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

Mar 29, 2018

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 $\frac{\pi}{3}$

#### 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 $\mathrm{dx}$

So the volume of one slice is

$\mathrm{dV} = \pi {y}^{2} \mathrm{dx}$

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

$\int$ $\mathrm{dV} = \int$ $\pi {y}^{2} \mathrm{dx}$

$V = \pi \int$ ${y}^{2} \mathrm{dx}$

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

$V = \pi \int$ ${\left(- x + 1\right)}^{2} \mathrm{dx}$

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

$\pi {\int}_{0}^{1} {\left(- x + 1\right)}^{2} \mathrm{dx}$

$\Rightarrow \pi {\int}_{0}^{1} \left({x}^{2} - 2 x + 1\right) \mathrm{dx}$

$\Rightarrow \pi {\left[{x}^{3} / 3 - {x}^{2} + x\right]}_{0}^{1}$

$\Rightarrow \pi \left[\left(\frac{1}{3} - 1 + 1\right) - \left(0 - 0 + 0\right)\right] = \frac{\pi}{3}$

So the volume of the cone obtained by revolving

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

$\frac{\pi}{3}$
$V = \pi {r}^{2} \frac{h}{3}$
$V = \pi \left({1}^{2}\right) \frac{1}{3} = \frac{\pi}{3}$