# How do you find the volume bounded by x^2=4y and the line x=2y revolved about the x-axis?

Jun 28, 2016

$= \frac{4 \pi}{15}$

#### Explanation:

the small element width dx shown will have volume once revolved around the x axis of

$\mathrm{dV} = \left(\pi {y}_{2}^{2} - \pi {y}_{1}^{2}\right) \mathrm{dx}$

where $x = 2 {y}_{2}$ and ${x}^{2} = 4 {y}_{1}$

$\setminus \implies V = \pi {\int}_{0}^{2} \setminus {y}_{2}^{2} - {y}_{1}^{2} \setminus \mathrm{dx}$

So $V = \pi {\int}_{0}^{2} \setminus {x}^{2} / 4 - {x}^{4} / 16 \setminus \mathrm{dx}$

$= \frac{\pi}{16} {\int}_{0}^{2} \setminus 4 {x}^{2} - {x}^{4} \setminus \mathrm{dx}$

$= \frac{\pi}{16} {\left[\frac{4 {x}^{3}}{3} - {x}^{5} / 5\right]}_{0}^{2}$

$= \frac{\pi}{16} \frac{64}{15}$

$= \frac{4 \pi}{15}$