Question

How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by `y=x^2`, x = 2, x = 7, y = 0 revolved about the x=8?

Answer 1

`3565pi/6`

Explanation:

shell method: volume = `int_a^b2pirhdx`

In this problem, `a=2` and `b=7` (from the boundaries x=2 and x=7).

r is the distance from a certain x value in the interval [2,7] to the axis of rotation, which is x=8. Since r must be a positive distance, it is `8-x`.

h is the height of each shell (parallel to axis of rotation). in this case it is the distance between the upper and lower bounds of the region, or `x^2-0=x^2`

plugging in: volume = `int_2^(7)2pi(8-x)(x^2)dx`
`=2piint_2^7(-x^3+8x^2)dx`
`=2pi * [F(7) - F(2)]`, where `F(x) = -1/4x^4+8/3x^3` (fundamental theorem of calculus).
`=2pi * [(-1/4(7)^4+8/3(7)^3) - (-1/4(2)^4+8/3(2)^3)]`

simplify: `=2pi * [-2401/4+2744/3+16/4-64/3]`
`=2pi * [-2385/4+2680/3]`
`=2pi * [3565/12]`
`= 3565pi/6`