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

1 Answer
Jim H
Oct 13, 2016

See below.

Explanation:

Draw a picture (sketch) of the region.
Note that the graphs intersect where #y=x# and #y^2=4x#, so that is where #x^2=4x#.
And that happens at #(0,0)# and at #(4,4)#.

enter image source here

We are asked to use shells, so we take thin representative rectangles parallel to the axis of rotation. In this case that is the #y#-axis.
The slice has solid black boundaries and the radius to the axis of rotation is shown as a dashed black line.

enter image source here

Since the thin part is #dx#, we need both equations as functions of #x#.

#y^2=4x# in the first quadrant (which is where the region is) gets us #y = 2sqrtx# for the upper function.

The lower function is #y=x#

The volume of a representative shell is #2pirh*"thickness"#

In this case, we have

radius #r = x# (the dashed black line),

height #h = upper - lower = 2sqrtx-x# and

#"thickness" = dx#.

#x# varies from #0# to #4#, so the volume of the solid is:

#int_0^4 2pi x(2sqrtx-x)dx =2 piint_0^4 (2x^(3/2)-x^2) dx#

(details left to the student)

# = (128pi)/15#

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