# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=sqrt(16-x^2) and the x axis rotated about the x axis?

Feb 12, 2017

Volume $= \frac{256 \pi}{3} \setminus {\text{unit}}^{3}$

#### 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:
graph{(y-sqrt(16-x^2))=0 [-8, 8, -2, 5]}

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

$\sqrt{16 - {x}^{2}} = 0 \implies x = \pm 4$, and the range of $y$ is $y \in \left[- 4 , 4\right]$

By symmetry we can restrict $y \in \left[0 , 4\right]$ and double the result.

$y = \sqrt{16 - {x}^{2}} \implies {y}^{2} = 16 - {x}^{2} \implies x = \sqrt{16 - {y}^{2}}$

The Volume of Revolution about $O x$ is given by:

$\frac{1}{2} V = {\int}_{y = a}^{y = b} 2 \pi \setminus y \setminus g \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus \setminus \setminus \setminus = {\int}_{y = 0}^{y = 4} 2 \pi \setminus y \setminus \sqrt{16 - {y}^{2}} \setminus \mathrm{dy}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 2 \pi \setminus {\int}_{0}^{4} y \setminus \sqrt{16 - {y}^{2}} \setminus \mathrm{dy}$

We can use a substitution to perform this integral;

Let $u = 16 - {y}^{2} \implies \frac{\mathrm{du}}{\mathrm{dy}} = - 2 y$
When $\left\{\begin{matrix}y = - 4 \\ y = 4\end{matrix}\right. \implies \left\{\begin{matrix}u = 16 \\ u = 0\end{matrix}\right.$

Substituting into the integral we get:

$\frac{1}{2} V = 2 \pi \setminus {\int}_{16}^{0} \setminus \sqrt{u} \left(- \frac{1}{2}\right) \setminus \mathrm{du}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \pi \setminus {\int}_{0}^{16} \setminus \sqrt{u} \setminus \mathrm{du}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{16}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{2}{3} \pi \setminus {\left[{u}^{\frac{3}{2}}\right]}_{0}^{16}$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{2}{3} \pi \setminus \left({16}^{\frac{3}{2}} - 0\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{128 \pi}{3}$
$\therefore V = \frac{256 \pi}{3}$

Note:
The volume is that of a sphere of radius $4$, so we can validate the volume as $V = \frac{4}{3} \pi {r}^{3} = \frac{256 \pi}{3}$