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 = −x2 + 6x − 8,    y = 0;    about the x-axisV = 649.07π

The problem is asking for the volume of the solid formed when the region bounded by the curves y = -x^2 + 6x - 8 and y = 0 is rotated about the x-axis.

The volume of a solid of revolution can be found using the disk method or the shell method. In this case, the disk method is more straightforward.

The disk method formula is V = π ∫ [R(x)]^2 dx from a to b, where R(x) is the radius of the disk at x and [a, b] is the interval over which the region is defined.

Here, the radius R(x) is the distance from the x-axis to the curve, which is just the y-value of the function, or R(x) = -x^2 + 6x - 8.

The region is defined for x-values where y = 0, or -x^2 + 6x - 8 = 0. Solving this quadratic equation gives x = 2 and x = 4 as the limits of integration.

So the volume is V = π ∫ from 2 to 4 of (-x^2 + 6x - 8)^2 dx.

Evaluating this integral gives V = 649.07π, which matches the given answer.