# 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=x, 0<=x<=1 rotated about the x-axis?

Volume $V = \frac{\pi}{3} \text{ }$cubic units

#### Explanation:

Using the cylindrical shell method. The differential is

$\mathrm{dV} = 2 \pi \cdot r \cdot h \cdot \mathrm{dr}$

$\mathrm{dV} = 2 \cdot \pi \cdot y \cdot \left(1 - x\right) \cdot \mathrm{dy}$

but $x = y$, therefore

$\mathrm{dV} = 2 \cdot \pi \cdot y \cdot \left(1 - y\right) \cdot \mathrm{dy}$

our limits for x are $0 \rightarrow 1$
our limits for y are $0 \rightarrow 1$

We solve the volume by integrating both sides with limits $y = 0$ to $y = 1$

$\int \mathrm{dV} = 2 \cdot \pi \cdot \int y \cdot \left(1 - y\right) \cdot \mathrm{dy}$

$V = 2 \cdot \pi \cdot {\int}_{0}^{1} \left(y - {y}^{2}\right) \cdot \mathrm{dy}$

$V = 2 \cdot \pi \cdot {\left[{y}^{2} / 2 - {y}^{3} / 3\right]}_{0}^{1}$

$V = 2 \cdot \pi \cdot \left[{1}^{2} / 2 - {1}^{3} / 3 - \left({0}^{2} / 2 - {0}^{3} / 3\right)\right]$

$V = 2 \cdot \pi \cdot \left[\frac{1}{2} - \frac{1}{3} - 0\right]$

$V = 2 \cdot \pi \cdot \frac{1}{6}$

$V = \frac{\pi}{3} \text{ }$cubic units

God bless....I hope the explanation is useful.