Question

How do you find the volume of the solid `y=-x+1` revolved about the x-axis?

Answer 1

If we restrict `(0` `<` `x<1)`, we will obtain a cone by revolving

`y=-x+1`

about the x-axis, and its volume will be `pi/3`

Explanation:

If you were to revolve the entire function `y=-x+1` about the x-axis, you would get a solid of infinite volume.

Let's instead choose to consider only the function on the interval

`(0` `<` `x<1)`

(the solid of revolution will be a cone)

Picture the integral by chopping the cone into a series of infinitely thin vertical slices.

Each slice is then a cylinder with its radius equal to `y` and thickness equal to `dx`

So the volume of one slice is

`dV=piy^2dx`

and the volume of the entire solid is the sum (integral) of all slices

`int` `dV=int` `piy^2dx`

`V=piint` `y^2dx`

In this problem, `y=-x+1`, so we write

`V=piint` `(-x+1)^2dx`

Now let's find the definite integral from 0 to 1.

`piint_0^1(-x+1)^2dx`

`rArrpiint_0^1(x^2-2x+1)dx`

`rArrpi[x^3/3-x^2+x]_0^1`

`rArrpi[(1/3-1+1)-(0-0+0)]=pi/3`

So the volume of the cone obtained by revolving

`y=-x+1` for `(0` `<` `x<1)`

about the x-axis is

`pi/3`

Note: This agrees with the formula for the volume of a cone

`V=pir^2h/3`

Our cone has radius of 1 and a height of 1, so

`V=pi(1^2)1/3=pi/3`