Question

Let R be the region bounded by y=x^2, x=1, and y=0. use the shell method to find the volume of the solid generated when R is revolved about the line x= 2

Answer 1

`V = (5\pi)/(6)` or 2.62

Explanation:

Since our region (on the left) is bounded by x = 1 and x = 0 (where
`y = x^2 = 0`, if we take center at x = 2 then our radius will range from 1 to 2 (x=1 to x = 0). We can use the following integration to calculate the volume using shell method

`V = \int\limits^2_1 {2\pi r h} \, dr`

where r = 2 - x so x = 2 - r and
`h = y = x^2 = (2-r)^2` for
`1 \leq r \leq 2`

`V = \int\limits^2_1 {2\pi r(2-r)^2} \, dr`

`V = \int\limits^2_1 {2\pi r(4 - 4r + r^2)} \, dr`

`V = 2\pi \int\limits^2_1 {r^3 - 4r^2 +4r} \, dr`

`V = 2\pi\left[(r^4)/(4) - (4r^3)/(3) + 2r^2\right]^2_1`

`V = 2\pi\left[\left((2^4)/(4) - (4*2^3)/(3) + 2*2^2\right) - \left((1^4)/(4) - (4*1^3)/(3) + 2*1^2\right)\right]`

`V = 2\pi(4 - 32/3 +8 - 1/4 + 4/3 - 2)`

`V = \pi(20 - 56/3 - 1/2)`

`V = \pi(120 - 112 - 3)/(6)`

`V = (5\pi)/(6)` or 2.62