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Math Help - Region enclosed by the graph

  1. #1
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    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.
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  2. #2
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    Do you have the answer to a)?
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  3. #3
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    Quote Originally Posted by downer View Post

    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?
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  4. #4
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    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...
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  5. #5
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    Quote Originally Posted by bhuang View Post
    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 View Post
    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.
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  6. #6
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    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.
    Attached Thumbnails Attached Thumbnails Region enclosed by the graph-msp7019a91h560b72166100002iii15hhaebb6fga.gif  
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