Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis.
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, we can use the method of cylindrical shells.
The region bounded by the graphs of the equations y = 9 - x^2, y = 0, x = 2, and x = 3 is a shape between the x-axis and the curve y = 9 - x^2 from x = 2 to x = 3.
First, find the limits of integration by setting the two equations equal to each other:
9 - x^2 = 0
x^2 = 9
x = ±3
The limits of integration are from x = 2 to x = 3.
The radius of the shells is x, and the height of the shells is the difference between the two functions y = 9 - x^2 and y = 0.
Volume = ∫[2,3] 2πx(9 - x^2) dx
= 2π ∫[2,3] (9x - x^3) dx
= 2π [4.5x^2 - 0.25x^4] | from 2 to 3
= 2π [(4.5(3)^2 - 0.25(3)^4) - (4.5(2)^2 - 0.25(2)^4)]
= 2π [(4.5(9) - 0.25(81)) - (4.5(4) - 0.25(16))]
= 2π [40.5 - 20.25 - 18 + 4]
= 2π [6.25]
= 12.5π
Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis is 12.5π cubic units.