# How do you find the volume bounded by y=1 and y=x^2 revolved about the y=1?

Jul 16, 2017

$\frac{16}{15} \pi$

#### Explanation:

$R = 1 - {x}^{2}$

R = 0 => x = ± 1

$V = {\int}_{-} {1}^{1} \pi {R}^{2} \textrm{d} x$

$V = \pi {\int}_{-} {1}^{1} {\left(1 - {x}^{2}\right)}^{2} \textrm{d} x$

$V = \pi {\int}_{-} {1}^{1} \left({x}^{4} - 2 {x}^{2} + 1\right) \textrm{d} x$

$V = \pi {\left[{x}^{5} / 5 - 2 {x}^{3} / 3 + x\right]}_{-} {1}^{1}$

$V = \pi \left[\frac{1}{5} - \frac{2}{3} + 1 - \left(- \frac{1}{5} + \frac{2}{3} - 1\right)\right]$

$V = \pi \left[\frac{2}{5} - \frac{4}{3} + 2\right] = \pi \frac{6 - 20 + 30}{15}$