Question

How do you find the volume of the solid obtained by rotating the region bounded by: `y=sqrt(x-1)`, y=0, x=5 rotated about y=7?

Answer 1

Please see below.

Explanation:

I would use shells. Here is a picture of the region with a slice taken parallel to the axis of rotation.
The slice is taken at a value of `y`, so we need to rewrite the curve `y=sqrt(x-1)` as `x = y^2+1`

The thickness of the slice and the shell is `dy`
The radius is `r = 7-y`
The height is `h = 5-(y^2+1) = 4-y^2`

The shell has volume `2pirhxx"thickness" = 2pi(7-y)(4-y^2)dy`

`y` varies from `0` to `2` (That is the `y` value on the curve at `x=5`.)

The resulting solid has volume:

`V = 2piint_0^2(7-y)(4-y^2)dy`

`= 2piint_0^2(28-4y-7y^2+y^3) dy`

`= 2pi[28y-2y^2 - 7/3y^3+y^4/4]_0^2`

`= (200pi)/3`

If you prefer washers

Then the thickness is `dx`,
the greater radius is `7`, and
the lesser radius is `sqrt(x-1)`.

`x` varies from `1` to `5`, so the volume of the solid is

`V = piint_1^5(7^2-(7-sqrt(x-1))^2)dx`

`= pi int_1^5(14sqrt(x-1) - (x-1))dx`

To integrate I would substitute `u = x-1` to get
`piint_0^4(14u^(1/2)-u)du`

Answer 2

See below.

Explanation:

Looking at the graph, we can see that the area A rotated around `y=7` is the volume we seek.

We can find this volume by finding the volume of the area (A+B). This is easy and doesn’t require integration. The radius of the cylinder that will be formed has a radius of 7, this is just the height from the x axis. We square this and multiply by `pi` x the length of the interval `[ 1,5]`

Length of interval is `5-1=4`

`:.`

Volume (A + B ):

`V=7^2pi*196pi`

From this we need to subtract the volume of B. From the graph we can see that the radius is `7-sqrt(x-1)`. This will have to be found by integration in the normal way.

Volume of B:

`(7-sqrt(x-1))^2=48-14sqrt(x-1)+x`

`V=pi*int_(1)^(5)(48-14sqrt(x-1)+x) dx=48x+1/2x^2-28/3(x-1)^(3/2)`

Area first:

`[48x+1/2x^2-28/3(x-1)^(3/2)]^(5)-[48x+1/2x^2-28/3(x-1)^(3/2)]_(1)`
..................................................................................................................................

`[48(5)+1/2(5)^2-28/3((5)-1)^(3/2)]^(5)-[48(1)+1/2(1)^2-28/3((1)-1)^(3/2)]_(1)`

`pi*[505/2-112/3sqrt(4)]-[97/2]=388/3pi`

Volume= `388/3pi`

Volume of A:

`196pi-388/3pi=200/3pi=66.667pi`

Volume of revolution: