Find the volume of the solid generated by revolving the shaded region about the x axis 5x+4y=40

I'll assume that the "shaded area" is defined by the x - and y-axes and the line 5x+4y=40 (and therefore is in Quadrant I only). It's triangular. One of the legs rests upon the x-axis, and x ranges from 0 to 8. The other leg rests upon the y-axis, and y ranges from 0 to 10.

Let's use the disk method to determine the volume of the solid generated if this area is revolved around the x-axis. Since 5x+4y=40, 4y=40-5x, and y = 10 - (5/4) x. Imagine "slices" of this solid that are perpendicular to the x-axis and that have the radius y = r = y = 10 - (5/4) x.

Then the area of any such slice is pi*r^2, or pi * (10-5x/4) ^2. The thickness of each slice is dx. Thus, the volume of each slice is dv = pi * (10-5x/4) ^2 dx.

Integrate this from x=0 to x=8 to determine the total volume of the solid. The result of integration is V = (266 2/3) pi, or 800pi/3. This is the vol. of the solid.

The answer is x=-4y/5 + 8