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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
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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
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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
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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
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@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

1 Answer 1

up vote 7 down vote accepted

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:

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

enter image description here

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

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