The area bounded by astroid `x^(2//3)+y^(2//3)=1` (in sq. units) is

Options

A. `(3pi)/(8)" sq. units"`
B. `(3pi^(2))/(8)" sq. units"`
C. `(3pi)/(16)" sq. units"`
D. None of these

Answer

To find the area bounded by the astroid given by the equation \( x^{2/3} + y^{2/3} = 1 \), we can use the parametric equations for the astroid and the formula for area in polar coordinates. Here’s a step-by-step solution:

### Step 1: Parametrize the Astroid
The astroid can be parametrized using the following equations:
\[ x = \cos^3 t, \quad y = \sin^3 t \]
where \( t \) ranges from \( 0 \) to \( 2\pi \).

### Step 2: Compute \( dx \) and \( dy \)
Next, we need to find the derivatives \( dx \) and \( dy \):
\[ dx = \frac{d}{dt}(\cos^3 t) = 3 \cos^2 t (-\sin t) dt = -3 \cos^2 t \sin t \, dt \]
\[ dy = \frac{d}{dt}(\sin^3 t) = 3 \sin^2 t \cos t \, dt \]

### Step 3: Set Up the Area Integral
The area \( A \) bounded by the astroid can be calculated using the formula:
\[ A = \frac{1}{2} \int_0^{2\pi} (x \, dy - y \, dx) \]
Substituting the values of \( x \), \( y \), \( dx \), and \( dy \):
\[ A = \frac{1}{2} \int_0^{2\pi} \left( \cos^3 t (3 \sin^2 t \cos t) - \sin^3 t (-3 \cos^2 t \sin t) \right) dt \]
This simplifies to:
\[ A = \frac{1}{2} \int_0^{2\pi} \left( 3 \cos^3 t \sin^2 t \cos t + 3 \sin^3 t \cos^2 t \sin t \right) dt \]
\[ = \frac{3}{2} \int_0^{2\pi} \left( \cos^4 t \sin^2 t + \sin^4 t \cos^2 t \right) dt \]

### Step 4: Simplify the Integral
Notice that:
\[ \cos^4 t \sin^2 t + \sin^4 t \cos^2 t = \cos^2 t \sin^2 t (\cos^2 t + \sin^2 t) = \cos^2 t \sin^2 t \]
Thus, we have:
\[ A = \frac{3}{2} \int_0^{2\pi} \cos^2 t \sin^2 t \, dt \]

### Step 5: Use the Identity for \( \cos^2 t \sin^2 t \)
Using the identity \( \cos^2 t \sin^2 t = \frac{1}{4} \sin^2(2t) \):
\[ A = \frac{3}{2} \int_0^{2\pi} \frac{1}{4} \sin^2(2t) \, dt \]
\[ = \frac{3}{8} \int_0^{2\pi} \sin^2(2t) \, dt \]

### Step 6: Evaluate the Integral
The integral \( \int \sin^2(kt) \, dt \) can be computed using the identity:
\[ \int \sin^2(kt) \, dt = \frac{t}{2} - \frac{\sin(2kt)}{4k} + C \]
Thus, for \( k = 2 \):
\[ \int_0^{2\pi} \sin^2(2t) \, dt = \left[ \frac{t}{2} - \frac{\sin(4t)}{8} \right]_0^{2\pi} = \frac{2\pi}{2} - 0 = \pi \]

### Step 7: Calculate the Area
Substituting back into the area formula:
\[ A = \frac{3}{8} \cdot \pi = \frac{3\pi}{8} \]

### Final Answer
The area bounded by the astroid \( x^{2/3} + y^{2/3} = 1 \) is:
\[ \boxed{\frac{3\pi}{8}} \text{ square units.} \]