Find the volume of the solid obtained by rotating the region enclosed by the graphs of y=15-x, y=3x+11 and x=0 about the y-axis.

1 answer

Find out where the curves y = 15-x and y = 3x + 11 intersect
15 -x = 3x + 11.
4x = 4
x = 1. At that point, y = 14.
The enclosed volume goes from the y axis to x = 1. You are rotating a triangle about the y axis. The limits of integration are from y = 11 to y = 15. Those are the maximum and minimum y values of the enclosed volume.

Do an integration of
2 pi x^2 dy from y=11 to y=14, using the y = 3x + 11 function, and add the integral of 2 pi x^2 dy from y=14 to y = 15, using the y = 15 -x function. You will have to convert x to y in those integrations, using the appropriate formula.

It's a lot easier to visualize if you draw a figure, but I can't do that here.