# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=3x^4, y=0, x=2 revolved about the x=4?

Jul 20, 2016

$\left(\frac{128}{15}\right) \pi$ cubic units

#### Explanation:

The volume of this annular solid, open at y = 0 and y = 48 and

appearing as an inverted funnel is

$\pi \int \left({4}^{2} - {\left(x - 4\right)}^{2}\right) d y$, from y=0 to y = 48.

$= \pi \int \left(2 x - {x}^{2}\right) d y$, from y=0 to y = 48..

$= \pi \int \left(2 {\left(\frac{y}{3}\right)}^{\frac{1}{4}} - {\left(\frac{y}{3}\right)}^{\frac{1}{2}}\right) d y$, from y=0 to y = 48..

$= \pi \left[2 {\left(\frac{y}{3}\right)}^{\frac{5}{4}} / \left(\frac{5}{4}\right) - {\left(\frac{y}{3}\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]$. between y = 0 and y = 48

$= . \pi \left[2 {\left(\frac{48}{3}\right)}^{\frac{5}{4}} / \left(\frac{5}{4}\right) - {\left(\frac{48}{3}\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]$

=.pi [256/5 - 128/3)]

$= \left(\frac{128}{15}\right) \pi$ cubic units.