Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
Y = 9 – x², y = 0, x =2, x =3
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis, we will use the method of cylindrical shells.
The region bounded by the graphs of the equations is the area between the curves y = 9 - x^2, y = 0, x = 2, and x = 3. This region is a quarter circle with radius 3 and height 9 (radius of the circle is 3 and height is 9 - 0 = 9).
The volume generated by revolving this region about the y-axis can be calculated using the following formula:
V = 2π ∫[a,b] x * [f(x)] dx
Where a = 2, b = 3, and f(x) = 9 - x^2.
V = 2π ∫[2,3] x(9 - x^2) dx
V = 2π [9x - x^3/3] from 2 to 3
V = 2π [(9(3) - 3^3/3) - (9(2) - 2^3/3)]
V = 2π [27 - 9 - 18 + 8/3]
V = 2π [27 - 9 - 54/3 + 8/3]
V = 2π [18 - 46/3]
V = 2π (54/3 - 46/3)
V = 2π (8/3)
V = 16π/3
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 16π/3 cubic units.