What is the volume of the solid produced by revolving #f(x)=e^x-xlnx, x in [1,5] #around the x-axis?

1 Answer
Jul 18, 2018

The volume is #29486.554# #"u"^3#, or the sort of exact result is:

#V = pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27# #"u"^3#


To do these kinds of solid of revolution problems around the #x# axis, just remember that you are projecting a circle whose radius varies as #r = f(x)#.

The volume is then given by:

#V = pi int_(a)^(b) f^2(x)dx#

The #f(x)# serves as the varying radius, and the integral over #[a,b]# gives the length of the object, sliced up into discs of thickness #dx#.

In this case, we have:

#V = pi int_(1)^(5) (e^x - xlnx)^2dx#

#= pi int_(1)^(5) e^(2x) - 2xe^xlnx + x^2ln^2xdx#

#= pi overbrace(int_(1)^(5) e^(2x)dx)^("Integral 1") - 2pi overbrace(int_(1)^(5) xe^xlnxdx)^"Integral 2" + pi overbrace(int_(1)^(5) x^2ln^2xdx)^"Integral 3"#

  • The first integral is #1/2e^(2x)# evaluated from #x = 1# to #5#, which is #1/2(e^(10) - e^2)#.
  • The second integral is not possible with elementary functions, and the numerical solution is #848.042#.
  • The third integral is hard but doable here.

#int x^2ln^2xdx = ?#

Let:

  • #u = ln^2x#
  • #dv = x^2dx#
  • #du = (2lnx)/xdx#
  • #v = x^3/3#

Thus,

#int x^2ln^2xdx#

#= (x^3ln^2x)/3 - int x^3/3 (2lnx)/xdx#

#= 1/3 x^3ln^2x - 2/3int x^2 lnxdx#

Repeat with:

  • #u = lnx#
  • #dv = x^2dx#
  • #du = 1/xdx#
  • #v = x^3/3#

And we get:

#=> 1/3 x^3ln^2x - 2/3[1/3x^3lnx - int x^3/3 1/xdx]#

#= 1/3 x^3ln^2x - 2/3[1/3x^3lnx - 1/3 int x^2dx]#

#= 1/3 x^3ln^2x - [2/9x^3lnx - 2/9 int x^2dx]#

#= 1/3 x^3ln^2x - 2/9x^3lnx + 2/9 int x^2dx#

#= {:[1/3 x^3ln^2x - 2/9x^3lnx + 2/9 x^3/3]|:}_(1)^(5)#

#= [1/3 (5)^3ln^2(5) - 2/9(5)^3ln(5) + 2/9(5)^3/3] - [cancel(1/3 (1)^3ln^2(1))^(0) - cancel(2/9(1)^3ln(1))^(0) + 2/9 (1)^3/3]#

#= [1/3 125ln^2(5) - 2/9(125ln5 - 125/3)] - 2/9(1/3)#

#= 125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27#

So, adding up all the results,

#V = pi { 1/2(e^(10) - e^2)} - 2pi {848.042} + pi {125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27}#

#= color(blue)(pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27)#

or #29486.554#.