# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=f(x)=3x-x^2 and x axis revolved about the x=-1?

Nov 11, 2017

$V = \frac{45 \pi}{2} u n i t {s}^{3}$

#### Explanation:

Imagine a cylindrical shell as a rectangular prism with width $f \left(x\right)$, length $2 \pi r$, and thickness $\delta x$. The reason for the length being $2 \pi r$ is that if we unravel a cylindrical shell into a rectangular prism, the length corresponds to the circumference of the circular cross-section of the original cylinder.

The volume of the cylindrical shell is width x length x thickness (height)

Now width $= f \left(x\right) = 3 x - {x}^{2}$

$\therefore$ Volume of shell$= 2 \pi r \left(3 x - {x}^{2}\right) \delta x$

If we sketch the parabola $y = 3 x - {x}^{2}$ we can see that the region bound by this parabola and the x-axis resides in the domain $0 \le x \le 3$

Also, the radius of the cylindrical shell is considered to be the distance from the axis of revolution $x = - 1$ and the edge of the shell (i.e. at a position $x$ within the domain $0 \le x \le 3$)

Hence, r$= 1 + x$

$\therefore$Volume of shell$= 2 \pi \left(1 + x\right) \left(3 x - {x}^{2}\right) \delta x$

i.e. the change in volume, $\delta v = 2 \pi \left(1 + x\right) \left(3 x - {x}^{2}\right) \delta x$

To find the volume, we limit the thickness of the shells, and find their summation within the domain $0 \le x \le 3$

$V = {\lim}_{\delta x \to 0} {\sum}_{x = 0}^{3} \left(2 \pi \left(1 + x\right) \left(3 x - {x}^{2}\right) \delta x\right)$

$= 2 \pi {\int}_{0}^{3} \left(1 + x\right) \left(3 x - {x}^{2}\right) \mathrm{dx}$

$= 2 \pi {\int}_{0}^{3} \left(3 x + 2 {x}^{2} - {x}^{3}\right) \mathrm{dx}$

$= 2 \pi {\left[\frac{3}{2} {x}^{2} + \frac{2}{3} {x}^{3} - \frac{1}{4} {x}^{4}\right]}_{0}^{3}$

$= 2 \pi \left[\frac{45}{4}\right]$

$= \frac{45 \pi}{2} u n i t {s}^{3}$