Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by `125 y = x^3` , y = 8 , x = 0 revolved about the x-axis?

Answer 1

See below

Explanation:

Here is the region:

A representative slice taken perpendicular to the axis of rotation has volume

`int \ pi(8^2 - (x^3/5^3)^2) dx` using `VOR = int pi y^2 dx`

`x` varies from `0` to `10`, so the solid will have volume:

`piint_0^10 (64 - x^6/5^6) dx = pi[64x-x^7/(7(5^6))]_0^10`

`= pi[640 - 10^7/(7(5^6))]`

`= pi[640 - (10 * 2^6)/7]`

`= pi[640 - 640/7]`

`= pi[(7*640)/7 - 640/7]`

`= pi (6 * 640)/7`

`= (3840pi)/7`

We can also use cylindrical shells to get the same answer:

`V =int_(y=a)^(y=b)2pi \ y \ g(y) \ dy`
`\ \ =2pi int_0^8 \ y \ root(3)(125y) \ dy`
`\ \ =2pi int_0^8 5 \ y \ y^(1/3) \ dy`
`\ \ =10pi int_0^8 y^(4/3) \ dy`
`\ \ =10pi [y^(7/3)/(7/3)]_0^8`
`\ \ =(30pi)/7 [y^(7/3)]_0^8`
`\ \ =(30pi)/7 (128-0)`
`\ \ =(3840pi)/7`