# 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(x), y = 0, y = 12 - x rotated about the x axis?

Jan 20, 2017

Volume = $\frac{99 \pi}{2}$

#### 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;

$12 - x = \sqrt{x} \implies {\left(12 - x\right)}^{2} = x$
$\therefore 144 - 24 x + {x}^{2} = x$
$\therefore {x}^{2} - 25 x + 144 = 0$
$\therefore \left(x - 9\right) \left(x - 16\right) = 0 \implies x = 9 , 16$

And our solution is $x = 9$, (as $x = 16$ is the solution for $- \sqrt{x}$)

$x = 9 \implies y = 12 - x = 3$

So the bounding curves $y = \sqrt{x}$ and $y = 12 - x$ meet at $\left(9 , 3\right)$

And we also need $x = g \left(y\right)$, rather than $y = f \left(x\right)$

$y = 12 - x \implies x = 12 - y$
$y = \sqrt{x} \setminus \setminus \setminus \setminus \setminus \implies x = {y}^{2}$

Then the required volume is given by:

$V = {\int}_{y = a}^{y = b} 2 \pi \setminus y \setminus g \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = 2 \pi {\int}_{y = 0}^{y = 3} \setminus y \left\{\left(12 - y\right) - \left({y}^{2}\right)\right\} \setminus \mathrm{dy}$
$\setminus \setminus \setminus = 2 \pi {\int}_{y = 0}^{y = 3} \setminus y \left(12 - y - {y}^{2}\right) \setminus \mathrm{dy}$
 \ \ \= 2pi int_(y=0)^(y=3) \ 12y-y^2 - y^3) \ dy
$\setminus \setminus \setminus = 2 \pi {\left[6 {y}^{2} - \frac{1}{3} {y}^{3} - \frac{1}{4} {y}^{4}\right]}_{0}^{3}$
$\setminus \setminus \setminus = 2 \pi \left(54 - 9 - \frac{81}{4}\right)$
$\setminus \setminus \setminus = \frac{99 \pi}{2}$