Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
use the ring method
y=9-x^2,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 using the ring method, we will integrate.
The region bounded by the graphs y = 9 - x^2, y = 0, x = 2, and x = 3 is a region between the curve y = 9 - x^2 and the x-axis, from x = 2 to x = 3.
The outer radius of the ring at x is 3, and the inner radius is 9 - x^2. The volume of each ring is π(outer radius^2 - inner radius^2)dx.
Therefore, the volume of the solid can be calculated as follows:
V = ∫[2, 3] π(3^2 - (9 - x^2)^2)dx
V = π∫[2, 3] (9 - 9 + 2x^2 - x^4)dx
V = π∫[2, 3] (2x^2 - x^4)dx
V = π[(2/3)x^3 - (1/5)x^5] | from 2 to 3
V = π[((2/3)(3)^3 - (1/5)(3)^5) - ((2/3)(2)^3 - (1/5)(2)^5)]
V = π[(2/3)(27) - (1/5)(243) - (2/3)(8) + (1/5)(32)]
V = π[18 - 48.6 - 5.333 + 6.4]
V = π[-29.933]
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is approximately -94.2π cubic units.