# How do you find the volume of the solid obtained by rotating the region bounded by the curves y=x, x=0, and y=(x^2)-6 rotated around the y=3?

Aug 2, 2015

$\pi {\int}_{0}^{3} {\left(3 - \left({x}^{2} - 6\right)\right)}^{2} - {\left(3 - x\right)}^{2} \mathrm{dx} = \frac{603 \pi}{5}$

#### Explanation:

The grey region is what we will be rotating around the horizontal line $y = 3$.

The outer radius is $3 - \left({x}^{2} - 6\right)$

The inner radius is $3 - x$

Using the method of washers

$\pi {\int}_{0}^{3} {\left(3 - \left({x}^{2} - 6\right)\right)}^{2} - {\left(3 - {x}^{2}\right)}^{2} \mathrm{dx}$

$\pi {\int}_{0}^{3} {\left(9 - {x}^{2}\right)}^{2} - {\left(3 - x\right)}^{2} \mathrm{dx}$

$\pi {\int}_{0}^{3} 81 - 18 {x}^{2} + {x}^{4} - \left(9 - 6 x + {x}^{2}\right) \mathrm{dx}$

$\pi {\int}_{0}^{3} 81 - 18 {x}^{2} + {x}^{4} - 9 + 6 x - {x}^{2} \mathrm{dx}$

$\pi {\int}_{0}^{3} 72 - 19 {x}^{2} + {x}^{4} + 6 x \mathrm{dx}$

Integrating

$\pi \left[72 x - \frac{19}{3} {x}^{3} + {x}^{5} / 5 + 3 {x}^{2}\right]$

$\pi \left[72 \left(3\right) - \frac{19}{3} {\left(3\right)}^{3} + {3}^{5} / 5 + 3 {\left(3\right)}^{2}\right]$

$\pi \left[216 - 171 + \frac{243}{5} + 27\right]$

$\pi \left[72 + \frac{243}{5}\right]$

$\pi \left[\frac{360}{5} + \frac{243}{5}\right] = \frac{603 \pi}{5}$