# How do you find the volume of the region enclosed by the curves y=x, y=-x, and x=1 rotated about y=1?

Aug 27, 2015

The volume is $2 \pi$

#### Explanation:

Using the method of washers

Let the outer radius be $1 - \left(- x\right) = 1 + x$

Let the inner radius be $1 - x$

The integral for the volume is

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

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

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

$\pi {\int}_{0}^{1} \left(4 x\right) \mathrm{dx}$

Integrating we get

$2 \pi {x}^{2}$

Evaluating from $0$ to $1$

$2 \pi {\left(1\right)}^{2} - 0 = 2 \pi$