# How do you use the changing of variable formula to solve this problem?

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

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

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

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

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

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