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.