
#1
Oct412, 10:39 AM

P: 44

I've been told this is a trick question, but I don't understand why:
How would I describe the solid generated by 2∏∫ x/(1+x^{2})dx on [0,2] How I would do it I would rewrite the intergal as 2∏∫ x * 1/(1+x^{2})dx and apply substitution. I would then use the volume of disks method and integrate the integral about the new bounds. Is this the right way to go about it? 



#2
Oct412, 12:07 PM

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#3
Oct412, 12:21 PM

P: 1,666

Consider the formula for a solid of revolution involving rotating the function f(x) around the xaxis. That formula is:
[tex]V=\int_a^b \pi \left(f(x)\right)^2 dx[/tex] Now, how could you rewrite your integral in that form to represent a solid of revolution? 



#4
Oct412, 12:24 PM

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Describe the solid generated by the integral
I wouldn't call it so much a "trick" question as an ambiguous one.
Firstly, the question doesn't appear to be asking you to actually evaluate the integral, but merely to describe the solid for which the volume calculation would generate this integral. The ambiguous part is "the solid". If the question were to ask for "a solid" then I'd consider it a legitimate question. Even if we restrict ourselves to volumes of rotation then I can think of at least two different solids that could lead to that integral. Perhaps you could find them both as an exercise. That is, find: 1. A volume of rotation around the "y" axis, calculated using the method of shells. 2. A volume of rotation around the "x" axis, calculated using the method of discs. Edit: I posted the above before seeing either of the two previous replies (due to slow typing and distractions ). I agree however with both of the above replies. The form on the integral expression lends itself most naturally to a method of shells rotation (because of the [itex]2 \pi[/itex] out front) as LCKurtz noted. There is however no real reason why we couldn't include a [itex]\sqrt{2}[/itex] in the function being evaluated by the disc method as per jackmels approach, resulting in the same given integral. This is why I call it an ambiguous question that could have been better worded. 



#5
Oct412, 01:34 PM

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Are we viewing the integral [itex]\displaystyle 2\pi\int_{0}^{2}\frac{x}{1+x^2}\,dx [/itex] as giving the volume of a solid formed by rotating the area bounded by some function and the coordinate axes about the yaxis? What is it that you are supposed to say about this integral or do with this integral? 



#6
Oct412, 02:31 PM

P: 1,666

[tex]2\pi \int_0^2 \frac{x}{1+x^2}dx=\int_0^2 \pi \left(\sqrt{\frac{2x}{1+x^2}}\right)^2dx[/tex] and so the solid is that rotated around the xaxis. Now put mine on top of the stack when you hand them out to class. Of course I'm not a HH soooooo, watch it. 



#7
Oct412, 09:09 PM

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What about this one. The volume below the function [itex]f(x,y) = \frac{x}{x^2+1}[/itex], and above the rectangle in the x,y plane given by [itex]0 \leq x \leq 2[/itex] and [itex]0 \leq y \leq 2 \pi[/itex]. 



#8
Oct512, 01:00 AM

P: 44

LCKurtz, that's exactly what I was looking for. I apologise to the others if I the question was too ambiguous, the textbook just worded it this way.
Nevertheless, thanks everyone! 



#9
Oct512, 03:41 PM

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