The area bounded by the curves `y^2=x^3 and |y|=2x` is

Options

A. `5/(16)` sq units
B. `(16)/5` sq units
C. `3/8` sq units
D. None of these

Answer

To find the area bounded by the curves \( y^2 = x^3 \) and \( |y| = 2x \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by \( y^2 = x^3 \), which can be rewritten as \( y = \pm \sqrt{x^3} \). This represents a cubic curve that opens upwards and downwards.

The second curve is \( |y| = 2x \), which can be split into two linear equations:
- \( y = 2x \) (upper line)
- \( y = -2x \) (lower line)

### Step 2: Find the points of intersection
To find the points where these curves intersect, we set \( y = 2x \) and \( y^2 = x^3 \):

1. Substitute \( y = 2x \) into \( y^2 = x^3 \):
\[ (2x)^2 = x^3 \]
\[ 4x^2 = x^3 \]
\[ x^3 - 4x^2 = 0 \]
\[ x^2(x - 4) = 0 \]
This gives us \( x = 0 \) and \( x = 4 \).

2. Now, find the corresponding \( y \) values:
- For \( x = 0 \): \( y = 2(0) = 0 \) → Point (0, 0)
- For \( x = 4 \): \( y = 2(4) = 8 \) → Point (4, 8)

Next, we check the intersection with \( y = -2x \):
1. Substitute \( y = -2x \) into \( y^2 = x^3 \):
\[ (-2x)^2 = x^3 \]
\[ 4x^2 = x^3 \]
\[ x^3 - 4x^2 = 0 \]
This gives the same solutions: \( x = 0 \) and \( x = 4 \).

2. Find the corresponding \( y \) values:
- For \( x = 0 \): \( y = -2(0) = 0 \) → Point (0, 0)
- For \( x = 4 \): \( y = -2(4) = -8 \) → Point (4, -8)

### Step 3: Set up the integral for the area
The area \( A \) bounded by the curves can be calculated by integrating the difference of the upper curve and the lower curve from \( x = 0 \) to \( x = 4 \).

The upper curve is \( y = 2x \) and the lower curve is \( y = \sqrt{x^3} \). The area is given by:
\[ A = \int_{0}^{4} (2x - \sqrt{x^3}) \, dx \]

### Step 4: Calculate the integral
1. Simplify the integral:
\[ A = \int_{0}^{4} (2x - x^{3/2}) \, dx \]

2. Compute the integral:
\[ A = \int_{0}^{4} 2x \, dx - \int_{0}^{4} x^{3/2} \, dx \]

- For \( \int 2x \, dx \):
\[ = 2 \cdot \frac{x^2}{2} \Big|_{0}^{4} = x^2 \Big|_{0}^{4} = 16 - 0 = 16 \]

- For \( \int x^{3/2} \, dx \):
\[ = \frac{x^{5/2}}{5/2} \Big|_{0}^{4} = \frac{2}{5} x^{5/2} \Big|_{0}^{4} = \frac{2}{5} (4^{5/2} - 0) = \frac{2}{5} \cdot 32 = \frac{64}{5} \]

3. Combine the results:
\[ A = 16 - \frac{64}{5} = \frac{80}{5} - \frac{64}{5} = \frac{16}{5} \]

### Final Area
The area bounded by the curves is:
\[ A = \frac{16}{5} \text{ square units} \]