# Area under curve bounded by rectangles

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?

• What do you mean by $x*y$? – Allawonder Sep 9 '19 at 8:51
• @Allawonder I mean $x$ multiplied by $y$ – resound Sep 9 '19 at 9:03
• 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. – Allawonder Sep 9 '19 at 9:06
• Calling something by a different name doesn't make it clearer -- it obscures the point. – Allawonder Sep 9 '19 at 9:19
• 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). – Wlod AA Sep 9 '19 at 10:14

## 1 Answer

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).

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 $$2$$ :-).

• 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. – resound Sep 9 '19 at 9:26
• @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. – Wlod AA Sep 9 '19 at 9:30
• @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. – Wlod AA Sep 9 '19 at 9:49