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April 3rd, 2014, 08:56 PM   #1
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Volume of a solid rotated about ...

Consider the solid obtained by rotating the region bounded by the given curves about the line y = -1.
y = 1/x, y = 0, x = 1, x = 3
Find the volume V of this solid.
V = ???

I graphed it. Now, A(x) = pi(-1-1/x)^2 -pi(-1-0)^2
= pi[1+2/x +1/x^2-1]

and at this point I am really confused. Integrating is not my problem it is setting up the problem correctly is what I guess I am asking help for.

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April 3rd, 2014, 09:06 PM   #2
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I would use the washer method. The volume of an arbitrary washer is:

$\displaystyle dV=\pi\left(R^2-r^2 \right)\,dx$

where:

$\displaystyle R=1+\frac{1}{x}$

$\displaystyle r=1$

Hence, we have:

$\displaystyle dV=\pi\left(\left(1+\frac{1}{x} \right)^2-1^2 \right)\,dx$

Expanding and collecting like terms, we obtain:

$\displaystyle dV=\pi\left(\frac{2}{x}+\frac{1}{x^2} \right)\,dx$

Now integrate from $x=1$ to $x=3$:

$\displaystyle V=\pi\int_1^3 \frac{2}{x}+\frac{1}{x^2}\,dx$

I will leave the integrating to you...
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April 4th, 2014, 09:11 AM   #3
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I would add a constant term to the equations for y to move the problem "up" so that we rotate around the x-axis as normal.

So the bounding curves become $y=\frac{1}{x}+1$, $y = 1$, $x = 1$ and $x = 3$.

We are then looking to subtract the volume obtained by rotating $y = 1$ about the axis, from the volume obtained by rotating $y = \frac{1}{x}+1$ about the axis.

All of this is simply MarkFL's post dressed up in different language. The resulting integral should be the same.
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April 4th, 2014, 04:51 PM   #4
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Both of you this is a great help to be able to go back to the drawing board, see the setup and then see where I was off with my thinking etc.

Quick question for either or both of you. the 2/x why am I having difficulty here with my thinking pertaining to the Integration?

I find myself wanting to take 2/x and re-write as 2X^-1 which is eqivelent but in Integrating, I would have 2X^0 not making x one but resulting in the natural log I think and then I'm confused again.


Anyways, I am pounding out Abstract Algebra homework now and looking forward to sitting down with this problem and you two's help to get on board with this problem. I think the confusion on the 2/x integration is the last of my confusions for this problem so Mark and V8 TY!

ps... who is the lead singer on your profile there Mark? Is that Bruce (lead singer of Iron Maiden?)
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April 4th, 2014, 05:39 PM   #5
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My Googling suggests that it is Jonathon Davis of the band Korn. I know it's not Bruce Dickinson.

You are correct in thinking that $$\int{\frac{2}{x}dx = 2\log{x}}$$(plus an additive constant which is often written as $\log{A}$ for convenience).
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Last edited by v8archie; April 4th, 2014 at 05:41 PM.
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April 4th, 2014, 05:41 PM   #6
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The natural log function is the way to go:

$\displaystyle \int\frac{2}{x}\,dx=2\int\frac{dx}{x}=2\ln|x|+C$

My avatar is of Jonathan Davis, the lead singer of KoЯn.
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April 4th, 2014, 05:42 PM   #7
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Google Image search is pretty amazing, huh?
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April 5th, 2014, 01:59 PM   #8
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I know Korn and saw them at Lalapolooza some yrs back. He is a very distinctive singer.
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April 5th, 2014, 02:02 PM   #9
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Hey guys one more question.


Do I use the constant of Integration when I am dealing with a definite integral??

I thought I would be integrating from a=1,b=3 ?
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April 5th, 2014, 03:22 PM   #10
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Quote:
Originally Posted by mr. jenkins View Post
Hey guys one more question.


Do I use the constant of Integration when I am dealing with a definite integral??

I thought I would be integrating from a=1,b=3 ?
No, you want to us use the anti-derivative form of the fundamental theorem of calculus for definite integrals:

$\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)$
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