y=6 - 3x,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=6 - 3x equal to y=0, we find the point of intersection as:
6 - 3x = 0
-3x = -6
x = 2
So the points of intersection are (2, 0) and (0, 6).
Now we need to set up the integral to find the volume. The radius of the solid at any point y is the x-coordinate of that point. Therefore, the radius of the solid at any point y ranges from x = 0 to x = 2.
The volume generated by revolving the region about the y-axis is given by the formula:
V = π * ∫[a, b] (r(y))^2 dy
V = π * ∫[0, 6] (2)^2 dy
V = π * ∫[0, 6] 4 dy
V = 4π * [y] [0, 6]
V = 4π * (6 - 0)
V = 24π cubic units
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 24π cubic units.