# How do we calculate the area of a region bounded by four different curves?

Calculate the area(express both respectively in integral with one variable) bounded by the following curves (i.e. the shape with one side corresponding to one curve): $$xy=1, \quad xy^2=3,\quad x^2-y^2=26,\quad x^2-y^3=11$$

This problem is created by myself, but it is beyond my knowledge to solve it.

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@Victor: There are two regions with the property that they have one side given by each curve (see this plot). Do you mean the upper or the lower one? –  Zev Chonoles Dec 30 '11 at 3:27
This is bound to be very messy -- if feasible at all. Let me ask you: Can you do easier problems, such as the area enclosed between $y = 2x$, $y = x$ and $y = 1/x$ with $x,y \gt 0$? –  t.b. Dec 30 '11 at 3:39
You can ask whatever you want... You didn't answer my question: would you be able to solve the problem I asked you about? –  t.b. Dec 30 '11 at 3:44
As I said: it is an easier instance of the same problem. You still didn't answer my question: would you be able to solve the one I asked? –  t.b. Dec 30 '11 at 3:57
@t.b.:+1, I like your keeping asking "would you be able to solve the one I asked". "Doing the easier problems" is exactly the philosophy of Polya. :-) –  Jack Dec 30 '11 at 4:09

As long as you're only asking for an expression as an integral, and not an actual number, we can calculate the area as follows:

Let

• $a$ be the positive real solution of $x^5-11x^3-1=0$
• $b$ be the positive real solution of $x^{7/2}-11x^{3/2}-3\sqrt{3}=0$
• $c$ be the positive real solution of $x^4-26x^2-1=0$
• $d$ be the positive real solution of $x^3-26x-3=0$

We have $a<b<c<d$, and the dashed lines in the picture below indicate their positions. The curves are colored as follows:

$$\color{red}{xy=1},\quad \color{green}{xy^2=3},\quad \color{blue}{x^2-y^2=26},\quad \color{black}{x^2-y^3=11}$$

As you can see, the equations for $a,b,c,d$ were obtained by solving for the $x$-coordinate of the relevant intersections of the curves.

In the upper right quadrant, we can re-express our four curves as $$\color{red}{y=\tfrac{1}{x}},\quad \color{green}{y=\sqrt{\tfrac{3}{x}}},\quad \color{blue}{y=\sqrt{x^2-26}},\quad \color{black}{y=(x^2-11)^{1/3}}$$

The area below the black curve and above the red curve, from $a$ to $b$, is $$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx$$ The area below the green curve and above the red curve, from $b$ to $c$, is $$\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx$$ The area below the green curve and above the blue curve, from $c$ to $d$, is $$\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$ Thus the area of the upper region is $$\int_a^b\left((x^2-11)^{1/3}-\tfrac{1}{x}\right)dx+\int_b^c\left(\sqrt{\tfrac{3}{x}}-\tfrac{1}{x}\right)dx+\int_c^d\left(\sqrt{\tfrac{3}{x}}-\sqrt{x^2-26}\right)dx$$

We can do a similar computation for the lower region.

Mathematica code:

NSolve[x^5 - 11x^3 - 1 == 0, x]

NSolve[x^(7/2) - 11x^(3/2) - 3*Sqrt[3] == 0, x]

NSolve[x^4 - 26x^2 - 1 == 0, x]

NSolve[x^3 - 26x - 3 == 0, x]

a = 3.320739129529704

b = 3.437347103656831

c = 5.102784025451723

d = 5.155761179910075

ContourPlot[{x*y == 1, x*y^2 == 3, x^2 - y^2 == 26, x^2 - y^3 == 11,
x == a, x == b, x == c, x == d}, {x, 2.5, 6}, {y, -2, 2},
ContourStyle -> {{Red, Thick}, {Green, Thick}, {Blue, Thick}, {Black,
Thick}, {Black, Dashed}, {Black, Dashed}, {Black,
Dashed}, {Black, Dashed}}]


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Why and how do you get a,b,c,d? –  Victor Dec 30 '11 at 4:07
I am writing up an explanation and will post shortly –  Zev Chonoles Dec 30 '11 at 4:13
Thanks in advance –  Victor Dec 30 '11 at 4:14
Also, Zev, can the change of variable formula help in solving this problem? –  Victor Dec 30 '11 at 4:24