1
$\begingroup$

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?

  • 0
    @BillCook- how to find both respectively?2012-01-10

1 Answers 1

2

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.

  • 5
    Well, good luck looking for it.2012-01-01