Question

How do you find the volume of a solid that is enclosed by `y=1/x`, x=1, x=3, y=0 revolved about the y axis?

Answer 1

`4pi`

Explanation:

The shape that we are integrating should look something like a big top for a circus. We may have to split up the top and the bottom halves at `y= 1/3`.

We can use an integration over `y` here.

`V = int_0^1 A(y)dy`

To find the area of a slice we look at points below `y=1/3` and points above.

Below
The outer radius is constant at 3. There is an inner radius cut out from the `x=1` boundary, i.e. the area is the annulus between `r = 1` and `r = 3`. We know this area:
`A = pi * 3^2 - pi * 1^2 = 8pi`

Above
The bound is a little more complicated above `y = 1/3`.
In this region, the inner radius is still a constant (hence a `-pi` term we will add) but the outer radius goes like `1/y`, i.e.
`A(y) = pi x^2 - pi = pi/y^2 - pi`

We can integrate this whole thing piecewise:
`V = int_0^1 A(y) dy = int_0^(1/3) A(y) dy + int_(1/3)^1 A(y)dy`
`int_0^(1/3) A(y) dy = 1/3 * 8pi = (8pi)/3`

`int_(1/3)^1 A(y) dy = -piy^-1 - pi y = -pi(1 - 3) - pi(1 - 1/3) = 2pi - 2/3 pi = (4pi)/3`

Therefore, the total volume is
`V = (8pi)/3 + (4pi)/3 = 4pi`