Find the volume of the solid of revolution generated by revolving the region bounded by y=2x, y=x^2 about the line y=-5.

To find the volume of the solid of revolution generated by revolving the region bounded by \( y = 2x \) and \( y = x^2 \) about the line \( y = -5 \), we will apply the method of cylindrical shells.

### Step 1: Find the points of intersection
We first need to determine the points of intersection of the curves \( y = 2x \) and \( y = x^2 \).

Set \( 2x = x^2 \):
\[
x^2 - 2x = 0
\]
\[
x(x - 2) = 0
\]
Thus, \( x = 0 \) and \( x = 2 \).

So, the region of interest lies between \( x = 0 \) and \( x = 2 \).

### Step 2: Set up the volume integral using cylindrical shells
The formula for the volume using the cylindrical shell method is:
\[
V = 2\pi \int_{a}^{b} ( \text{radius of shell} ) ( \text{height of shell} ) \, dx
\]
Here:
- The **radius** is the distance from the line of revolution \( y = -5 \) to a point on the curve, which is \( |y + 5| \).
- The **height** is the vertical distance between the curves \( y = 2x \) and \( y = x^2 \).

We will split the integral into two parts based on the two curves that bound the region.

#### For \( y = 2x \) (upper curve):
The radius of the cylindrical shell at a point \( x \) is:
\[
r(x) = |2x + 5|
\]
The height is \( 2x - x^2 \).

Thus, the integral becomes:
\[
V = 2\pi \int_0^2 (2x + 5)(2x - x^2) \, dx
\]

### Step 3: Evaluate the integral
Now we expand the integrand:
\[
(2x + 5)(2x - x^2) = 4x^2 - 2x^3 + 10x - 5x^2 = -2x^3 - x^2 + 10x
\]
Thus, the volume integral is:
\[
V = 2\pi \int_0^2 (-2x^3 - x^2 + 10x) \, dx
\]
We can now evaluate this integral term by term.

\[
V = 2\pi \left[ \frac{-2x^4}{4} - \frac{x^3}{3} + \frac{10x^2}{2} \right]_0^2
\]
\[
V = 2\pi \left[ \frac{-2(2^4)}{4} - \frac{(2^3)}{3} + \frac{10(2^2)}{2} \right]
\]
\[
V = 2\pi \left[ \frac{-32}{4} - \frac{8}{3} + 20 \right]
\]
\[
V = 2\pi \left[ -8 - \frac{8}{3} + 20 \right]
\]
\[
V = 2\pi \left[ 12 - \frac{8}{3} \right]
\]
\[
V = 2\pi \left[ \frac{36}{3} - \frac{8}{3} \right]
\]
\[
V = 2\pi \left[ \frac{28}{3} \right]
\]
\[
V = \frac{56\pi}{3}
\]

Thus, the volume of the solid is \( \frac{56\pi}{3} \) cubic units.

Would you like further details or have any questions?

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Here are five related questions you can explore:
1. How does the method of disks differ from the method of cylindrical shells?
2. What is the volume if the region is rotated about the x-axis instead of \( y = -5 \)?
3. How would the problem change if the bounds were different?
4. Can this problem be solved using a different coordinate system, such as polar coordinates?
5. What is the significance of the points of intersection in finding the volume?

**Tip:** For problems involving rotation around a line other than the axis, always visualize the distance (radius) to ensure correct setup of the integral.

Find the volume of the solid of revolution generated by revolving the region bounded by y=2x, y=x^2 about the line y=-5.

Find the volume of the solid generated by revolving the region bounded by y=2x and y=x^2 about the line y=-5. This solution involves the cylindrical shell method, integration, and solving for the volume. Suitable for advanced high school or college-level students.

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