y=3(2-x),y=0,x=0
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis, we first need to find the points of intersection between the two equations.
Setting y = 0 in the first equation gives:
0 = 3(2-x)
0 = 6 - 3x
3x = 6
x = 2
So the points of intersection are at (2,0) and (0,0).
The region bounded by the graphs is between y = 3(2-x) and the x-axis. To find the volume of the solid generated by revolving this region about the y-axis, we will integrate the area of cross-sections perpendicular to the y-axis.
The radius of each cross-section is the x-value of the function, so the radius at a height y is x = 2 - y/3.
Thus, the volume is given by the integral of π(radius)^2 dy from y = 0 to y = 3:
V = π∫[0,3] (2 - y/3)^2 dy
V = π∫[0,3] (4 - 4y/3 + y^2/9) dy
V = π[4y - 2y^2/3 + y^3/27] from 0 to 3
V = π[12 - 6 + 27/27]
V = 6π
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 6π cubic units.