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Is it possible to find the common area between curve $xy>b^2$ (rectangular hyperbola) and the rectangle $2\leq x\leq A$ and $2\leq y\leq B$ where $A$ and $B$ are much larger than $b^2$?

From what I think, it should be something like area of rectangle subtracted the area of hyperbola with x-axis. But doing this we also remove the area that is not in the rectangle(for some values of $b^2$), so we add that small part again.

However, how to calculate the area between the hyperbola and x-axis. If I integrate the hyperbola from $2\leq x\leq A$ what about the bound on $y$, how to consider both the bounds?

I don't know if it's really the right approach, but it should not be undefined because $x$ and $y$ are bounded. So how do I calculate the area?

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  • $\begingroup$ What do you mean by $x*y$? $\endgroup$
    – Allawonder
    Sep 9, 2019 at 8:51
  • $\begingroup$ @Allawonder I mean $x$ multiplied by $y$ $\endgroup$
    – resound
    Sep 9, 2019 at 9:03
  • $\begingroup$ I see. Shouldn't that be better written as $xy$? Also, the inequality $xy>b^2$ defines a region bounded by a hyperbola, not a hyperbola. $\endgroup$
    – Allawonder
    Sep 9, 2019 at 9:06
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    $\begingroup$ Calling something by a different name doesn't make it clearer -- it obscures the point. $\endgroup$
    – Allawonder
    Sep 9, 2019 at 9:19
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    $\begingroup$ People react with a knee jerk but long years ago I've spent time on notation. Mathematical notation follows its traditions often by inertia. It's hard to improve things (as your, @Allawonder, reaction proves). $\endgroup$
    – Wlod AA
    Sep 9, 2019 at 10:14

1 Answer 1

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Let's assume that $\ b>0\ $ (why not?! :-)

Then, the only non-trivial case is $\ b>2;\ $ thus let's assume this $\ b>2.$

Now let's solve the problem, i.e. let's do the computation.

Let $$\ \alpha:=\frac Ab\qquad \beta:=\frac Bb \qquad d:=\frac 2b $$

The common area between curve $x*y>1$ (rectangular hyperbola) and the rectangle $\ d\leq x\leq \alpha\ $ and $d\leq y\leq \beta\ $ is

$$ (\alpha-d)\cdot(\beta-d) - \int_d^\frac 1d \frac 1x\cdot dx\quad =\\ \ \\ \ (\alpha-d)\cdot(\beta-d)\ +\ 2\cdot \log(d) $$

(of course $\ \log(d)<0\ $ since $\ 0<d<1$).

Now, the original area is the above times $\ b^2,$ i.e.

$$ (A-2)\cdot(B-2)\ +\ 2\cdot(\log(2)-\log(b)) $$

I think that that's it.

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PS. You may consider $\ 2\ $ as an arbitrary positive real parameter, just imagine that it is arbitrary $\ z>0.\ $ The result will hold under this general notation. (If you squeeze your eyes then you'll see "$z$" in place of $2$ :-).

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  • $\begingroup$ By interesting, I mean it's interesting that you solved it in a manner I didn't approach it, and I thanked you because rather than posting a hint or a half thought answer, you tried to make it a complete answer. I am sorry, I never meant to disrespect you. $\endgroup$
    – resound
    Sep 9, 2019 at 9:26
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    $\begingroup$ @resound "I never meant to disrespect you.". Well, you've succeeded. Even now, you've written that I "tried". I didn't try. I simply have provided an answer. You may simply acknowledge the correctness of my solution or you may invalidate it if I were wrong. (And one may also ask straightforward questions, of course). That's all. $\endgroup$
    – Wlod AA
    Sep 9, 2019 at 9:30
  • $\begingroup$ @resound, you have asked about the area (remember?) and not about the integer points. These questions are related though, and this relationship forms a classical research topic. $\endgroup$
    – Wlod AA
    Sep 9, 2019 at 9:49

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