How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #x=y-y^2# and the y axis rotated about the y axis?

1 Answer
Oct 25, 2015

See the explanation section, below.

Explanation:

Here is a picture of the region. Because we have been asked to ue shells, a representative slice has been taken parallel to the axis of revolution. This make the thickness of our slice #dx#

enter image source here

The volume of the representative shell is:

#2 pixx "radius"xx"height"xx"thickness"#

As mentioned, #"thickness" = dx#
and we can see that #"radius"=x#

The #"height"# will be the greater #y# value minus the lesser #y# value. We are working in terms of #x#, so we need to write these two #y# values as functions of #x#.

We need to solve #x=y-y^2# for #y# (in terms of #x#).

There are a couple of ways of doing this. (1) complete the square or (2) use the quadratic formula to solve #y^2-y+x=0# for #y#.
(There are other ways as well. For example, you can use the vertex formula to write the equation in standard form for a sideways opening parabola.)

I'll complete the square.

#y^2-y = -x#

#y^2-y +1/4 = 1/4-x#

#(y-1/2)^2 = (1-4x)/4#

#y-1/2 = +-sqrt((1-4x)/4)#

#y = 1/2+-sqrt(1-4x)/2 = (1+-sqrt(1-4x))/2 #

(Yes, this is the same as the answer you'll get by using the quadratic formula.)

The greater #y# (the one on top) is
#y_"top" = (1+sqrt(1-4x))/2#

and the lesser #y# (the one on the bottom) is
#y_"bottom" = (1-sqrt(1-4x))/2#

So, finally, we can write the volume of the representative cylindrical shell:

#2 pixx "radius"xx"height"xx"thickness" = 2pix((1+sqrt(1-4x))/2-(1-sqrt(1-4x))/2) dx#

Don't Panic. We can simplify this to get

#2pixsqrt(1-4x)dx#

We still haven't found the bounds on #x#.

In the region we have #0 <= x#, and, examining the solution for #y#, we see that if #x > 1/4# we'll get imaginary values for #y#. So, we get #x# varies from #0# to #1/4#.

The volume of the solid of interest is

#V = int_0^(1/4) 2 pi x sqrt(1-4x) dx#

Which can be evaluated by parts or by the substitution #u = 1-4x# so that #du = -4 dx# and #x = 1/4(1-u)#.