Using the disc method, find the volume of the solid formed by revolving the region bounded by $$ y = x^2 $$ and $$ y = 0 $$ from $$ x = 0 $$ to $$ x = 2 $$ about the x-axis.

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To find the volume of the solid formed by revolving the region bounded by $$ y = x^2 $$ and $$ y = 0 $$ from $$ x = 0 $$ to $$ x = 2 $$ about the $$ x $$-axis using the **disk method**, follow these steps:

### Disk Method Formula
When rotating around the $$ x $$-axis, the volume $$ V $$ is given by:
$$
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
$$
where:
- $$ f(x) $$ is the function representing the boundary of the region.
- $$ [a, b] $$ is the interval of integration.

### Applying to the Given Problem
1. **Identify $$ f(x) $$:**
   $$
   f(x) = x^2
   $$
   
2. **Set up the integral:**
   $$
   V = \pi \int_{0}^{2} (x^2)^2 \, dx = \pi \int_{0}^{2} x^4 \, dx
   $$
   
3. **Evaluate the integral:**
   $$
   \int x^4 \, dx = \frac{x^5}{5} + C
   $$
   Applying the limits from 0 to 2:
   $$
   \left. \frac{x^5}{5} \right|_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5}
   $$
   
4. **Multiply by $$ \pi $$ to get the volume:**
   $$
   V = \pi \times \frac{32}{5} = \frac{32}{5} \pi
   $$

### Final Answer
$$
V = \frac{32}{5} \pi
$$
The volume of the solid is $$ \frac{32}{5} \pi $$.

Using the disc method, find the volume of the solid formed by revolving the region bounded by and from to about the x-axis.

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