Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
y=3(2-x),y=0,x=0
First, we need to find the points of intersection of the two graphs: y = 3(2-x) and y = 0.
Setting the two equations equal to each other:
3(2-x) = 0
6 - 3x = 0
3x = 6
x = 2
So the x-coordinate of the intersection point is x = 2. Now we need to set up the integral to find the volume of the solid generated by revolving the region about the y-axis.
The volume V of the solid formed by rotating the region bounded by the graphs of y = 3(2-x), y = 0, and x = 0, around the y-axis can be found using the shell method:
V = ∫[a,b] 2πx*f(x) dx, where a = 0 and b = 2
V = ∫[0,2] 2πx*3(2-x) dx
V = 6π ∫[0,2] (2x - x^2) dx
V = 6π [x^2 - (x^3)/3] |[0,2]
V = 6π [(2^2 - (2^3)/3) - (0 - (0^3)/3)]
V = 6π [(4 - 8/3) - 0]
V = 6π (4/3)
V = 8π cubic units
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of y = 3(2-x), y = 0, and x = 0 about the y-axis is 8π cubic units.