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

By using changing of variable formula to express those area into a integral with 4 different variable, that is, mapping the curves into another plane(when you parametrize one curve you with get one number, you get 4 different number in total with four curves)

I know you may think this question may be the duplicate of that question, but that question only ask for using only one variable integral:

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

i know the change of variable formula only work up to 3-dimentional, so does changing of the variable formula help to solving my problem?

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@Victor: Is there any reason to think that a helpful change of variables exists? It is very fine for you to go and experiment with this toy - you will learn a lot! One of the things you are about to learn concerns the limitations of the technique. – Jyrki Lahtonen Jan 1 '12 at 18:42
Can you provide a plot? – draks ... Jan 4 '12 at 11:20
@Victor I would have to agree with Jyrki that there really isn't any reason to think there is a "nice" change of variables for this problem. But there is a more fundamental issue at hand, what region are you referring to? Here is a plot of these curves. You can see several regions which are bound by them. Which region are you looking to find the area of? – Bill Cook Jan 4 '12 at 22:29
@BillCook- how to find both respectively? – Victor Jan 10 '12 at 23:14
up vote 3 down vote accepted

Had you wrote

$$ xy=1, \quad xy=3,\quad x^2-y^2=26,\quad x^2-y^2=11 \ , $$

the answer would be easier: you could use the following change of variables

$$ u = xy, \quad v = x^2 - y^2 \ . $$

Then your area would be

$$ \iint_D dxdy = \int_1^3 \int_{11}^{26}\vert JT(u,v) \vert dvdu \ , $$

where $D$ is the area enclosed by the curves, and $JT(u,v)$ the jacobian of the change of coordinates $(x,y) = T(u,v)$. Unfortunately, this is the inverse of the change of variables you actually know. Namely,

$$ (u,v) = T^{-1}(x,y) = (xy, x^2 - y^2) \ , $$

but you could resort to the fact that

$$ JT(u,v) = JT(x,y)^{-1}\circ T(u,v) = \frac{-1}{2(x^2+y^2)\circ T(u,v)} \ . $$

Still, that $T(u,v)$ insists to appear. So, in fact, this kind of exercise usually goes like this: compute

$$ \iint_D (x^2+y^2)dxdy \ . $$

In this situation, you cancel out both $(x^2 + y^2)\circ T(u,v)$ and you're happy as a clam:

$$ \iint_D (x^2+y^2)dxdy = \int_1^3 \int_{11}^{26} ((x^2 + y^2)\circ T(u,v)) \frac{1}{2(x^2+y^2)\circ T(u,v)}dvdu = \frac{1}{2} \int_1^3 \int_{11}^{26} dvdu \ . $$

So, it's just usually a prefabricated exercise in order to practise the change of variables and the Jacobian of the inverse function.

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i need a answer that match with my original problem – Victor Jan 1 '12 at 17:37
Well, good luck looking for it. – a.r. Jan 1 '12 at 17:41

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