The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method.
y = −x^2 + 10x − 16,    y = 0;    about the x-axis

We are tasked with finding the volume of the solid formed by rotating the region bounded by the curve \( y = -x^2 + 10x - 16 \) and the line \( y = 0 \) (the x-axis) about the x-axis.

### Step 1: Find the bounds
We first need to determine where the curve intersects the x-axis. This is done by solving for \( x \) when \( y = 0 \):

\[
0 = -x^2 + 10x - 16
\]

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

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

For the equation \( -x^2 + 10x - 16 = 0 \), we have \( a = -1 \), \( b = 10 \), and \( c = -16 \). Plugging these values into the quadratic formula:

\[
x = \frac{-10 \pm \sqrt{10^2 - 4(-1)(-16)}}{2(-1)}
\]
\[
x = \frac{-10 \pm \sqrt{100 - 64}}{-2}
\]
\[
x = \frac{-10 \pm \sqrt{36}}{-2}
\]
\[
x = \frac{-10 \pm 6}{-2}
\]

This gives two solutions:

\[
x_1 = \frac{-10 + 6}{-2} = 2, \quad x_2 = \frac{-10 - 6}{-2} = 8
\]

So, the curve intersects the x-axis at \( x = 2 \) and \( x = 8 \).

### Step 2: Set up the volume integral (Disk method)
The volume of the solid formed by rotating the region about the x-axis can be found using the disk method. The formula for the volume is:

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

In our case, \( f(x) = -x^2 + 10x - 16 \), and the limits of integration are from \( x = 2 \) to \( x = 8 \). So the volume is:

\[
V = \pi \int_{2}^{8} \left[ -x^2 + 10x - 16 \right]^2 \, dx
\]

### Step 3: Simplify the integrand
We now expand \( (-x^2 + 10x - 16)^2 \):

\[
\left(-x^2 + 10x - 16\right)^2 = (x^4 - 20x^3 + 132x^2 - 320x + 256)
\]

### Step 4: Perform the integration
We integrate each term:

\[
\int_{2}^{8} \left( x^4 - 20x^3 + 132x^2 - 320x + 256 \right) \, dx
\]

Each term is integrated as follows:

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

Thus, the integral becomes:

\[
V = \pi \left[ \frac{x^5}{5} - 5x^4 + 44x^3 - 160x^2 + 256x \right]_{2}^{8}
\]

### Step 5: Compute the definite integral
We now evaluate this expression at \( x = 8 \) and \( x = 2 \) and subtract the results:

For \( x = 8 \):

\[
\frac{8^5}{5} - 5(

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. y = −x^2 + 10x − 16, y = 0; about the x-axis

Find the volume of the solid formed by rotating the region bounded by the curve y = −x^2 + 10x − 16 and the x-axis using the disk method. This problem involves solving a quadratic equation and applying the disk method to compute the volume of a solid of revolution. Suitable for Grades 10-12.

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