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 April 3rd, 2014, 08:56 PM #1 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 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. Thanks.
 April 3rd, 2014, 09:06 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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... Thanks from mr. jenkins
 April 4th, 2014, 09:11 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 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. Thanks from MarkFL and mr. jenkins
 April 4th, 2014, 04:51 PM #4 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 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?) Thanks from MarkFL
 April 4th, 2014, 05:39 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 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). Thanks from MarkFL and mr. jenkins Last edited by v8archie; April 4th, 2014 at 05:41 PM.
 April 4th, 2014, 05:41 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.
 April 4th, 2014, 05:42 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Google Image search is pretty amazing, huh? Thanks from mr. jenkins
 April 5th, 2014, 01:59 PM #8 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 I know Korn and saw them at Lalapolooza some yrs back. He is a very distinctive singer.
 April 5th, 2014, 02:02 PM #9 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 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 ?
April 5th, 2014, 03:22 PM   #10
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Quote:
 Originally Posted by mr. jenkins 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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