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

Feb 26, 2017

Volume $= \frac{768 \pi}{7} \setminus \setminus \setminus u n i {t}^{3}$

#### Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness $\delta x$ between the $x$-axis and the curve at some particular $x$-coordinate it would have an area:

$\delta A \approx \text{width" xx "height} = y \delta x = f \left(x\right) \delta x$

If we then rotated this infinitesimally thin vertical line about $O y$ then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume $\delta V$ would be given by:

$\delta V \approx 2 \pi \times \text{radius" xx "thickness} = 2 \pi x \delta A = 2 \pi x f \left(x\right) \delta x$

If we add up all these infinitesimally thin cylinders then we would get the precise total volume $V$ given by:

$V = {\int}_{x = a}^{x = b} 2 \pi \setminus x \setminus f \left(x\right) \setminus \mathrm{dx}$

Similarly if we rotate about $O x$ instead of $O y$ then we get the volume as:

$V = {\int}_{y = a}^{y = b} 2 \pi \setminus y \setminus g \left(y\right) \setminus \mathrm{dy}$

So for this problem we have:

We need the point of intersection for the bounds of integration;

$x \setminus = 0 \implies y = 0$
${x}^{3} = 8 \implies x = 2$

And we are integrating wrt $y$ so we need $x = g \left(y\right)$ (ie ${f}^{-} 1 \left(x\right)$)

$y = {x}^{3} \implies x = {y}^{\frac{1}{3}}$

Then the required volume is given by:

$V = {\int}_{y = a}^{y = b} 2 \pi \setminus y \setminus g \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = 2 \pi {\int}_{0}^{8} \setminus y \cdot {y}^{\frac{1}{3}} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = 2 \pi {\int}_{0}^{8} \setminus {y}^{\frac{4}{3}} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = 2 \pi {\left[{y}^{\frac{7}{3}} / \left(\frac{7}{3}\right)\right]}_{0}^{8}$
$\setminus \setminus \setminus = \frac{6 \pi}{7} {\left[{y}^{\frac{7}{3}}\right]}_{0}^{8}$
$\setminus \setminus \setminus = \frac{6 \pi}{7} \left({8}^{\frac{7}{3}} - 0\right)$
$\setminus \setminus \setminus = \frac{6 \pi}{7} \cdot 128$
$\setminus \setminus \setminus = \frac{768 \pi}{7} \setminus \setminus \setminus u n i {t}^{3}$

Washer Method
We call also use the "washer" method which gives the volume of revolution about $O x$ (this is the non-shaded area under the curve) as:

$V = \pi \setminus {\int}_{x = a}^{x = b} \setminus {\left(f \left(x\right)\right)}^{2} \setminus \mathrm{dx}$

Which in this problem gives:

 V="(volume bounded by " y=8) - ("volume bounded by " y=x^3)

Volume bounded by $y = {x}^{3}$ is:

$V = \pi \setminus {\int}_{0}^{2} \setminus {\left({x}^{3}\right)}^{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\int}_{0}^{2} \setminus {x}^{6} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\left[{x}^{7} / 7\right]}_{0}^{2}$
$\setminus \setminus \setminus = \pi \setminus \left(\frac{128}{7} - 0\right)$
$\setminus \setminus \setminus = \frac{128 \pi}{7}$

The required volume of revolution is then the volume of a cylinder ($\pi {r}^{2} h$) of height $2$, radius $8$ minus this result:

$V = \left(\pi\right) \left(64\right) \left(2\right) - \frac{128 \pi}{7}$
$\setminus \setminus \setminus = 128 \pi - \frac{128 \pi}{7}$
$\setminus \setminus \setminus = \frac{768 \pi}{7}$, as before

Or if you prefer, we can perform the calculation in a single integral, as:

 V="(volume bounded by " y=8) - ("volume bounded by " y=x^3)
$\setminus \setminus \setminus = \pi \setminus {\int}_{0}^{2} \setminus {\left(8\right)}^{2} \setminus \mathrm{dx} - \pi \setminus {\int}_{0}^{2} \setminus {\left({x}^{3}\right)}^{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\int}_{0}^{2} \setminus {\left(8\right)}^{2} - {\left({x}^{3}\right)}^{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\int}_{0}^{2} \setminus 64 - {x}^{6} \mathrm{dx}$
$\setminus \setminus \setminus = \pi \setminus {\left[64 x - {x}^{7} / 7\right]}_{0}^{2}$
$\setminus \setminus \setminus = \pi \setminus \left(128 - \frac{128}{7} - 0\right)$
$\setminus \setminus \setminus = \frac{768 \pi}{7}$, as before