# Region enclosed by the graph

• Apr 22nd 2010, 07:11 PM
downer
Region enclosed by the graph
A problem I had recently on a test has been bothering me for quite some time now and would like to seek assistance for it. There were 3 questions that went along with it.

The problem goes:
Let R be the region enclosed by the graph of y = 2lnx and y = x/2, and the lines x = 2 and x = 8.

a) Find the area of R
b) Set up an integral expression, in terms of a single variable,for the volume of the solid generated when R is revolved about the x-axis.
c)Set up an integral expression, in terms of a single variable,for the volume of the solid generated when R is revolved about the line x = -1.

Any possible help regarding any of the 3 questions is greatly appreciated.
• Apr 22nd 2010, 07:33 PM
bhuang
Do you have the answer to a)?
• Apr 22nd 2010, 07:52 PM
pickslides
Quote:

Originally Posted by downer

The problem goes:
Let R be the region enclosed by the graph of y = 2lnx and y = x/2, and the lines x = 2 and x = 8.

For part a) solve $\int_2^82\ln x - \frac{x}{2}~dx$

What do you get?
• Apr 22nd 2010, 07:59 PM
bhuang
I did x/2 - 2lnx instead because y=x/2 is graph that gives the larger values. I end up getting -7.49, but I think some of my algebra may have been wrong...
• Apr 22nd 2010, 08:06 PM
pickslides
Quote:

Originally Posted by bhuang
I did x/2 - 2lnx instead because y=x/2 is graph that gives the larger values.

I don't agree with this.

In my understanding $2\ln x > \frac{x}{2}, \forall x \in [2,8]$

Quote:

Originally Posted by bhuang
I end up getting -7.49, but I think some of my algebra may have been wrong...

This answer is not correct. If anything it supports my statement above.
• Apr 22nd 2010, 08:12 PM
drumist
I've attached a graph of the two curves so you can see which one is above and which is below. We are only concerned with the region between x=2 and x=8.