# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=5x^2, y=5x revolved about the x-axis?

Aug 22, 2017

#### Explanation:

In order to use shells, we need to take our representative slice parallel to the axis of rotation. So we will take a horizontal slice at a value of $y$ and of thickness $\mathrm{dy}$

Below is a partial picture. The curves intersect at $\left(0 , 0\right)$ and 1,5).

The horizontal slice is shown in black and the radius of revolution is the dotted black line. Its length is $y$.

The volume of a representative shell is $2 \pi r h \cdot \text{thickness}$

In this problem, $\text{thickness} = \mathrm{dy}$ and $r = y$

$h$ is the greater $x$ value - the lesser $x$ value.
The greater $x$ value is the one on the right.

We need to rewrite the equation $y = 5 {x}^{2}$ so that we have $x$ as a function of $y$. ($y$ is the variable we'll be using for the integration.)

$x = \sqrt{\frac{y}{5}}$ (We ignore the negative square root because $x$ it positive in this set-up.)

The lesser $x$ is on the left where $x = \frac{y}{5}$

So the reprentative shell has volume

$2 \pi y \left(\sqrt{\frac{y}{5}} - \frac{y}{5}\right) \mathrm{dy}$

$y$ varies from $0$ to $5$, so the solid has volume

$V = {\int}_{0}^{5} \pi y \left(\sqrt{\frac{y}{5}} - \frac{y}{5}\right) \mathrm{dy}$

Details omitted

$= \frac{5 \pi}{3}$