find the volume revolved around y-axis
y=9-x^2,y=0,x=2,x=3
To find the volume revolved around the y-axis of the region between the curves y = 9 - x^2, y = 0, x = 2, and x = 3, we will use the method of cylindrical shells.
First, we need to find the limits of integration. It is evident from the given curves that the region of interest lies between x = 2 and x = 3.
The formula for the volume revolved around the y-axis using cylindrical shells is:
V = 2π ∫[a,b] x*f(x)dx
Where f(x) is the height of the shell and x is the radius of the shell.
The height f(x) = 9 - x^2 and the radius x. Therefore the integral is:
V = 2π ∫[2,3] x(9 - x^2)dx
V = 2π ∫[2,3] (9x - x^3)dx
Now integrate using the limits of integration x = 2 and x = 3:
V = 2π [ (9/2)x^2 - (1/4)x^4 ] from 2 to 3
V = 2π [ (9/2)(3^2) - (1/4)(3^4) ] - 2π [ (9/2)(2^2) - (1/4)(2^4) ]
V = 2π [ (81/2) - 27/4 ] - 2π [ (36/2) - 4 ]
V = 2π [ 81/2 - 27/4 - 18 + 4 ]
V = 2π [ 81/2 - 27/4 - 14 ]
V = 2π [ 81/2 - 27/2 ]
V = 2π [ 54/2 ]
V = 54π
Therefore, the volume of the solid obtained by revolving the region around the y-axis is 54π cubic units.