Sketch the common region included between the curve y+8 = x^2 - 2x and the x-axis find the volume of the solid generated by revolving the common area about the line x+5=0.

Let's break down the problem into steps:

### 1. **Find the points of intersection between the curve and the x-axis:**

The curve is given by:

\[
y + 8 = x^2 - 2x
\]

To find where it intersects the x-axis, set \( y = 0 \):

\[
0 + 8 = x^2 - 2x
\]

\[
x^2 - 2x - 8 = 0
\]

This is a quadratic equation. We solve it using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = -2\), and \(c = -8\). Substituting these into the quadratic formula:

\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}
\]

\[
x = \frac{2 \pm \sqrt{4 + 32}}{2}
\]

\[
x = \frac{2 \pm \sqrt{36}}{2}
\]

\[
x = \frac{2 \pm 6}{2}
\]

So, the solutions are:

\[
x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{-4}{2} = -2
\]

### 2. **Determine the volume of the solid formed by revolving the region about the line \( x + 5 = 0 \):**

The line \( x + 5 = 0 \) simplifies to \( x = -5 \). The volume \( V \) of the solid generated by revolving the region around this line can be found using the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

\[
V = 2\pi \int_{a}^{b} (r(x) \cdot h(x)) \, dx
\]

Where:
- \( r(x) \) is the radius of the shell, which is the distance from the line \( x = -5 \).
- \( h(x) \) is the height of the shell, given by \( y \) (the function \( y(x) = x^2 - 2x - 8 \)).

So:

\[
r(x) = |x + 5|
\]

\[
h(x) = x^2 - 2x - 8
\]

The limits of integration are from \( x = -2 \) to \( x = 4 \).

### 3. **Set up and evaluate the integral:**

The volume is:

\[
V = 2\pi \int_{-2}^{4} (x + 5)(x^2 - 2x - 8) \, dx
\]

Now, I'll expand and integrate this expression:

\[
V = 2\pi \int_{-2}^{4} \left[ (x^3 + 5x^2 - 2x^2 - 10x - 8x - 40) \right] dx
\]

\[
V = 2\pi \int_{-2}^{4} \left[ x^3 + 3x^2 - 18x - 40 \right] dx
\]

Now, integrate term by term:

\[
V = 2\pi \left[ \frac{x^4}{4} + x^3 - 9x^2 - 40x \right]_{-2}^{4}
\]

Substituting the limits into the antiderivative:

At \( x = 4 \):

\[
\frac{4^4}{4} + 4^3 - 9(4^2) - 40(4)
\]

At \( x = -2 \):

\[
\frac{(-2)^4}{4} + (-2)^3 - 9(-2)^2 - 40(-2)
\]

Finally, subtract the values to get the volume.

I'll calculate this step.The volume of the solid generated by revolving the region bounded by the curve \( y + 8 = x^2 - 2x \) and the x-axis about the line \( x + 5 = 0 \) is approximately \(-1357.17\) cubic units.

However, the negative sign indicates that the region might have been considered incorrectly in terms of limits or orientation. Typically, we take the absolute value when calculating volumes. So, the absolute value of this result is:

\[
V \approx 1357.17 \text{ cubic units}
\]

Do you have any questions or would you like further details? Here are 5 related questions:

1. How would the volume change if the region was revolved around the y-axis?
2. What if the region was revolved around the x-axis instead of \(x = -5\)?
3. How does changing the curve affect the volume of the solid?
4. Can you generalize the method of cylindrical shells for any arbitrary axis of rotation?
5. How would the problem change if the curve was a cubic function instead of a quadratic?

**Tip:** When dealing with volumes of solids of revolution, ensure you carefully consider the axis of rotation and orientation of the region.

Calculate the volume of the solid generated by revolving the area bounded by the curve y + 8 = x^2 - 2x and the x-axis around the line x + 5 = 0. Step-by-step solution using the method of cylindrical shells. Suitable for advanced high school students and calculus learners.

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