Jacobi Matrix and Multiple Intgrals

Kork
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Homework Statement



Let D be the set of points (x,y) in R^2 for which 0 is ≤ x ≤ 1 and 0 ≤ y ≤ 1. Find a function g: R^2 --> R for which:

∫_0^1 ∫_0^1 h(x,y)dxdy = ∫_0^1∫_0^1 h(y^5, x^3) * g(x,y)dxdy

is true for all functions h: D--> R integrable over D

In the question before this I was asked to find a jacobi matrix and determinant for f(x,y) = (y^5,x^3)

I found that the determinant is -15y^4x^3

Homework Equations




The Attempt at a Solution



∬_D f(x,y)dxdy= ∬_S g(v^5,u^3) * -15y^4x^2

Is this my final result? If not, can I get some help on how to go on?
 
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Kork said:

Homework Statement



Let D be the set of points (x,y) in R^2 for which 0 is ≤ x ≤ 1 and 0 ≤ y ≤ 1. Find a function g: R^2 --> R for which:

∫_0^1 ∫_0^1 h(x,y)dxdy = ∫_0^1∫_0^1 h(y^5, x^3) * g(x,y)dxdy

is true for all functions h: D--> R integrable over D

In the question before this I was asked to find a jacobi matrix and determinant for f(x,y) = (y^5,x^3)

I found that the determinant is -15y^4x^3

Homework Equations




The Attempt at a Solution



∬_D f(x,y)dxdy= ∬_S g(v^5,u^3) * -15y^4x^2

Is this my final result? If not, can I get some help on how to go on?

Your answer shouldn't have 4 variables and needs the integration variables.

Your problem certainly suggests the change of variables ##x=u^5,\ y=v^3##. You need to check that maps the square to the square. Then use your change of variables theorem to express your integral in terms of ##u## and ##v##. Your result should have just ##u## and ##v## variables in it. Of course, they are dummy variables in the final result.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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