# Find the volume of the solid by using cylindrical shells?

## May 28, 2018

$\textcolor{b l u e}{V = 2 \pi {\int}_{0}^{1} \left(y\right) \left(\sqrt{y} - {y}^{3}\right) \cdot \mathrm{dy} = \frac{2 \cdot \pi}{5} {\left(u n i t e\right)}^{3}}$

#### Explanation:

the volume by cylindrical shell method when the curve rotating about the $\text{x-axis}$ is given by:

$\textcolor{red}{V = 2 \pi {\int}_{a}^{b} \left(y\right) \left({x}_{2} - {x}_{1}\right) \cdot \mathrm{dy}}$

${x}_{2} = \pm \sqrt{y}$
but the area enclosed by the curve lies in the first quadrant.

so ${x}_{2} = \sqrt{y}$

${x}_{1} = {y}^{3}$

lets find the interval of integral.

$\sqrt{y} = {y}^{3}$ square both sides

${y}^{6} = y \Rightarrow {y}^{6} - y = 0$

$y \cdot \left({y}^{5} - 1\right) = 0$

$y = 0 \mathmr{and} y = 1$

the interval of the integral $y \in \left[0 , 1\right]$

now let set up the integral:

$V = 2 \pi {\int}_{0}^{1} \left(y\right) \left(\sqrt{y} - {y}^{3}\right) \cdot \mathrm{dy}$

$V = 2 \pi {\int}_{0}^{1} \left({y}^{\frac{3}{2}} - {y}^{4}\right) \cdot \mathrm{dy} = {\left[- \frac{2 \cdot \pi \cdot \left({y}^{5} - 2 \cdot {y}^{\frac{5}{2}}\right)}{5}\right]}_{0}^{1}$

$= \frac{2 \cdot \pi}{5} {\left(u n i t e\right)}^{3}$

show below the region revolving (shaded region) : show the link below that will help you to understand how to find the volume by cylindrical shell method:
cylindrical shell method