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
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis, we need to use the method of cylindrical shells.
First, we need to find the points of intersection of the curves y = 3(2 - x) and y = 0.
Setting y = 0 in the first equation:
0 = 3(2 - x)
0 = 6 - 3x
3x = 6
x = 2
So the points of intersection are (2,0) and (0,0).
Next, we need to find the height of the shell at a general x value. The height is given by the difference between the equations: y = 3(2 - x) - 0 = 6 - 3x.
The radius of the shell is the x value of the shell.
The volume of a cylindrical shell is given by V = 2πrh*dx, where r is the radius, h is the height, and dx is a small change in x.
Integrating V from 0 to 2 to cover the whole region, we get:
V = ∫[0,2] 2π(2π(6 - 3x))dx
= 2π ∫[0,2] (12π - 6πx)dx
= 2π [12πx - 3πx^2] [0,2]
= 2π(24π - 12π)
= 24π^2
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 24π^2 cubic units.