# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by 125 y = x^3 , y = 8 , x = 0 revolved about the x-axis?

Jan 23, 2017

See below

#### Explanation:

Here is the region: A representative slice taken perpendicular to the axis of rotation has volume

$\int \setminus \pi \left({8}^{2} - {\left({x}^{3} / {5}^{3}\right)}^{2}\right) \mathrm{dx}$ using $V O R = \int \pi {y}^{2} \mathrm{dx}$

$x$ varies from $0$ to $10$, so the solid will have volume:

$\pi {\int}_{0}^{10} \left(64 - {x}^{6} / {5}^{6}\right) \mathrm{dx} = \pi {\left[64 x - {x}^{7} / \left(7 \left({5}^{6}\right)\right)\right]}_{0}^{10}$

$= \pi \left[640 - {10}^{7} / \left(7 \left({5}^{6}\right)\right)\right]$

$= \pi \left[640 - \frac{10 \cdot {2}^{6}}{7}\right]$

$= \pi \left[640 - \frac{640}{7}\right]$

$= \pi \left[\frac{7 \cdot 640}{7} - \frac{640}{7}\right]$

$= \pi \frac{6 \cdot 640}{7}$

$= \frac{3840 \pi}{7}$

We can also use cylindrical shells to get the same answer:

$V = {\int}_{y = a}^{y = b} 2 \pi \setminus y \setminus g \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus = 2 \pi {\int}_{0}^{8} \setminus y \setminus \sqrt{125 y} \setminus \mathrm{dy}$
$\setminus \setminus = 2 \pi {\int}_{0}^{8} 5 \setminus y \setminus {y}^{\frac{1}{3}} \setminus \mathrm{dy}$
$\setminus \setminus = 10 \pi {\int}_{0}^{8} {y}^{\frac{4}{3}} \setminus \mathrm{dy}$
$\setminus \setminus = 10 \pi {\left[{y}^{\frac{7}{3}} / \left(\frac{7}{3}\right)\right]}_{0}^{8}$
$\setminus \setminus = \frac{30 \pi}{7} {\left[{y}^{\frac{7}{3}}\right]}_{0}^{8}$
$\setminus \setminus = \frac{30 \pi}{7} \left(128 - 0\right)$
$\setminus \setminus = \frac{3840 \pi}{7}$