Expert: Alon Mandes - 12/8/2008

Question
QUESTION: Hi,

The function f is denoted by f(x)= (ln(x))/(x^3), x >or= 1.

The region enclosed by the x-axis, the graph of f and the line x=3 is denoted by R.

Find the volume of the solid of revolution obtained when R is rotated through 360 degrees about the x-axis.

I am completely lost. I have tried the volume of revolution equations, but nothing seems to work out. I constantly get ln(0) which is undefined...thus leaving me with no answer. Any guidance or help is most appreciated!

ANSWER: The equation for volume of the solid of revolution is :
V=π∫[f(x)]^2 dx. Where x goes from 1 to 3. In our case f(x)=ln(x)/x^3
So, the volume will be :
V=π∫[ln(x)/x^3]^2 dx. x=1 -> x=3.
You may proceed from here, all you have to do is calculating the
integral.

Alon.


---------- FOLLOW-UP ----------

QUESTION: Thank you for the help - I completely understand that portion now. However, I am not sure how to go about integrating  V=π∫[ln(x)/x^3]^2 dx. x=1 -> x=3. I know that I can compute it on the calculator, however I need to show my work and do not know how to integrate it. I can integrate lnx/x^3 but not (lnx/x^3)^2...

Answer
No Problem Kaela, here's the method of integration :
I will use the substitute method, where u=lnx. that gives us
1. x=e^u
2. du=dx/x=dx/(e^u) -> dx=e^u du
Now  
π∫[ln(x)/x^3]^2 * dx. x=1 -> x=3 becomes :
π∫[u/e^(3u)]^2  * e^u du. u=ln(1) -> u=ln(3).
π∫u^2/e^(6u) * e^u du. u=ln(1) -> u=ln(3).
π∫u^2/e^(5u) * du. u=ln(1) -> u=ln(3).
π∫(u^2)e^(-5u) du. u=ln(1) -> u=ln(3).
From table of integrals we get that :
∫t^2/e^(-at) * dt = -e^(-at)[t^2/a + 2t/a^2 + 2/a^3].
So let's apply that on our integral :
π∫(u^2)e^(-5u) du =
-πe^(-5u)[u^2/5 + 2u/25 + 2/125].u=ln(1) -> u=ln(3).
=-πe^(-5ln(3))[0.24+0.08+0.016]+π[2/125]
=-π*1.84+π*0.016.

Alon.