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

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

 

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

 

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]

 

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

 

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]

 

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

 

Nope. The hyperbola in the 1st quadrant does have an infinite area.
What makes you think that it doesn't?

 

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.

 

I'm afraid not. Your limit approaches infinity instead of 0.

 

Which questions?
The questions I see, have already been answered by dimension10.
He did not make any mistakes.

 

Oh, I didn't see these questions before. I assume you edited your post later.

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.

 

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.

 

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.