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This problem involves calculating the volume of a solid with a base bounded by two curves and rectangular cross-sections perpendicular to the \(x\)-axis. The longer sides of these rectangles lie in the \(xy\)-plane, and their lengths are twice their widths.

### Step-by-Step Solution

1. **Set up the region of the base**:
   The region is bounded by the curves:
   \[
   y = x^2 \quad \text{and} \quad y = 2x
   \]
   To find the limits of integration, solve for the points where these curves intersect:
   \[
   x^2 = 2x
   \]
   \[
   x^2 - 2x = 0
   \]
   \[
   x(x - 2) = 0
   \]
   So, \(x = 0\) and \(x = 2\). Therefore, the region of integration is from \(x = 0\) to \(x = 2\).

2. **Width and length of the rectangles**:
   The width of each rectangle is determined by the vertical distance between the two curves:
   \[
   \text{Width} = (2x - x^2)
   \]
   The length is twice the width:
   \[
   \text{Length} = 2(2x - x^2)
   \]

3. **Volume of the rectangular cross-sections**:
   The area \(A(x)\) of a rectangle at each \(x\)-position is:
   \[
   A(x) = \text{Length} \times \text{Width} = 2(2x - x^2)(2x - x^2)
   \]
   Simplifying:
   \[
   A(x) = 2(2x - x^2)^2
   \]
   Expand \((2x - x^2)^2\):
   \[
   (2x - x^2)^2 = 4x^2 - 4x^3 + x^4
   \]
   So, the area is:
   \[
   A(x) = 2(4x^2 - 4x^3 + x^4) = 8x^2 - 8x^3 + 2x^4
   \]

4. **Set up the integral for the volume**:
   The volume \(V\) is the integral of the area function over the interval from 0 to 2:
   \[
   V = \int_0^2 (8x^2 - 8x^3 + 2x^4) \, dx
   \]

5. **Evaluate the integral**:
   Now, integrate each term:
   \[
   \int 8x^2 \, dx = \frac{8x^3}{3}, \quad \int -8x^3 \, dx = -2x^4, \quad \int 2x^4 \, dx = \frac{2x^5}{5}
   \]
   Therefore, the volume is:
   \[
   V = \left[ \frac{8x^3}{3} - 2x^4 + \frac{2x^5}{5} \right]_0^2
   \]
   Substituting the limits \(x = 2\) and \(x = 0\):
   \[
   V = \left( \frac{8(2)^3}{3} - 2(2)^4 + \frac{2(2)^5}{5} \right) - \left( \frac{8(0)^3}{3} - 2(0)^4 + \frac{2(0)^5}{5} \right)
   \]
   Simplifying:
   \[
   V = \left( \frac{64}{3} - 32 + \frac{64}{5} \right)
   \]
   Find a common denominator to simplify:
   \[
   V = \frac{64}{3} - \frac{96}{3} + \frac{192}{15}
   \]
   Simplifying the fractions:
   \[
   V = \frac{64}{3} - \frac{96}{3} + \frac{64}{5} = -\frac{32}{3} + \frac{64}{5}
   \]
   To combine these, find a common denominator (15):
   \[
   V = \frac{-160}{15} + \frac{192}{15} = \frac{32}{15}
   \]

So, the volume of the solid is:
\[
\boxed{\frac{32}{15}}
\]

---

Do you have any further questions on this? Here are five related questions to deepen understanding:

1. What would change if the cross-sections were squares instead of rectangles?
2. How would you set up the problem if the cross-sections were triangles?
3. How does the slicing method differ from the disk or washer methods in calculating volumes?
4. What would happen if the bounds were changed, say from \(x = 1\) to \(x = 2\)?
5. How do different types of cross-sections affect the final volume of a solid?

**Tip**: Always visualize the problem's geometry to better understand how the cross-sections relate to the solid's overall shape!

Use the slicing method to find the volume of a solid whose base is the region bounded by y = x^2 and y = 2x. Cross-sections perpendicular to the x-axis are rectangles whose lengths are double their widths, and the longer sides (lengths) are in the xy-plane.

Learn how to use the slicing method to calculate the volume of a solid with a base bounded by the curves y = x^2 and y = 2x. Cross-sections are rectangles with lengths double their widths. Step-by-step solution for calculus students.

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