# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 15e−x^2, y = 0, x = 0, x = 1 revolved about the y-axis?

Dec 8, 2016

$V = 2 \pi \left[\frac{15}{2e-1} / 4\right]$

#### 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:

$\setminus \setminus \setminus \setminus \setminus V = {\int}_{0}^{1} 2 \pi x \left(15 e - {x}^{2}\right) \mathrm{dx}$
$\therefore V = 2 \pi {\int}_{0}^{1} \left(15 e x - {x}^{3}\right) \mathrm{dx}$
$\therefore V = 2 \pi {\left[15 e {x}^{2} / 2 - {x}^{4} / 4\right]}_{0}^{1}$
$\therefore V = 2 \pi \left[\frac{15}{2e-1} / 4\right]$