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^topic

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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)?...There are a few approximations for the area, but they all result in the same answer. The most accurate approximation is the integral. There are a few approximations for the area, but they all result in the same answer. The most accurate approximation is the integral.

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^topic

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- #2

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What are limits of the region you need to get the area of?

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Just, what is the general process for finding the area? Say for a=3.

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Rewrite your equation to the form y=f(x) and take the integral.

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joebohr said:^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]

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dimension10 said: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).

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joebohr said: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?

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dimension10 said: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.

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joebohr said: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]

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dimension10 said: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.

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joebohr said: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)= -∞

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dimension10 said: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?

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joebohr said: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?

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- #15

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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.

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[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.

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I like Serena said: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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joebohr said: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.

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joebohr said: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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joebohr said:

[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.

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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 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.

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- #25

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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.

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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)?

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joebohr said: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).

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r=L/(1-e cos theta)" what is L?

- #29

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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

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Great, thanks!

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