What is the volume of the solid produced by revolving #f(x)=x^2-x+1, x in [1,3] #around #x=1#?

2 Answers
Mar 30, 2017

#(32pi)/3#

Explanation:

The given function is a parabola opening upwards, as shown in the figure. The portion to revolved lies between x=1 and 3 and the axis of revolution is x=1

Now consider an elementary strip of thickness #Delta y# and length (x-1) at a distance y from the x-axis. On revolving it around x=1, the volume of the solid so generated would be #pi (x-1)^2 Delta y#. Pl see attached figure.

The volume of the whole solid so generated would be

#int_(y=1)^ (y=7) pi (x-1)^2 dy#

From the given eq y= #x^2-x+1# it would be dy= (2x-1)dx. Using this substitution, the volume integral would become

#pi int_(x=1)^(x=3) (x-1)^2 (2x-1)dx#

=#pi int_1^3 (2x^3-5x^2 +4x -1)dx#

=#pi[2x^4 /4 -5 x^3 /3 +4x^2 /2 -x]_1^3#

=#pi[81/2-135/3 +18-3-(1/2 -5/3+2-1)]#

=#pi[81/2 -135/3+15 -3/2 +5/3]#

=#pi[39+15-130/3]# =#(32pi)/3#

enter image source here

Apr 1, 2017

The region is not fully specified. I'll answer both possible questions.

Explanation:

# f(x)=x^2-x+1# for #x in [1,3]# around the line #x=1#.

Upper Region
If the region is above the parabola and below #y=7#, then we have:
The region bounded by the functions is shown below in red.
If we take a representative slice parallel to the axis of revolution we have a slice at #x# and thickness of #dx#.

enter image source here

When we revolve the slice, we get a cylindrical shell with volume

#2pirh * "thickness"#

In this case #r = x-1#
(Because we're revolving around #x=1# the radius is the distance between the #x# value of the slice and #1#.)

and #h = 7-(x^2-x+1)#
(The height of the slice is the upper #y# value minus the lower.)

The thickness is #dx#

The volume of the representative shell is: #2pi(x-1)(7-(x^2-x+1))dx#

We will be integrating with respect to #x#, so we observe that #x# varies from #1# to #3#.

The volume of the solid is
#V=int_1^3 2pi(x-1)(7-(x^2-x+1))dx#

# = 2pi int_1^3 (x-1)(7-(x^2-x+1))dx#

# = 2pi(16/3) = (32pi)/3#

LowerRegion
If the region is belowthe parabola and above the #x#-axis, then we have:
The region bounded by the functions is shown below in red.
If we take a representative slice parallel to the axis of revolution we have a slice at #x# and thickness of #dx#.

enter image source here

When we revolve the slice, we get a cylindrical shell with volume

#2pirh * "thickness"#

In this case #r = x-1#
(Because we're revolving around #x=1# the radius is the distance between the #x# value of the slice and #1#.)

and #h = (x^2-x+1)#
(The height of the slice is the upper #y# value minus the lower. value, which is #0#)

The thickness is #dx#

The volume of the representative shell is: #2pi(x-1)(x^2-x+1)dx#

We will be integrating with respect to #x#, so we observe that #x# varies from #1# to #3#.

The volume of the solid is
#V=int_1^3 2pi(x-1)(x^2-x+1)dx#

# = 2pi int_1^3 (x-1)(x^2-x+1)dx#

# = 2pi(26/3) = (52pi)/3#