# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by 27y=x^3, y=0 , x=6 revolved about the y=8?

Nov 19, 2016

The volume is $= \frac{960 \pi}{7}$

#### Explanation:

The volume of a small shell

$\mathrm{dV} = \pi \left({8}^{2} - {\left(8 - y\right)}^{2}\right) \mathrm{dx}$

As, $y = {x}^{3} / 27$

$\mathrm{dV} = \pi \left(64 - {\left(8 - {x}^{3} / 27\right)}^{2}\right) \mathrm{dx}$

$\mathrm{dV} = \pi \left(64 - 64 + 16 {x}^{3} / 27 - {x}^{6} / 729\right) \mathrm{dx}$

$V = \pi {\int}_{0}^{6} \left(16 {x}^{3} / 27 - {x}^{6} / 729\right) \mathrm{dx}$

$= \pi {\left[16 {x}^{4} / 4 \cdot \frac{1}{27} - {x}^{7} / 7 \cdot \frac{1}{729}\right]}_{0}^{6}$

$= \pi \left(4 \cdot {6}^{6} / 27 - {6}^{7} / \left(7 \cdot 729\right) - 0\right)$

$= \pi \left(\frac{5184}{27} - \frac{384}{7}\right)$

$V = \frac{960 \pi}{7}$

graph{(y-x^3/27)(y-8)=0 [-19.15, 16.9, -4.32, 13.7]}