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To solve this problem, we need to compute the volume of the solid generated when the region enclosed by the curves \( y = 2x \) and \( y = x^2 \) is rotated around the \(x\)-axis. We can do this using the *method of disks* or *washers*.

### Step 1: Determine the points of intersection
First, we find where the two curves intersect by setting \( y = 2x \) equal to \( y = x^2 \):

\[
2x = x^2
\]

Rearranging:

\[
x^2 - 2x = 0
\]
\[
x(x - 2) = 0
\]

So, the points of intersection are \( x = 0 \) and \( x = 2 \). These will be the limits of integration.

### Step 2: Set up the washer method
For a rotation around the \(x\)-axis, the volume is given by the washer formula:

\[
V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] dx
\]

Here:
- \( R_{\text{outer}}(x) = 2x \) (the curve farther from the x-axis, \(y = 2x\))
- \( R_{\text{inner}}(x) = x^2 \) (the curve closer to the x-axis, \(y = x^2\))

Thus, the volume integral becomes:

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

### Step 3: Compute the integral
Now, we evaluate the integral:

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

First, integrate term by term:

\[
\int 4x^2 \, dx = \frac{4x^3}{3}
\]
\[
\int x^4 \, dx = \frac{x^5}{5}
\]

Now apply the limits:

\[
V = \pi \left[ \frac{4x^3}{3} \Big|_0^2 - \frac{x^5}{5} \Big|_0^2 \right]
\]
\[
V = \pi \left[ \frac{4(2)^3}{3} - \frac{(2)^5}{5} \right]
\]
\[
V = \pi \left[ \frac{32}{3} - \frac{32}{5} \right]
\]

### Step 4: Simplify
To simplify, find a common denominator for the fractions:

\[
V = \pi \left[ \frac{160}{15} - \frac{96}{15} \right]
\]
\[
V = \pi \cdot \frac{64}{15}
\]

Thus, the volume is:

\[
V = \frac{64\pi}{15}
\]

The volume of the solid generated by rotating the region around the x-axis is \( \frac{64\pi}{15} \).

Would you like more details or have any questions?

Here are 5 questions related to this concept:
1. How would the volume change if the region were rotated around the y-axis instead?
2. What if we rotated around the line \( y = -1 \) instead of the x-axis?
3. Can you use the shell method to find the volume for this problem?
4. How would the integral change if one of the curves were shifted?
5. How do the limits of integration change if the region is bounded by different functions?

**Tip:** When dealing with volumes of revolution, always carefully identify the outer and inner functions before setting up the integral.

A region is enclosed by the curves y = 2x and y = x^2. What is the volume of the solid generated when this region is rotated around the x-axis?

This problem involves finding the volume of a solid generated when the region enclosed by y = 2x and y = x^2 is rotated around the x-axis using the washer method. Learn how to set up the integral and compute the volume with detailed steps. Suitable for advanced high school or early college calculus students.

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