How do you find the volume bounded by #y=ln(x)# and the lines y=0, x=2 revolved about the y-axis?

2 Answers
Aug 7, 2017

For the solution by cylindrical shells, see below.

Explanation:

Here is a picture of the region and a representative slice taken parallel to the axis of rotation.

enter image source here

The slice is taken at some value of #x# and has thickness #dx#. So our functions will need to be functions of #x#

Revolving about the #y# axis will result in a cylindrical shell.

The volume of this representative shell is

#2pirh " thickness"#

The radius is shown as a dashed black line in the picture and has length #r = x#

The height of the shell will be the great #y# value minus the lesser #y# value. Since the lesser #y# value is #0#, we have #h = lnx#
As already mentioned, the thickness is #dx#

The representative volume is
#2pixlnxdx#

#x# varies from #1# to #2#, so the solid has volume

#V = int_1^2 2pixlnx dx = 2 pi int_1^2 xlnx dx#

Use integration by parts to get

# = 2pi(2ln2-3/4)#

(Rewrite the answer to taste.)

Aug 7, 2017

For the solution by washers see below.

Explanation:

Here is a picture of the region and a representative slice taken perpendicular to the axis of rotation.

enter image source here

The slice is taken at some value of #y# and has thickness #dy#. So our functions will need to be functions of #y#

The curve #y = lnx# can also be represented by #x = e^y#

Revolving about the #y# axis will result in a washer.

The volume of this representative washer is

#piR^2 *" thickness" - pir^2 " thickness" = pi(R^2-r^2) *" thickness"#

Where #R# is the greater radius -- shown as a dashed line. And #r# is the lesser radius -- shown as a dotted line.

In this case #R = "the "x" value on the right" = 2#
and #R = "the "x" value on the left" = e^y#

The representative volume is
#pi(2^2-(e^y)^2) dy = pi(4-e^(2y))dy#

#x# varies from #1# to #2#,

so #y# varies from #0# to #ln2#

And the volume is

#V = int_0^(ln2) pi(4-e^(2y))dy = piint_0^(ln2) (4-e^(2y))dy#

Use integration to get

# = pi[(4ln2-1/2e^(2ln2))-(-1/2e^0)] = pi(4ln2-3/2)#

(Rewrite the answer to taste.)