Calculate the volume of the solid generated by revolving the region bounded by the curve $$ y = x^2 $$ and the line $$ y = 4 $$ around the x-axis.

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To calculate the volume of the solid generated by revolving the region bounded by the curve $$ y = x^2 $$ and the line $$ y = 4 $$ around the x-axis, we'll use the **washer method**. Here's a step-by-step solution:

### 1. **Identify the Points of Intersection**
First, determine where the curves $$ y = x^2 $$ and $$ y = 4 $$ intersect:

$$
x^2 = 4 \implies x = \pm 2
$$

So, the region of interest is between $$ x = -2 $$ and $$ x = 2 $$.

### 2. **Set Up the Washer Method Integral**
When revolving around the x-axis, the volume $$ V $$ can be calculated using the washer method:

$$
V = \pi \int_{a}^{b} \left[ (R(x))^2 - (r(x))^2 \right] dx
$$

Where:
- $$ R(x) $$ is the outer radius (distance from the x-axis to the upper curve).
- $$ r(x) $$ is the inner radius (distance from the x-axis to the lower curve).

In this case:
- The **outer curve** is $$ y = 4 $$, so $$ R(x) = 4 $$.
- The **inner curve** is $$ y = x^2 $$, so $$ r(x) = x^2 $$.

### 3. **Express the Integral**
Substitute $$ R(x) $$ and $$ r(x) $$ into the integral:

$$
V = \pi \int_{-2}^{2} \left[ 4^2 - (x^2)^2 \right] dx = \pi \int_{-2}^{2} \left[ 16 - x^4 \right] dx
$$

Since the integrand is even, we can simplify the calculation by integrating from 0 to 2 and then doubling the result:

$$
V = 2\pi \int_{0}^{2} \left[ 16 - x^4 \right] dx
$$

### 4. **Compute the Integral**
Evaluate the integral:

$$
\int_{0}^{2} \left[ 16 - x^4 \right] dx = \left[ 16x - \frac{x^5}{5} \right]_0^2 = \left( 16 \times 2 - \frac{32}{5} \right) = 32 - \frac{32}{5} = \frac{160 - 32}{5} = \frac{128}{5}
$$

### 5. **Calculate the Volume**
Multiply by $$ 2\pi $$:

$$
V = 2\pi \times \frac{128}{5} = \frac{256}{5} \pi
$$

### **Final Answer**
The volume of the solid is:

$$
V = \frac{256}{5} \pi \text{ cubic units}
$$

Answer: Its volume is 256⁄5 multiplied by π, that is V = (256⁄5) · π cubic units
The volume of the solid is $$ \frac{256}{5} \pi $$ cubic units.

Calculate the volume of the solid generated by revolving the region bounded by the curve and the line around the x-axis.

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