Jacobian Transformations for Evaluating Integrals with Inequalities

• bodensee9
In summary, after incorrectly evaluating the Jacobian, the student is seeking pointers on how to solve the equation for change of variables.
bodensee9
Can someone help me with the following?
I am supposed to evaluate
∫∫ e^(x+y)dA where the area of integration is given by the inequality |x|+|y|≤1.
So, suppose I do one of these Jacobians, and I set u = |x| and v = |y|, so wouldn’t the equation have to satisfy the inequality u+v≤1, and u≥0, v≥0? So, would wouldn’t the Jacobian be 1? But clearly I’m doing something wrong here, so any hints would be greatly appreciated! Thanks!

The function in a change of variable should one-to-one (as your example shows - it simply omits regions of the integration domain). Differentiability is also handy. Absolute value is neither.

I'm not sure if I understand what you mean. Can you clarify more or point out where I'm going wrong with this? Thanks.

I thought I did? You can SEE where it's going wrong, right? The formula for change of variables involving Jacobians has premises. You can't blindly use any old functions. |-1|=|1|=1. It's not one-to-one. Hence you can't blindly use it in change of variable. Jacobian isn't even defined at (0,0).

Note that absolute values are non-negative, so for example, |x|<=1, that is $-1\leq{x}\leq{1}$

An equivalent inequality to the one you have been given is:
$$|x|-1\leq{y}\leq{1}-|x|$$
In the region $x\geq{0}$, this translates to:
$$x-1\leq{y}\leq{1-x}, 0\leq{x}\leq{1}$$
Make a similar translation for negative x's!

Last edited:
Thanks, that is very helpful. So, can I just ask, since |x|+|y| ≤1, so this means that |x|≤1-|y|. |y| - 1 ≤ |x| ≤ 1-|y|. So this means that y – 1 ≤ x ≤ 1-y, where 0 ≤ y ≤ 1. But I don’t see why I can’t just use those parameters to integrate? So, could I do ∫∫ e^(x+y)dxdy where y – 1 ≤ x ≤ 1-y and where 0 ≤ y ≤ 1?

Of course you could use y rather than x as the "outer" variable.

Yes. That's exactly what you should do.

Dick said:
Yes. That's exactly what you should do.

Eeh?
What's wrong about using x as the outer variable instead?

arildno said:
Eeh?
What's wrong about using x as the outer variable instead?

Nothing! Either is just fine. I was replying to bodensee.

Thanks very much for all your help!

1. What is a Jacobian transformation?

A Jacobian transformation is a mathematical method used to express a set of variables in terms of a different set of variables. It is commonly used in multivariate calculus and differential equations to simplify complex problems.

2. What is the purpose of a Jacobian transformation?

The main purpose of a Jacobian transformation is to simplify mathematical expressions and make them easier to solve. It can also be used to convert between different coordinate systems, such as Cartesian and polar coordinates.

3. How is a Jacobian transformation calculated?

A Jacobian transformation is calculated by taking the partial derivatives of a set of variables with respect to another set of variables. These partial derivatives are then arranged in a matrix, known as the Jacobian matrix, which is used to perform the transformation.

4. What is the relationship between a Jacobian transformation and the determinant?

The determinant of the Jacobian matrix is used to determine if a Jacobian transformation is reversible. If the determinant is non-zero, then the transformation is reversible. Additionally, the determinant can also be used to calculate the change in volume or area after a transformation.

5. How is a Jacobian transformation used in real-world applications?

Jacobian transformations have many applications in fields such as physics, engineering, and economics. They are used to solve problems involving multiple variables, such as optimization and differential equations. They are also used in computer graphics to transform objects in three-dimensional space.

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