# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by x=sqrt(y), x=0, y=4 about the x-axis?

Jul 12, 2016

$\frac{128 \pi}{5}$ cubic units.

#### Explanation:

The ends of this first-quadrant segment of the parabola $y = \sqrt{x}$ are

(0, 0) and (2, 4).

The shell has a paraboloid-hole in the middle. The base is circular,

with radius ( range of y ) 4 units and the height is ( range of x, from 0

to 2 ) 2 units.

The hole volume has to be subtracted from the volume

of a right circular cylinder, of radius 4 and height 2 units..

So, the volume of the hollow solid = $\pi \int \left({4}^{2} - {y}^{2}\right) d x$, with

the limits, from x = 0 to x = 2

$= \pi \int \left(16 - {x}^{4}\right) d x$, with the limits, from x = 0 to x = 2

$= \pi \left[16 x - {x}^{5} / 5\right]$, between the limits

$= \left(32 - \frac{32}{5}\right) \pi$

$= \frac{128 \pi}{5}$ cubic units.