Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
y=6 - 3x,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 can use the method of cylindrical shells.
The region bounded by the graphs of y = 6 - 3x, y = 0, and x = 0 lies in the first quadrant between the y-axis and the line y = 6. This region is a right triangle with base 2 and height 6.
The volume of the solid generated by revolving this region about the y-axis can be calculated using the formula for the volume of a solid of revolution:
V = ∫[a,b] 2πx * h(x) dx
Where a and b are the limits of integration (in this case, a = 0 and b = 2), h(x) is the height of the shell at x, and 2πx is the circumference of the shell at x.
The height of the shell at x is given by h(x) = 6 - 3x, and the circumference of the shell at x is given by 2πx.
Therefore, the volume V can be calculated as:
V = ∫[0,2] 2πx * (6 - 3x) dx
= 2π ∫[0,2] (6x - 3x^2) dx
= 2π [3x^2 - x^3] |[0,2]
= 2π [(3*2^2 - 2^3) - (3*0^2 - 0^3)]
= 2π [12 - 8]
= 2π * 4
= 8π
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 8π cubic units.