Let R be the region bounded by the curves y = -x^2+6x-5 and y = 0. Let S be the solid formed by revolving R about the x axis. Select and use the most appropriate method to find the volume of S. If necessary, round to 4 decimal places.

To find the volume of the solid \( S \), we use the **disk method** because the region \( R \) is revolved around the x-axis. The volume of a solid of revolution formed by revolving a region around the x-axis is given by:

\[
V = \pi \int_{a}^{b} \left[f(x)\right]^2 \, dx
\]

### Step 1: Identify the region boundaries
The curve is \( y = -x^2 + 6x - 5 \), and we are revolving the region bounded by this curve and \( y = 0 \) (the x-axis).

To find the limits of integration \( a \) and \( b \), we need to determine where the curve intersects the x-axis (i.e., where \( y = 0 \)):

\[
-x^2 + 6x - 5 = 0
\]

This is a quadratic equation, which we can solve using the quadratic formula:

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

For \( -x^2 + 6x - 5 = 0 \), we have \( a = -1 \), \( b = 6 \), and \( c = -5 \). Plugging these values into the quadratic formula:

\[
x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-5)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 20}}{-2} = \frac{-6 \pm \sqrt{16}}{-2}
\]

\[
x = \frac{-6 \pm 4}{-2}
\]

Thus, the two solutions are:

\[
x = \frac{-6 + 4}{-2} = 1 \quad \text{and} \quad x = \frac{-6 - 4}{-2} = 5
\]

So the region is bounded by \( x = 1 \) and \( x = 5 \).

### Step 2: Set up the volume integral
We will revolve the curve \( y = -x^2 + 6x - 5 \) around the x-axis. Using the disk method, the volume is:

\[
V = \pi \int_{1}^{5} \left[(-x^2 + 6x - 5)\right]^2 \, dx
\]

### Step 3: Compute the integral
We now need to expand and integrate the function:

\[
(-x^2 + 6x - 5)^2 = (x^4 - 12x^3 + 46x^2 - 60x + 25)
\]

Thus, the volume integral becomes:

\[
V = \pi \int_{1}^{5} \left(x^4 - 12x^3 + 46x^2 - 60x + 25\right) \, dx
\]

Now, integrate each term:

\[
\int x^4 \, dx = \frac{x^5}{5}, \quad \int x^3 \, dx = \frac{x^4}{4}, \quad \int x^2 \, dx = \frac{x^3}{3}, \quad \int x \, dx = \frac{x^2}{2}, \quad \int 1 \, dx = x
\]

Thus, the integral becomes:

\[
V = \pi \left[\frac{x^5}{5} - 12 \cdot \frac{x^4}{4} + 46 \cdot \frac{x^3}{3} - 60 \cdot \frac{x^2}{2} + 25x \right]_{1}^{5}
\]

### Step 4: Evaluate the definite integral
Substitute \( x = 5 \) and \( x = 1 \) into the expression:

For \( x = 5 \):

\[
\frac{5^5}{5} - 12 \cdot \frac{5^4}{4} + 46 \cdot \frac{5^3}{3} - 60 \cdot \frac{5^2}{2} + 25 \cdot 5 = 625 - 1875 + 1916.67 - 750 + 125 = 41.67
\]

For \( x = 1 \):

\[
\frac{1^5}{5} - 12 \cdot \frac{1^4}{4} + 46 \cdot \frac{1^3}{3} - 60 \cdot \frac{1^2}{2} + 25 \cdot 1 = 0.2 - 3 + 15.33 - 30 + 25 = 7.53
\]

Now, subtract the results:

\[
41.67 - 7.53 = 34.14
\]

### Step 5: Multiply by \( \pi \)
Finally, multiply by \( \pi \):

\[
V = \pi \cdot 34.14 \approx 107.2488
\]

### Final Answer:
The volume of the solid is approximately \( \boxed{107.2488} \) cubic units.

Would you like further details or explanations?

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**Related Questions:**
1. How would the volume change if the region was revolved around the y-axis?
2. What is the area of the region bounded by the curves before the revolution?
3. How would the volume calculation change if the curve was revolved around the line \( y = -1 \)?
4. Can you apply the washer method in this scenario? Why or why not?
5. How do you determine the volume if the region is revolved around \( y = 5 \) instead?

**Tip:** Always verify your limits of integration by checking where the curve intersects the axis of revolution.

Learn how to calculate the volume of a solid formed by revolving the region bounded by the curve y = -x^2 + 6x - 5 and y = 0 around the x-axis using the disk method. This detailed step-by-step solution involves integration and solving quadratic equations, making it suitable for advanced high school or early college calculus students.

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