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 using the ring method, we first need to find the equation for the area of the circular cross-section created by revolving the region around the y-axis.
The outer radius (R) of the ring is the distance from the y-axis to the curve y = 9 - x^2.
The inner radius (r) of the ring is the distance from the y-axis to the curve y = 0 (the x-axis).
The outer radius, R, is given by:
R = 3 - x
The inner radius, r, is just the distance from the y-axis to the x-axis, which is 0.
The equation for the area of the ring is:
A = π(R^2 - r^2)
Substitute the expressions for R and r:
A = π((3 - x)^2 - 0^2)
A = π(9 - 6x + x^2)
Now integrate this area from x = 2 to x = 3 to find the volume:
V = ∫[2,3] π(9 - 6x + x^2) dx
V = π∫[2,3] (9 - 6x + x^2) dx
V = π(9x - 3x^2 + x^3) ∣ [2,3]
V = π((27 - 9 + 27) - (18 - 12 + 8))
V = π(45 - 14)
V = 31π
Therefore, the volume of the solid generated by revolving the region about the y-axis is 31π cubic units.