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?

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Answer 1 · 2016-07-12T09:38:49

$(128pi)/5$ cubic units.

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 = $pi int (4^2-y^2) d x$ , with

the limits, from x = 0 to x = 2

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

$=pi[16x-x^5/5]$ , between the limits

$=(32-32/5)pi$

$=(128pi)/5$ cubic units.

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=sqrt(y), x=0, y=4 about the x-axis?