Find the volume of the solid generated by revolving the region delimited by 5x+5y=40, y=0, and x=0 about the x-axis.

To find the volume of the solid generated by revolving a region around the x-axis, we can use the method of cylindrical shells.

First, let's graph the given region:

We have the equation 5x + 5y = 40, which can be rearranged as y = (40 - 5x)/5, and the lines y = 0 and x = 0.

The region is a triangle formed by the points (0, 0), (0, 8), and (8, 0).

To find the volume, we need to integrate the area of the cylindrical shells that make up the solid.

The volume of an individual cylindrical shell is given by the formula V = 2πrhΔx, where r is the distance from the axis of revolution to the outer edge of the shell, h is the height of the shell, and Δx is the infinitesimal width of the shell.

In this case, the radius r is equal to x, the height h is equal to the difference between the corresponding y-values on the curve and the y-value at the axis of revolution, and Δx is an infinitesimally small change in x.

For each value of x, the height h can be calculated as follows: h = (40 - 5x)/5 - 0 = (40 - 5x)/5.

To find the volume of the entire solid, we need to integrate the volume of all the cylindrical shells from x = 0 to x = 8. The integral is given by:

V = ∫(0 to 8) 2πx(40 - 5x)/5 dx.

Simplifying the integrand, we get:

V = 2π/5∫(0 to 8) x(40 - 5x) dx.

Evaluating the integral:

V = 2π/5 [∫(0 to 8) 40x - 5x^2 dx].

Using the power rule of integration, we get:

V = 2π/5 [(20x^2 - (5/3)x^3)] (0 to 8).

Evaluating the integral bounds:

V = 2π/5 [(20(8)^2 - (5/3)(8)^3) - (20(0)^2 - (5/3)(0)^3)].

Simplifying further:

V = 2π/5 [(1280 - 5120/3) - (0 - 0)].

V = 2π/5 [1280 - 5120/3].

Calculating the final value:

V ≈ 2π/5 [1280 - 1706.67].

V ≈ 2π/5 [-426.67].

V ≈ -3413.36π/25.

So, the volume of the solid generated by revolving the given region about the x-axis is approximately -3413.36π/25 cubic units.