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.