Find the volume of the solid generated by revolving the region bounded by
y = 5x-x^2 and y = 0
about the line x = -3

To find the volume of the solid generated by revolving the region bounded by \( y = 5x - x^2 \) and \( y = 0 \) about the line \( x = -3 \), we will use the **method of cylindrical shells**.

### Step-by-step solution:

#### 1. **Set up the problem**:
The region is bounded by \( y = 5x - x^2 \) and the x-axis, \( y = 0 \). This curve is a parabola opening downwards, with its vertex at \( x = 2.5 \). The limits of integration will be from the x-intercepts of the parabola.

#### 2. **Find the x-intercepts**:
Set \( y = 5x - x^2 = 0 \) to solve for \( x \).

\[
x(5 - x) = 0
\]

So, \( x = 0 \) and \( x = 5 \). Thus, the region extends from \( x = 0 \) to \( x = 5 \).

#### 3. **Formula for the volume using cylindrical shells**:
The formula for the volume using cylindrical shells when rotating around a vertical line \( x = a \) (here \( a = -3 \)) is:

\[
V = 2\pi \int_{x_1}^{x_2} (x - a) f(x) \, dx
\]

Where:
- \( x_1 = 0 \) and \( x_2 = 5 \) (the bounds of the region),
- \( a = -3 \) (the axis of rotation),
- \( f(x) = 5x - x^2 \) (the height of the shell at \( x \)).

#### 4. **Set up the integral**:
The distance from a point \( x \) to the line \( x = -3 \) is \( (x - (-3)) = x + 3 \). Therefore, the volume integral becomes:

\[
V = 2\pi \int_0^5 (x + 3)(5x - x^2) \, dx
\]

#### 5. **Simplify the integrand**:
First, expand \( (x + 3)(5x - x^2) \):

\[
(x + 3)(5x - x^2) = x(5x - x^2) + 3(5x - x^2)
\]
\[
= 5x^2 - x^3 + 15x - 3x^2
\]
\[
= 2x^2 - x^3 + 15x
\]

Now the volume integral becomes:

\[
V = 2\pi \int_0^5 (2x^2 - x^3 + 15x) \, dx
\]

#### 6. **Evaluate the integral**:
Now integrate each term:

\[
\int_0^5 2x^2 \, dx = \left[ \frac{2x^3}{3} \right]_0^5 = \frac{2(5)^3}{3} - 0 = \frac{250}{3}
\]

\[
\int_0^5 -x^3 \, dx = \left[ \frac{-x^4}{4} \right]_0^5 = \frac{-5^4}{4} = -\frac{625}{4}
\]

\[
\int_0^5 15x \, dx = \left[ \frac{15x^2}{2} \right]_0^5 = \frac{15(5)^2}{2} - 0 = \frac{375}{2}
\]

#### 7. **Combine the results**:
Now, sum the results of the integrals:

\[
V = 2\pi \left( \frac{250}{3} - \frac{625}{4} + \frac{375}{2} \right)
\]

To combine these, first get a common denominator (12):

\[
V = 2\pi \left( \frac{1000}{12} - \frac{1875}{12} + \frac{2250}{12} \right)
\]

\[
V = 2\pi \left( \frac{1000 - 1875 + 2250}{12} \right) = 2\pi \left( \frac{1375}{12} \right)
\]

\[
V = \frac{2750\pi}{12}
\]

#### 8. **Simplify the final result**:

\[
V = \frac{1375\pi}{6}
\]

Thus, the volume of the solid is \( \frac{1375\pi}{6} \) cubic units.

---

Would you like further details on any step or another example?

Here are 5 related questions to deepen your understanding:
1. How would the method change if the region were revolved around the y-axis?
2. What if the region were revolved around a different vertical line, such as \( x = 2 \)?
3. Can you derive the volume using the disk or washer method for this problem?
4. What happens if we revolve the region around a horizontal line, like \( y = 5 \)?
5. How does the choice of axis of rotation affect the cylindrical shell method?

**Tip**: When setting up a cylindrical shell problem, always ensure the radius and height of the shell are expressed in terms of the same variable (usually \( x \) or \( y \)).

Find the volume of the solid generated by revolving the region bounded by y = 5x - x^2 and y = 0 about the line x = -3

Explore how to calculate the volume of the solid generated by revolving the region bounded by y = 5x - x^2 and y = 0 around the line x = -3. This problem involves using the method of cylindrical shells, integral calculus, and quadratic functions. Detailed step-by-step solution provided for calculus students.

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