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
Use the ring method
To find the volume using the ring method, we need to integrate the cross-sectional area of each ring formed by revolving the region bounded by the equations about the y-axis.
The outer radius of each ring is the distance from the y-axis to the curve Y = 9 – x^2, which is x = sqrt(9 - y).
The inner radius of each ring is 2, as the region is bounded by x = 2.
The cross-sectional area of each ring is A = π(outer radius)^2 - π(inner radius)^2
= π((sqrt(9-y))^2 - 2^2)
= π(9-y - 4)
= π(5-y)
Thus, the volume V is given by the integral of A with respect to y:
V = ∫[0, 5] π(5-y) dy
V = π[5y - (y^2)/2] | from 0 to 5
V = π[25/2 - 25/2]
V = 12.5π
Therefore, the volume of the solid generated by revolving the region bounded by the given equations about the y-axis is 12.5π cubic units.