# How do you find the volume when the region bounded by y = x+3, y = 0, x = -3 and x = 3 is revolved around the x-axis?

Apr 14, 2017

$72 \pi$

#### Explanation:

Use the disk method: $A = \pi \int {r}^{2}$

$r$ is $y = x + 3$ because it's bound by $y = 0$

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

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

$\pi \left(\left({3}^{3} / 3 + 3 \left({3}^{2}\right) + 9 \left(3\right)\right) - \left({\left(- 3\right)}^{3} / 3 + 3 {\left(- 3\right)}^{2} + 9 \left(- 3\right)\right)\right) = 72 \pi$