# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=6sin(5x^2), between 0 and sqrt(pi/5) revolved about Ox?

Dec 10, 2016

$2.4 \pi$ ($7.54$ 2 dp)

#### 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:

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

So for this problem we have:

$\setminus \setminus \setminus \setminus \setminus V = {\int}_{0}^{\sqrt{\frac{\pi}{5}}} 2 \pi x \left(6 \sin \left(5 {x}^{2}\right)\right) \mathrm{dx}$
$\therefore V = 12 \pi {\int}_{0}^{\sqrt{\frac{\pi}{5}}} x \sin \left(5 {x}^{2}\right) \mathrm{dx}$
$\therefore V = 12 \pi {\left[- \cos \frac{5 {x}^{2}}{10}\right]}_{0}^{\sqrt{\frac{\pi}{5}}}$
$\therefore V = \frac{- 12 \pi}{10} {\left[\cos \left(5 {x}^{2}\right)\right]}_{0}^{\sqrt{\frac{\pi}{5}}}$
$\therefore V = \frac{- 12 \pi}{10} \left(\cos \pi - \cos 0\right)$
$\therefore V = \frac{- 12 \pi}{10} \left(- 1 - 1\right)$
$\therefore V = \frac{24 \pi}{10}$
$\therefore V = 2.4 \pi$ ($\approx 7.5398 \ldots$)