The area enclosed by the curve `y^(2)=x^(4)(1-x^(2))` is

Options

A. `(pi)/(6)" sq. units"`
B. `(pi)/(4)" sq. units"`
C. `(pi)/(3)" sq. units"`
D. `(pi)/(2)" sq. units"`

Answer

To find the area enclosed by the curve given by the equation \( y^2 = x^4 (1 - x^2) \), we can follow these steps:

### Step 1: Understand the curve
The equation \( y^2 = x^4 (1 - x^2) \) indicates that the curve is symmetric about both the x-axis and y-axis because \( y^2 \) is always non-negative. The curve will intersect the x-axis when \( y = 0 \), which occurs when \( x^4 (1 - x^2) = 0 \).

### Step 2: Find the points of intersection
Set \( y^2 = 0 \):
\[ x^4 (1 - x^2) = 0 \]
This gives us:
1. \( x^4 = 0 \) → \( x = 0 \)
2. \( 1 - x^2 = 0 \) → \( x = \pm 1 \)

Thus, the points of intersection are \( x = -1, 0, 1 \).

### Step 3: Set up the area integral
To find the area enclosed by the curve, we can calculate the area in the first quadrant and then multiply it by 4 (due to symmetry). The area \( A \) in the first quadrant can be expressed as:
\[ A = \int_0^1 y \, dx \]
From the equation \( y^2 = x^4 (1 - x^2) \), we have:
\[ y = \sqrt{x^4 (1 - x^2)} = x^2 \sqrt{1 - x^2} \]
Thus, the area becomes:
\[ A = \int_0^1 x^2 \sqrt{1 - x^2} \, dx \]

### Step 4: Solve the integral
To solve the integral \( A = \int_0^1 x^2 \sqrt{1 - x^2} \, dx \), we can use the substitution \( x = \sin \theta \), which gives \( dx = \cos \theta \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = 1 \), \( \theta = \frac{\pi}{2} \)

Now substituting:
\[ A = \int_0^{\frac{\pi}{2}} (\sin^2 \theta) \sqrt{1 - \sin^2 \theta} \cos \theta \, d\theta \]
Since \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we have:
\[ A = \int_0^{\frac{\pi}{2}} \sin^2 \theta \cos^2 \theta \, d\theta \]

### Step 5: Simplify the integral
Using the identity \( \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta) \):
\[ A = \frac{1}{4} \int_0^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta \]
Using the integral formula \( \int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C \):
\[ A = \frac{1}{4} \left[ \frac{2\theta}{2} - \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{2}} = \frac{1}{4} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{8} \]

### Step 6: Calculate the total area
Since the area in the first quadrant is \( \frac{\pi}{8} \), the total area enclosed by the curve is:
\[ \text{Total Area} = 4A = 4 \times \frac{\pi}{8} = \frac{\pi}{2} \]

### Final Answer
The area enclosed by the curve \( y^2 = x^4(1 - x^2) \) is:
\[ \boxed{\frac{\pi}{2}} \]