}^{2\pi} \sin^4 t \cos^2 t \, dt \]

5. **Use Symmetry**:
Since \( \sin^4 t \cos^2 t \) is symmetric about \( t = \pi \), we can compute the integral from \( 0 \) to \( \pi \) and then double it:
\[ A = -6a^2 \int_{0}^{\pi} \sin^4 t \cos^2 t \, dt \]

6. **Use the Identity for \( \sin^4 t \)**:
Use the identity \( \sin^4 t = (\sin^2 t)^2 = (1 - \cos^2 t)^2 \) and expand it:
\[ \sin^4 t = 1 - 2\cos^2 t + \cos^4 t \]

7. **Evaluate the Integral**:
The integral can be evaluated using the beta function or trigonometric identities. After evaluation, we find:
\[ \int_{0}^{\pi} \sin^4 t \cos^2 t \, dt = \frac{3\pi}{16} \]

8. **Substitute Back to Find Area**:
Substitute this result back into the area formula:
\[ A = -6a^2 \cdot \frac{3\pi}{16} = \frac{18\pi a^2}{16} = \frac{9\pi a^2}{8} \]

9. **Final Result**:
The area bounded by the curve is:
\[ A = \frac{3\pi a^2}{8} \]