How do you find the area under a diagonal hyperbola xy=a where a is a constant?

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  • #1
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^topic
 

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  • #2
rock.freak667
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What are limits of the region you need to get the area of?
 
  • #3
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Just, what is the general process for finding the area? Say for a=3.
 
  • #4
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Rewrite your equation to the form y=f(x) and take the integral.
 
  • #5
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^topic


I'll assume your limits are 1 and e.

[tex]xy=a[/tex]

[tex]y=\frac{a}{x}[/tex]

[tex]\mbox{Area under the curve}=\int\limits_{1}^{e}\frac{a}{x}\mbox{d}x[/tex]

[tex]\mbox{Area under the curve}=a\left(\mbox{ln}(e)-\mbox{ln}(1)\right)[/tex]

[tex]\mbox{Area under the curve}=a(1-0)[/tex]

[tex]\mbox{Area}=a[/tex]
 
  • #6
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I'll assume your limits are 1 and e.

[tex]xy=a[/tex]

[tex]y=\frac{a}{x}[/tex]

[tex]\mbox{Area under the curve}=\int\limits_{1}^{e}\frac{a}{x}\mbox{d}x[/tex]

[tex]\mbox{Area under the curve}=a\left(\mbox{ln}(e)-\mbox{ln}(1)\right)[/tex]

[tex]\mbox{Area under the curve}=a(1-0)[/tex]

[tex]\mbox{Area}=a[/tex]


Perhaps I should have specified. I knew how to find the are under the curve from a to b, where a is not equal to zero, but wanted to find the area of the curve in either the 1st or 3rd quadrant (that is, from 0 to b or from -b to 0).
 
  • #7
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Perhaps I should have specified. I knew how to find the are under the curve from a to b, where a is not equal to zero, but wanted to find the area of the curve in either the 1st or 3rd quadrant (that is, from 0 to b or from -b to 0).

Is the a you are talking about the same as the a in xy=a?
 
  • #8
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Is the a you are talking about the same as the a in xy=a?

Nope, sorry about that, I should have used different variables. The second a (as in from a to b) just represents that I wasn't asking about the integral from c to b where c≠0 and was concerned with the integral from 0 to b.
 
  • #9
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Nope, sorry about that, I should have used different variables. The second a (as in from a to b) just represents that I wasn't asking about the integral from c to b where c≠0 and was concerned with the integral from 0 to b.

Ok. So,

[tex]\mbox{Area under the curve}=\int_{c}^{b}\frac{a}{x}\mbox{d}x[/tex]

[tex] \mbox{Area under the curve}= a \left(\mbox{ln} |b|-\mbox{ln} |c| \right) [/tex]
 
  • #10
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Ok. So,

[tex]\mbox{Area under the curve}=\int_{c}^{b}\frac{a}{x}\mbox{d}x[/tex]

[tex] \mbox{Area under the curve}= a \left(\mbox{ln} |b|-\mbox{ln} |c| \right) [/tex]

Right, as I said I got this result when I tried myself. But I wanted to know what to do when c=0.
 
  • #11
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Right, as I said I got this result when I tried myself. But I wanted to know what to do when c=0.

The area is infinite because ln(0)= -∞
 
  • #12
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The area is infinite because ln(0)= -∞

That's the problem, and it's what made me think I needed a new approach. Because obviously a hyperbola in the 1st quadrant doesn't have infinite area as it has asymptotes at the x and y axis. Is there perhaps another way to figure this out?
 
  • #13
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That's the problem, and it's what made me think I needed a new approach. Because obviously a hyperbola in the 1st quadrant doesn't have infinite area as it has asymptotes at the x and y axis. Is there perhaps another way to figure this out?

Nope. The hyperbola in the 1st quadrant does have an infinite area.
What makes you think that it doesn't?
 
  • #14
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Because it has a finite area except at the values x=0 and y=0 which approach a negligible area along the axes. If you graph it, the area is significantly less than a squared. Are there at least any approximation methods (maybe perturbation) out there that would work?
 
  • #15
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Negligible area it may be, but it extends to infinity, making it an infinite area in this case.

I can give you 2 examples that demonstrate the difference and the method how to calculate it (integration).


Example 1: x^2 y = 1

The corresponding area from x=1 to x=infinity is:
[tex]\textit{Area-from-1} = \int_1^\infty {1 \over x^2} dx = \left( - {1 \over x} \right) \large|_1^\infty = 1[/tex]

This is, as you can see, a finite value of only 1.


Example 2: x y = 1

The corresponding area from x=1 to x=infinity is:
[tex]\textit{Area-from-1} = \int_1^\infty {1 \over x} dx = \left( \ln x \right) \large|_1^\infty = \infty[/tex]

As you can see, this is not a finite value.
 
  • #16
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Wait, but for the second example, isn't the following true:
[itex]\int[/itex][itex]^{\infty}_{1}[/itex][itex]\frac{1}{x}[/itex]dx[itex]\equiv[/itex]-ln(1)+[itex]lim_{x->{\infty}}[/itex]ln(x)=0

?


Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.
 
  • #17
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I'm afraid not. Your limit approaches infinity instead of 0.
 
  • #18
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I'm afraid not. Your limit approaches infinity instead of 0.

Yeah, your right, but Wolfram was giving me a different result a minute ago, or maybe I made a typing mistake. Anyway, can you help with my other questions (sorry for so many)? There has to be some way to approximate it, especially considering that the hyperbola all but approaches the axes at an upper limit of 10*a.
 
  • #19
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Yeah, your right, but Wolfram was giving me a different result a minute ago, or maybe I made a typing mistake. Anyway, can you help with my other questions (sorry for so many)?

Which questions?
The questions I see, have already been answered by dimension10.
He did not make any mistakes.
 
  • #20
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Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.

These questions, you may not have noticed them because they were on the first page. Thank you for clearing up the first matter, I did not mean to imply that anyone other than myself was making a mistake.
 
  • #21
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Oh, I didn't see these questions before. I assume you edited your post later.

Wait, but for the second example, isn't the following true:
[itex]\int[/itex][itex]^{\infty}_{1}[/itex][itex]\frac{1}{x}[/itex]dx[itex]\equiv[/itex]-ln(1)+[itex]lim_{x->{\infty}}[/itex]ln(x)=0

?


Also, are there any clever approximations for this area? Would translating the figure into say (x-3)(y-3)=a make any difference (no more singularity)? How would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=1)? Thanks.

Clever approximations? The area is infinite. What's there to approximate?
Or do you mean an approximation for the natural logarithm for some value?

The translation you mention shifts the singularity to x=3, meaning there is still a singularity at x=3, and anyway, integrating to +infinity still yields infinity.

You can rotate the hyperbola.
The classical hyperbola formula is [itex]{x^2 \over a^2} - {y^2 \over b^2}=1[/itex].
But if you integrate between the axes and the hyperbola you'll still get infinity.
 
  • #22
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Don't think I edited it, maybe I just posted at the same time as you.


Anyway, for approximation methods I have a few ideas, would any of them work? They are as follows:

- cut off the area above some point on the x and y axis (say 10a)
- use a series expansion and take the limit of that instead of ln (you could have it approach 10a if necessary)
- integrate from some number sufficiently close to 0 instead of 0

Any thoughts?
 
  • #23
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If you cut of the area at some x and y, you get a finite value given by the natural logarithm.

If you make a series expansion instead of ln, you'll approximate the same result (if the series is convergent).

If you integrate from some number close to 0, you'll get a large value.
The closer the number is to 0, the closer the result is to infinity.
 
  • #24
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Which of these do you think would provide the best approximation (How would you go about cutting off the area at x=10a, y=10a?) Would perturbation methods help at all with this? Also, how would I find the area under a different type of hyperbola (is there a method that works for all hyperbolas that aren't in the form xy=a)? Thanks for all your help.
 
  • #25
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I'm starting to feel I'm patronizing, which I don't want to do.
But the best approximation for a finite part of the area is the natural logarithm.

If you want to do it for a different hyperbola formula, first you need to define which part of the area you want to know, and then you can set up a formula for that part (as long as that part is finite).

If you want you can set it up in polar coordinates:
r=L/(1-e cos theta)
This is useful if you want to use Kepler's 2rd law, that says the area a mass sweeps per unit of time, is constant.
 
  • #26
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So if I were to cut of the area at the point 10a, the area would be approximately equal to ln(10a)?
 
  • #27
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So if I were to cut of the area at the point 10a, the area would be approximately equal to ln(10a)?

If you were to cut the area given by the hyperbola xy=a, at x=10a and at y=10a, it will be:

1 + 2a ln(10a).
 
  • #28
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Thanks, that's exactly what I wanted! Just curious as to how you got this result though. Also, when you said "If you want you can set it up in polar coordinates:
r=L/(1-e cos theta)" what is L?
 
  • #29
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It starts with the integral from 1 to 10a from a/x dx.
This is the area between x=1 and x=10a, which is a ln(10a).

Due to symmetry this has the same area as the one betwee y=1 and y=10a.

Add the square block between 0 and 1, and you get 1 + 2a ln(10a).


L is the semi-latus rectum.
See for instance:
http://en.wikipedia.org/wiki/Conic_section
 
  • #30
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Great, thanks!
 

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